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# A Novel Strategy of Calculating Unit Hypercube Integrals

## Introduction

On this perception article, we’ll construct all of the equipment essential to judge unit hypercube integrals by a novel approach. We’ll first state a theorem on Dirichlet integrals, second develop a sequence of nested units that point-wise converges to a unit hypercube, and thirdly make these two items appropriate by way of a Dominated Convergence Theorem, and lastly show the strategy of integration. Notice: The identical approach is printed (in the identical method) within the expanded perception article entitled A Path to Fractional Integral Representations of Some Particular Capabilities.

## The Integrals of Dirichlet

Dirichlet integrals as I discovered them from an Superior Calculus guide are simply that method evaluating the integral to Gamma capabilities, they aren’t a sort of integral like Riemann integral, extra only a method that might go on a desk of integrals. Content material is the 4+-dimensional model of quantity (some writers use hypervolume as a substitute of content material).

For the proof of this Dirichlet Integrals Theorem, I refer the reader to the textual content Particular Capabilities by Askey, Andrews, and Roy. The end result attributable to Dirichlet is given by

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#### Theorem 1.1: Dirichlet Integrals

If ##t,{alpha _p},{beta _q},Re left[ {{gamma _r}} right] > 0forall p,q,r## and ##V_t^n: = left{ {left( {{z_1},{z_2}, ldots ,{z_n}} proper) in {mathbb{R}^n}|{z_j} geq 0forall j,sumlimits_{ok = 1}^n {{{left( {frac{{{z_k}}}{{{alpha _k}}}} proper)}^{{beta _k}}} leq t} } proper}##, then

\$\$iint {mathopÂ  cdots limits_{V_t^n} int {prodlimits_{lambdaÂ  = 1}^n {left( {z_lambda ^{{gamma _lambda } â€“ 1}} proper)} d{z_n} ldots d{z_2}d{z_1}} } = {t^{sumlimits_{p = 1}^n {frac{{{gamma _p}}}{{{beta _p}}}} }}{{prodlimits_{q = 1}^n {left[ {frac{{alpha _q^{{gamma _q}}}}{{{beta _q}}}Gamma left( {frac{{{gamma _q}}}{{{beta _q}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {frac{{alpha _q^{{gamma _q}}}}{{{beta _q}}}Gamma left( {frac{{{gamma _q}}}{{{beta _q}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {frac{{{gamma _k}}}{{{beta _k}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {frac{{{gamma _k}}}{{{beta _k}}}} } proper)}}\$\$

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#### Corollary 1.2: Content material Integral

If ##{alpha _p},{beta _q} > 0forall p,q## and ##{V^n}: = left{ {left( {{z_1},{z_2}, ldots ,{z_n}} proper) in {mathbb{R}^n}|{z_j} geq 0forall j,sumlimits_{ok = 1}^n {{{left( {frac{{{z_k}}}{{{alpha _k}}}} proper)}^{{beta _k}}} leq 1} } proper}##, then

##{textual content{Content material}}left( {{V^n}} proper): = iint {mathopÂ  cdots limits_{{V^n}} int {d{z_n} ldots d{z_2}d{z_1}} } = {{prodlimits_{q = 1}^n {left[ {frac{{{alpha _p}}}{{{beta _q}}}Gamma left( {frac{1}{{{beta _q}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {frac{{{alpha _p}}}{{{beta _q}}}Gamma left( {frac{1}{{{beta _q}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {frac{1}{{{beta _k}}}} } proper)}}} proper.} {Gamma left( {1 + sumlimits_{ok = 1}^n {frac{1}{{{beta _k}}}} } proper)}}##. The place content material refers to hypervolume when ##n geq 4##.

## The Orthotope

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The orthotope is the generalization of the oblong parallelepiped to ##{mathbb{R}^n}##. Let ##{b_k} > 0forall ok leq n## and think about the orthotope with polytope vertices of the shape ##left( { pm {b_1}, pm {b_2}, ldots , pm {b_n}} proper)## oriented with facet-centered axes decided by the set ##{P^n}: = left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}|left| {{x_k}} proper| leq {b_k}forall ok leq n} proper}##. It is going to be handy to develop a definition of the orthotope which can appear overly difficult however will make good use of it within the subsequent part. Let ##S_N^n: = left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}|sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N}}}Â  leq n} proper}##. Then we proceed to show by bivariate induction on n and N that

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#### Lemma 2.1: Nesting Property of ##S_N^n##

If ##S_N^n: = left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}|sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N}}}Â  leq n} proper}##, then ##S_{N + 1}^n subset S_N^n##.

##### Proof:

By bivariate induction on n and N (this has three components)[1]:

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First, we outline the assertion to be proved ##Pleft( {n,N} proper):S_{N + 1}^n subset S_N^n##.

(i) Show the bottom case ##Pleft( {a,b} proper)## the place ##a,b in {mathbb{Z}^ + }## are the bottom or smallest values of a and b such that ##Pleft( {a,b} proper)## holds. For this drawback meaning proving ##Pleft( {2,1} proper):S_2^2 subset S_1^2##. Now we have ##S_2^2: = left{ {left( {{x_1},{x_2}} proper) in {mathbb{R}^2}|sumlimits_{ok = 1}^2 {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^4}}Â  leq 2} proper}## implies the next inequality holds ##sumlimits_{ok = 1}^2 {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^4}}Â  leq 2##. Now to work given inequality down we’ll use calculus to search out the values that maximize the constraint ##fleft( {{x_1},{x_2}} proper) = sumlimits_{ok = 1}^2 {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^4}}Â  leq 2## and the advert hoc constraint ##gleft( {{x_1},{x_2}} proper) = tfrac{{{x_1}}}{{{b_1}}} + tfrac{{{x_2}}}{{{b_2}}} = c## for some optimistic fixed c (which is actually true). We proceed by the tactic of Lagrange multipliers

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Instance 2-d S_N^2 units for extrema

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\$\$nabla f = leftlangle {4tfrac{{x_1^3}}{{b_1^4}},4tfrac{{x_2^3}}{{b_2^4}}} rightrangleÂ  = lambda leftlangle {tfrac{1}{{{b_1}}},tfrac{1}{{{b_2}}}} rightrangleÂ  = lambda nabla g Rightarrow lambdaÂ  = 4{left( {tfrac{{{x_1}}}{{{b_1}}}} proper)^3} = 4{left( {tfrac{{{x_2}}}{{{b_2}}}} proper)^3} Leftrightarrow {x_2} = tfrac{{{b_2}}}{{{b_1}}}{x_1}\$\$

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Plug this relation into the previous inequality, specifically

##fleft( {{x_1},tfrac{{{b_2}}}{{{b_1}}}{x_1}} proper) = {left( {tfrac{{{x_1}}}{{{b_1}}}} proper)^4} + {left( {tfrac{{{x_1}}}{{{b_1}}}} proper)^4} leq 2 Rightarrow {left( {tfrac{{{x_1}}}{{{b_1}}}} proper)^2} leq 1 = :{max _{1 leq i leq 2}}left{ {{{left( {frac{{{x_i}}}{{{b_i}}}} proper)}^2}} proper}##

the place ##{max _{1 leq i leq n}}left{ {{a_i}} proper}: = max left{ {{a_1},{a_2}, ldots ,{a_n}} proper}##. Checking to see if ##S_2^2 subset S_1^2## by plugging in maximal values we get ##sumlimits_{ok = 1}^2 {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^2}}Â  leq 2{max _{1 leq i leq 2}}left{ {{{left( {frac{{{x_i}}}{{{b_i}}}} proper)}^2}} proper} = 2##, and therefore ##S_2^2 subset S_1^2## the bottom case is confirmed to carry.

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(ii) Now we show induction over n: assume that ##Pleft( {n,b} proper)## holds for some mounted ##n in {mathbb{Z}^ + }## and the place b is as partly (i). Then we have to show that ##Pleft( {n + 1,b} proper)## is subsequently true. For this drawback meaning proving ##Pleft( {n,1} proper):S_2^n subset S_2^n Rightarrow Pleft( {n + 1,1} proper):S_2^{n + 1} subset S_2^{n + 1}##.

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Assume \$\$Pleft( {n,1} proper) :S_2^n: = left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}|sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^4}}Â  leq n} proper} subset left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}| sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^2}}Â  leq n} proper} = :S_1^n\$\$ (it seems we won’t want this assumption to show what must be confirmed, just a few easy calculus). Now letâ€™s take a look at

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\$\$Pleft( {n + 1,1} proper):S_2^{n + 1}: = left{ {left( {{x_1},{x_2}, ldots ,{x_{n + 1}}} proper) in {mathbb{R}^{n + 1}}|sumlimits_{ok = 1}^{n + 1} {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^4}}Â  leq n + 1} proper} subset left{ {left( {{x_1},{x_2}, ldots ,{x_{n + 1}}} proper) in {mathbb{R}^{n + 1}}|sumlimits_{ok = 1}^{n + 1} {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^2}}Â  leq n + 1} proper} = :S_1^{n + 1}\$\$

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Allow us to simply use calculus to maximise the perform ##Fleft( {vec x} proper): = sumlimits_{ok = 1}^{n + 1} {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^2}} ## topic to the constraintÂ  ##Gleft( {vec x} proper): = sumlimits_{ok = 1}^{n + 1} {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^4}}Â  leq n + 1 ##. Utilizing the tactic of Lagrange Multipliers once more we’ve (in element kind): \$\${left( {nabla F} proper)_i} = 2tfrac{{{x_i}}}{{b_i^2}} = 4lambda tfrac{{x_i^3}}{{b_i^4}} = lambda {left( {nabla G} proper)_i} Rightarrow left( {{x_i} = 0} proper) vee left[ {left( {{x_i} ne 0} right) wedge left( {lambdaÂ  = tfrac{1}{2}{{left( {tfrac{{{b_i}}}{{{x_i}}}} right)}^2}} right)} right] \$\$ \$\$Rightarrow {z^2}: = {left( {tfrac{{{x_i}}}{{{b_i}}}} proper)^2} = {left( {tfrac{{{x_j}}}{{{b_j}}}} proper)^2},forall i ne j leq n + 1\$\$ is when the utmost happens as a result of clearly the zero was a minimal. Plugging into the constraint to get a sure for ##{z^2}## offers ##sumlimits_{ok = 1}^{n + 1} {{z^4}}Â  = left( {n + 1} proper){z^4} leq n + 1 Rightarrow {z^2} leq 1## after which to search out the precise most we get ##Fleft( {vec x} proper) leq sumlimits_{ok = 1}^{n + 1} {{z^2}}Â  = left( {n + 1} proper){z^2} = n + 1## and therefore ##S_2^{n + 1} subset S_1^{n + 1}##.

(iii) Now for the induction on N step: Assume ##Pleft( {n,N} proper)##Â  holds for some mounted N. We should show that ##Pleft( {n,N + 1} proper)## is subsequently true. Notably we won’t use the belief on this step. Now we have ##S_{N + 2}^n: = left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}|sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N + 4}}}Â  leq n} proper}##. From this the inequality ##sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N + 4}}}Â  leq n## holds for mounted N, so we should always maximize ##sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N + 2}}}##Â  topic to that constraint: continuing by Lagrange Multipliers methodology in element kind we’ve ##left( {2N + 2} proper)tfrac{{x_k^{2N + 1}}}{{b_k^{2N + 2}}} = left( {2N + 4} proper)lambda tfrac{{x_k^{2N + 3}}}{{b_k^{2N}}} Rightarrow frac{{N + 1}}{{N + 2}}tfrac{{b_k^2}}{{x_k^2}} = lambdaÂ  Rightarrow w: = tfrac{{x_k^2}}{{b_k^2}} = tfrac{{x_j^2}}{{b_j^2}}forall j ne ok leq n## now plug this worth into the constraint to get ##sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N + 4}}}Â  = sumlimits_{ok = 1}^n {{w^{2N + 4}}}Â  = n{w^{2N + 4}} leq n Rightarrow w = 1## which worth we plug into the perform being maximized to find out the utmost to be ##sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N + 2}}}Â  leq sumlimits_{ok = 1}^n {{w^{2N + 2}}}Â  = n## thus ##S_{N + 2}^n subset S_{N + 1}^n##. Therefore the lemma is true.

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#### Theorem 2.2: Equivalence of ##S^n## to ##P^n##

If ##S_N^n: = left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}|sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N}}}Â  leq n} proper}## and if ##{P^n}: = left{ {left( {{x_1},{x_2}, ldots ,{x_n}} proper) in {mathbb{R}^n}|left| {{x_k}} proper| leq {b_k}forall ok leq n} proper}##. Then ##{S^n} subsetÂ  cdotsÂ  subset S_{N + 1}^n subset S_N^n subsetÂ  cdotsÂ  subset S_1^n## the place ##{S^n}: = bigcaplimits_{j = 1}^inftyÂ  {S_j^n} ## and ##{S^n} = {P^n}##.

##### Proof:

That ##{S^n} subsetÂ  cdotsÂ  subset S_{N + 1}^n subset S_N^n subsetÂ  cdotsÂ  subset S_1^n## we’ve lemma 2.1 and that if ##vec x in {S^n}: = bigcaplimits_{j = 1}^inftyÂ  {S_j^n} ##, then ##vec x in S_N^nforall N in {mathbb{Z}^ + }##. For the equality ##{S^n} = {P^n}##, suppose ##exists vec y in Sleft( {{S^n},{P^n}} proper) = left( {{S^n} â€“ {P^n}} proper) cup left( {{P^n} â€“ {S^n}} proper)## , the place ##Sleft( {A,B} proper)## is the symmetric distinction of units A and B. Then both

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(i) \$\$start{gathered}exists vec y in left( {{S^n} â€“ {P^n}} proper) = left{ {vec x in {mathbb{R}^n}|left( {vec x in {S^n}} proper) wedge left( {vec x notin {P^n}} proper)} proper} = left{ {vec x in {mathbb{R}^n}|vec x in {S^n} = mathop {lim }limits_{M to infty } ,bigcaplimits_{j = 1}^M {S_j^n}Â  = mathop {lim }limits_{M to infty } S_M^n,} proper} cap left{ {vec x in {mathbb{R}^n}|exists ok in {mathbb{Z}^ + } mathrelbackepsilonÂ  left( {1 leq ok leq n} proper) wedge left( {tfrac{{left| {{x_k}} proper|}}{{{b_k}}} > 1} proper)} proper} finish{gathered}\$\$

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the place we’ve utilized \$\${S^n} subsetÂ  cdotsÂ  subset S_{N + 1}^n subset S_N^n subsetÂ  cdotsÂ  subset S_1^n\$\$ within the analysis of the restrict. Moreover \$\$exists ok in {mathbb{Z}^ + } mathrelbackepsilonÂ  left( {1 leq ok leq n} proper) wedge left( {tfrac{{left| {{x_k}} proper|}}{{{b_k}}} > 1} proper) Rightarrow exists varepsilonÂ  > 0 mathrelbackepsilonÂ  tfrac{{left| {{x_k}} proper|}}{{{b_k}}} = 1 + varepsilonÂ  Rightarrow exists M in {mathbb{Z}^ + } mathrelbackepsilonÂ  N geq M Rightarrow sumlimits_{j = 1}^n {{{left( {frac{{{x_j}}}{{{b_j}}}} proper)}^{2N}}}Â  > n\$\$

Even for the minimal situation that ##{x_j} = 0forall j ne ok leq n## we’ve by the binomial theorem that \$\$sumlimits_{j = 1}^n {{{left( {frac{{{x_j}}}{{{b_j}}}} proper)}^{2N}}}Â  = {left( {1 + varepsilon } proper)^{2N}} = sumlimits_{j = 1}^{2N} {left( {left. {start{array}{*{20}{c}}{2N} j finish{array}} proper)} proper.} ,{varepsilon ^j} = 1 + 2NvarepsilonÂ  + Nleft( {2N â€“ 1} proper){varepsilon ^2} +Â  cdots\$\$Â  so select ##N in {mathbb{Z}^ + } mathrelbackepsilonÂ  2NvarepsilonÂ  geq n Rightarrow N = leftlceil {tfrac{{n â€“ 1}}{{2varepsilon }}} rightrceil##Â  Â and therefore ##left( {{S^n} â€“ {P^n}} proper) = emptyset##Â  a contradiction.

Or (ii)

\$\$start{gathered}exists vec y in left( {{P^n} â€“ {S^n}} proper) = left{ {vec x in {mathbb{R}^n}|left( {vec x in {P^n}} proper) wedge left( {vec x notin {S^n}} proper)} proper} = left{ {vec x in {mathbb{R}^n}|tfrac{{left| {{x_k}} proper|}}{{{b_k}}} leq 1{textual content{ for }}ok = 1,2, ldots ,n,} proper} cap left{ {vec x in {mathbb{R}^n}|vec x notin bigcaplimits_{j = 1}^inftyÂ  {S_j^n} } proper} finish{gathered}\$\$

we’ve from the left-hand set ##tfrac{{left| {{x_k}} proper|}}{{{b_k}}} leq 1## for ##ok = 1,2, ldots ,n##, which means that \$\$forall N in {mathbb{Z}^ + },sumlimits_{j = 1}^n {{{left( {frac{{{x_j}}}{{{b_j}}}} proper)}^{2N}}}Â  leq n cdot {1^{2N}} = n Rightarrow vec x in bigcaplimits_{j = 1}^inftyÂ  {S_j^n}\$\$ in order that ##left( {{P^n} â€“ {S^n}} proper) = emptyset ##, a contradiction. Having discovered contradictions in each instances (i) and (ii) we conclude \$\$Sleft( {{S^n},{P^n}} proper) = left( {{S^n} â€“ {P^n}} proper) cup left( {{P^n} â€“ {S^n}} proper) = emptyset \$\$ a contradiction to the previous assumption proving ##{S^n} = {P^n}##.

## Unit Hypercube Integrals

For reference, we state right here the Lebesgue Dominated Convergence Theorem,

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#### Theorem 3.1 (DCT): Lebesgue Dominated Convergence Theorem

Suppose ##E in mathfrak{M}## (the household of measurable units). Let ##left{ {{f_n}} proper}##Â  be a sequence of measurable capabilities such that ##{f_n}left( x proper) to fleft( x proper)## nearly in all places on E as ##n to infty ##. If there exists a perform ##g in mathfrak{L}left( muÂ  proper)## on E (g is Lebesgue integrable with respect to ##mu ## on E), such that ##left| {{f_n}left( x proper),} proper| leq gleft( x proper)##, for ##n = 1,2,3, ldots## and ##forall x in E##, then \$\$mathop {lim }limits_{n to infty } ,int_E {{f_n}dmu }Â  = int_E {fdmu } \$\$[2]

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#### Theorem 3.2: Dominated Convergence Theorem with Nesting Property

Let ##{A_n},A subset {mathbb{R}^n}## and ##{A_n},A in mathfrak{M}## for ##n in {mathbb{Z}^ + }## such that \$\$A subsetÂ  cdotsÂ  subset {A_{n + 1}} subset {A_n} subsetÂ  cdotsÂ  subset {A_1}\$\$ and let ##A: = bigcaplimits_{j = 1}^inftyÂ  {{A_j}} ##. Then for Lebesgue measurable ##f:{mathbb{R}^n} to mathbb{C}## outline the set perform ##phi :{A_1} to mathbb{C}## by

\$\$phi left( E proper): = int_E {fdmu } ,forall E subset {A_1}\$\$ Then \$\$mathop {lim }limits_{n to infty } ,phi left( {{A_n}} proper) = phi left( A proper)\$\$ which is to say explicitly that \$\$mathop {lim }limits_{n to infty } ,int_{{A_n}} {fdmu }Â  = int_A {fdmu }\$\$

##### Proof:

Outline ##{chi _E}left( x proper): = left{ {start{array}{*{20}{c}}{1,}&{x in E} {0,}&{x notin E}finish{array}} proper.##

the attribute perform of the set E. Then let ##{f_n}: = f circ {chi _{bigcaplimits_{j = 1}^n {{A_j}} }} = left{ {start{array}{*{20}{c}}{fleft( x proper),}&{x in {A_n}} {0,}&{x notin {A_n}}finish{array}} proper.##

the place the nesting property has been used. Additionally let ##f: = f circ {chi _A} = left{ {start{array}{*{20}{c}}{fleft( x proper),}&{x in A} {0,}&{x notin A}finish{array}} proper.##

Then \$\${f_1}left( x proper) geq {f_2}left( x proper) geqÂ  cdotsÂ  geq {f_n}left( x proper) geq {f_{n + 1}}left( x proper) geqÂ  cdotsÂ  geq fleft( x proper)\$\$

therefore by the Lebesgue Dominated Convergence Theorem with \$\${f_1}left( x proper) geq left| {{f_n}left( x proper)} proper|,,forall x in {mathbb{R}^n},forall n in {mathbb{Z}^ + }\$\$

we’ve

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\$\$mathop {lim }limits_{n to infty } ,phi left( {{A_n}} proper) = mathop {lim }limits_{n to infty } ,int_{{A_n}} {fdmu }Â  = int_A {fdmu }Â  = phi left( A proper)\$\$

the required end result.

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#### A Wanted Restrict 3.3:

Weâ€™ll use this end result so much, almost each drawback, so right here it goes:

3.3) Use the Euler product type of the gamma perform to point out that ##mathop {lim }limits_{N to infty } {{{Gamma ^n}left( {1 + frac{ok}{N}} proper)} mathord{left/ {vphantom {{{Gamma ^n}left( {1 + frac{ok}{N}} proper)} {Gamma left( {1 + frac{{kn}}{N}} proper)}}} proper. } {Gamma left( {1 + frac{{kn}}{N}} proper)}} = 1## .

\$\$start{gathered}mathop {lim }limits_{N to infty } tfrac{{{Gamma ^n}left( {1 + tfrac{ok}{N}} proper)}}{{Gamma left( {1 + tfrac{{kn}}{N}} proper)}} = mathop {lim }limits_{N to infty } tfrac{{{{left( {1 + tfrac{ok}{N}} proper)}^{ â€“ n}}prodlimits_{lambdaÂ  = 1}^inftyÂ  {tfrac{{{{left( {1 + tfrac{1}{lambda }} proper)}^{nleft( {1 + tfrac{ok}{N}} proper)}}}}{{{{left( {1 + tfrac{{1 + tfrac{{kn}}{N}}}{lambda }} proper)}^n}}}} }}{{{{left( {1 + tfrac{{kn}}{N}} proper)}^{ â€“ 1}}prodlimits_{j = 1}^inftyÂ  {tfrac{{{{left( {1 + tfrac{1}{j}} proper)}^{1 + tfrac{{kn}}{N}}}}}{{left( {1 + tfrac{{1 + tfrac{{kn}}{N}}}{j}} proper)}}} }} = mathop {lim }limits_{N to infty } {left( {1 + tfrac{ok}{N}} proper)^{ â€“ n}}left( {1 + tfrac{{kn}}{N}} proper)prodlimits_{lambdaÂ  = 1}^inftyÂ  {left[ {tfrac{{{{left( {1 + tfrac{1}{lambda }} right)}^{n â€“ 1}}left( {1 + tfrac{{1 + tfrac{{kn}}{N}}}{lambda }} right)}}{{{{left( {1 + tfrac{{1 + tfrac{k}{N}}}{lambda }} right)}^n}}}} right]}Â  = prodlimits_{lambdaÂ  = 1}^inftyÂ  {left[ {{{left( {1 + tfrac{1}{lambda }} right)}^{n â€“ 1}}mathop {lim }limits_{N to infty } tfrac{{left( {1 + tfrac{{1 + tfrac{{kn}}{N}}}{lambda }} right)}}{{{{left( {1 + tfrac{{1 + tfrac{k}{N}}}{lambda }} right)}^n}}}} right]}Â  = prodlimits_{lambdaÂ  = 1}^inftyÂ  {left[ {{{left( {1 + tfrac{1}{lambda }} right)}^{n â€“ 1}}tfrac{{left( {1 + tfrac{1}{lambda }} right)}}{{{{left( {1 + tfrac{1}{lambda }} right)}^n}}}} right]}Â  = 1 finish{gathered}\$\$

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#### Instance 3.4: Content material of the Orthotope

For instance of Theorem 4.2 think about the orthotope \$\${S^{nleft(Â  +Â  proper)}}: = bigcaplimits_{j = 1}^inftyÂ  {S_j^{nleft(Â  +Â  proper)}} \$\$ the place we outline \$\$S_N^{nleft(Â  +Â  proper)}: = left{ {vec x in {mathbb{R}^n}|{x_i} geq 0forall i,sumlimits_{ok = 1}^n {{{left( {frac{{{x_k}}}{{{b_k}}}} proper)}^{2N}}}Â  leq n} proper}\$\$

Notice that by lemma 2.1 ##S_N^{nleft(Â  +Â  proper)}## and ##{S^{nleft(Â  +Â  proper)}}## fulfill the nesting property within the hypotheses of theorem 2.2 in order that we will consider the content material integral by corollary 1.2,

\$\$start{gathered}{textual content{content material}}left( {{S^{nleft(Â  +Â  proper)}}} proper) = mathop {lim }limits_{N to infty } ,iint {mathopÂ  cdots limits_{S_N^{nleft(Â  +Â  proper)}} int {d{z_n} ldots d{z_2}d{z_1}} } = mathop {lim }limits_{N to infty } ,left{ {{{{n^{sumlimits_{ok = 1}^n {tfrac{1}{{2N}}} }}prodlimits_{q = 1}^n {left[ {frac{{{b_p}}}{{2N}}Gamma left( {frac{1}{{2N}}} right)} right]} } mathord{left/{vphantom {{{n^{sumlimits_{ok = 1}^n {tfrac{1}{{2N}}} }}prodlimits_{q = 1}^n {left[ {frac{{{b_p}}}{{2N}}Gamma left( {frac{1}{{2N}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {frac{1}{{2N}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {frac{1}{{2N}}} } proper)}}} proper} = mathop {lim }limits_{N to infty } ,left[ {{n^{tfrac{n}{{2N}}}}tfrac{{{Gamma ^n}left( {frac{1}{{2N}}} right)prodlimits_{q = 1}^n {{b_p}} }}{{{{left( {2N} right)}^n}Gamma left( {1 + tfrac{n}{{2N}}} right)}}} right] = prodlimits_{q = 1}^n {left( {{b_p}} proper) cdot } underbrace {mathop {lim }limits_{N to infty } ,left( {{n^{tfrac{n}{{2N}}}}} proper)}_{ = {n^0}} cdot underbrace {mathop {lim }limits_{N to infty } left[ {tfrac{{{Gamma ^n}left( {1 + frac{1}{{2N}}} right)}}{{Gamma left( {1 + tfrac{n}{{2N}}} right)}}} right]}_{ = 1{textual content{ Restrict 3.3}}} = prodlimits_{q = 1}^n {{b_p}}Â  finish{gathered}\$\$

because it should.

Instance 3.4: The Zeta Perform

Present that \$\$Zleft( n proper): = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{ok = 1}^n {{x_k}} } proper)}^{ â€“ 1}}d{x_n} ldots d{x_2}d{x_1}} } }Â  = zeta left( n proper)forall n in {mathbb{Z}^ + } â€“ left{ 1 proper}\$\$

the place ##zeta left( {, cdot ,} proper)## is the Riemann zeta perform. The area of integration is a unit hypercube with one vertex on the origin and the vertex which is farthest from mentioned level is at ##left( {1,1, ldots ,1} proper)## suggesting using the sequence of units right here outlined all through the remnant of this Perception \$\$C_N^n: = left{ {vec x in {mathbb{R}^n}|{x_i} geq 0forall i,sumlimits_{ok = 1}^n {x_k^{2N}}Â  leq n â€“ 1} proper}\$\$ and likewise outline the set ##{C^n}: = bigcaplimits_{j = 1}^inftyÂ  {C_j^n} ##. Then we’ve

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\$\$start{gathered}Zleft( n proper) = mathop {lim }limits_{N to infty } ,int {int_{C_N^n} { cdots int {tfrac{{prodnolimits_{i = 1}^n {d{x_i}} }}{{1 â€“ prodnolimits_{ok = 1}^n {{x_k}} }}} } }Â  = mathop {lim }limits_{N to infty } ,int {int_{C_N^n} { cdots int {sumlimits_{ok = 1}^inftyÂ  {prodlimits_{i = 1}^n {left( {x_i^{ok â€“ 1}d{x_i}} proper)} } } } }Â  = mathop {lim }limits_{N to infty } ,sumlimits_{ok = 1}^inftyÂ  {int {int_{C_N^n} { cdots int {prodlimits_{i = 1}^n {left( {x_i^{ok â€“ 1}d{x_i}} proper)} } } } }Â  = mathop {lim }limits_{N to infty } ,sumlimits_{ok = 1}^inftyÂ  {{{left( {n â€“ 1} proper)}^{sumlimits_{p = 1}^n {frac{ok}{{2N}}} }}{{prodlimits_{q = 1}^n {left[ {frac{{{1^k}}}{{2N}}Gamma left( {frac{k}{{2N}}} right)} right]} } mathord{left/ {vphantom {{prodlimits_{q = 1}^n {left[ {frac{{{1^k}}}{{2N}}Gamma left( {frac{k}{{2N}}} right)} right]} } {Gamma left( {1 + sumlimits_{i = 1}^n {frac{ok}{{2N}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{i = 1}^n {frac{ok}{{2N}}} } proper)}}} = mathop {lim }limits_{N to infty } ,left{ {sumlimits_{ok = 1}^inftyÂ  {left[ {{{left( {n â€“ 1} right)}^{tfrac{{nk}}{{2N}}}} cdot tfrac{1}{{{k^n}}} cdot tfrac{{{Gamma ^n}left( {1 + tfrac{k}{{2N}}} right)}}{{Gamma left( {1 + tfrac{{nk}}{{2N}}} right)}}} right]} } proper} = sumlimits_{ok = 1}^inftyÂ  {tfrac{1}{{{ok^n}}} cdot underbrace {mathop {lim }limits_{N to infty } ,{{left( {n â€“ 1} proper)}^{tfrac{{nk}}{{2N}}}}}_{ = {{left( {n â€“ 1} proper)}^0}} cdot underbrace {mathop {lim }limits_{N to infty } ,tfrac{{{Gamma ^n}left( {1 + tfrac{ok}{{2N}}} proper)}}{{Gamma left( {1 + tfrac{{nk}}{{2N}}} proper)}}}_{ = 1{textual content{ Restrict 3.3}}}}Â  = sumlimits_{ok = 1}^inftyÂ  {tfrac{1}{{{ok^n}}}}Â  = zeta left( n proper) finish{gathered}\$\$

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This was a well known integral, however probably demonstrated a brand new solution to obtain a recognized end result. We will use this system of evaluating improper a number of integrals over the unit hypercube on many capabilities on this part. However first,

#### An alternate analysis of the above instance:

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\$\$start{gathered}Zleft( n proper) = int {int_{{C^n}} { cdots int {tfrac{{prodnolimits_{i = 1}^n {d{x_i}} }}{{1 â€“ prodnolimits_{ok = 1}^n {{x_k}} }}} } }Â  = int {int_{{C^{n â€“ 1}}} { cdots int {ln left( {tfrac{1}{{1 â€“ prodnolimits_{ok = 1}^{n â€“ 1} {{x_k}} }}} proper) cdot prodlimits_{i = 1}^{n â€“ 1} {tfrac{{d{x_i}}}{{{x_i}}}} } } }Â  = int {int_{{C^{n â€“ 1}}} { cdots int {ln left[ {prodlimits_{q = 0}^infty Â {left( {1 + prodlimits_{k = 1}^{n â€“ 1} {x_k^{{2^q}}} } right)} } right] cdot prodlimits_{i = 1}^{n â€“ 1} {tfrac{{d{x_i}}}{{{x_i}}}} } } }Â  = sumlimits_{q = 0}^inftyÂ  {int {int_{{C^{n â€“ 1}}} { cdots int {ln left( {1 + prodlimits_{ok = 1}^{n â€“ 1} {x_k^{{2^q}}} } proper) cdot prodlimits_{i = 1}^{n â€“ 1} {tfrac{{d{x_i}}}{{{x_i}}}} } } } }Â  = mathop {lim }limits_{N to infty } ,sumlimits_{q = 0}^inftyÂ  {sumlimits_{ok = 1}^inftyÂ  {tfrac{{{{left( { â€“ 1} proper)}^{ok â€“ 1}}}}{ok}int {int_{C_N^{n â€“ 1}} { cdots int {prodlimits_{ok = 1}^{n â€“ 1} {x_k^{ok{2^q} â€“ 1}d{x_k}} } } } } }Â  = mathop {lim }limits_{N to infty } ,sumlimits_{q = 0}^inftyÂ  {sumlimits_{ok = 1}^inftyÂ  {tfrac{{{{left( { â€“ 1} proper)}^{ok â€“ 1}}}}{ok}{{left( {n â€“ 2} proper)}^{sumlimits_{p = 1}^n {frac{{ok{2^q}}}{{2N}}} }}{{prodlimits_{i = 1}^{n â€“ 1} {left[ {frac{1}{{2N}}Gamma left( {frac{{k{2^q}}}{{2N}}} right)} right]} } mathord{left/ {vphantom {{prodlimits_{i = 1}^{n â€“ 1} {left[ {frac{1}{{2N}}Gamma left( {frac{{k{2^q}}}{{2N}}} right)} right]} } {Gamma left( {1 + sumlimits_{j = 1}^{n â€“ 1} {frac{{ok{2^q}}}{{2N}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{j = 1}^{n â€“ 1} {frac{{ok{2^q}}}{{2N}}} } proper)}}} }Â  = sumlimits_{q = 0}^inftyÂ  {sumlimits_{ok = 1}^inftyÂ  {tfrac{{{{left( { â€“ 1} proper)}^{ok â€“ 1}}}}{{{ok^n}{2^{left( {n â€“ 1} proper)q}}}} cdot underbrace {mathop {lim }limits_{N to infty } ,{{left( {n â€“ 2} proper)}^{frac{{left( {n â€“ 1} proper)ok{2^q}}}{{2N}}}}}_{ = {{left( {n â€“ 2} proper)}^0}} cdot underbrace {mathop {lim }limits_{N to infty } ,tfrac{{{Gamma ^{n â€“ 1}}left( {1 + frac{{ok{2^q}}}{{2N}}} proper)}}{{Gamma left( {1 + tfrac{{left( {n â€“ 1} proper)ok{2^q}}}{{2N}}} proper)}}}_{ = 1{textual content{ Restrict 3.3}}}} } Â  = sumlimits_{q = 0}^inftyÂ  {sumlimits_{ok = 1}^inftyÂ  {tfrac{{{{left( { â€“ 1} proper)}^{ok â€“ 1}}}}{{{ok^n}}} cdot {{left( {tfrac{1}{{{2^{n â€“ 1}}}}} proper)}^q}} }Â  = sumlimits_{ok = 1}^inftyÂ  {left[ {tfrac{{{{left( { â€“ 1} right)}^{k â€“ 1}}}}{{{k^n}}}sumlimits_{q = 0}^inftyÂ  {{{left( {tfrac{1}{{{2^{n â€“ 1}}}}} right)}^q}} } right]}Â  = {left( {1 â€“ {2^{1 â€“ n}}} proper)^{ â€“ 1}}sumlimits_{ok = 1}^inftyÂ  {tfrac{{{{left( { â€“ 1} proper)}^{ok â€“ 1}}}}{{{ok^n}}}}Â  = sumlimits_{ok = 1}^inftyÂ  {tfrac{1}{{{ok^n}}}}Â  = zeta left( n proper) finish{gathered}\$\$

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The reader might use this system of evaluating a number of integrals over the unit hypercube to confirm the nextÂ

### Abstract of Capabilities Represented as A number of Integrals Over the Unit Hypercube:

The Lerch Transcendent: ##Phi left( {z,n,y} proper): = sumlimits_{q = 0}^inftyÂ  {frac{{{z^q}}}{{{{left( {q + y} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ zprodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {lambda _k^{y â€“ 1}d{lambda _k}} } } } }##

Legendre Chi Perform: ##{chi _n}left( z proper): = sumlimits_{q = 0}^inftyÂ  {frac{{{z^{2q + 1}}}}{{{{left( {2q + 1} proper)}^n}}} = zint_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ {z^2}prodlimits_{q = 1}^n {lambda _q^2} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

Polygamma Perform: ##{psi _n}left( z proper): = sumlimits_{q = 0}^inftyÂ  {frac{{{{left( { â€“ 1} proper)}^{n + 1}}n!}}{{{{left( {z + q} proper)}^{n + 1}}}} = {{left( { â€“ 1} proper)}^{n + 1}}n!int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{q = 1}^{n + 1} {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^{n + 1} {lambda _k^{z â€“ 1}d{lambda _k}} } } } }##

Polylogarithm of Order n: ##{textual content{L}}{{textual content{i}}_n}left( z proper): = sumlimits_{q = 1}^inftyÂ  {frac{{{z^q}}}{{{q^n}}} = zint_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ zprodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

Hurwitz Zeta Perform: ##zeta left( {n,z} proper): = sumlimits_{q = 0}^inftyÂ  {frac{1}{{{{left( {q + z} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {lambda _k^{z â€“ 1}d{lambda _k}} } } } }##

Riemann Zeta Perform: ##zeta left( n proper): = sumlimits_{q = 1}^inftyÂ  {frac{1}{{{q^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

Dirichlet Beta Perform: ##beta left( n proper): = sumlimits_{q = 0}^inftyÂ  {frac{{{{left( { â€“ 1} proper)}^q}}}{{{{left( {2q + 1} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 + prodlimits_{q = 1}^n {lambda _q^2} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

Dirichlet Eta Perform: ##eta left( n proper): = sumlimits_{q = 1}^inftyÂ  {frac{{{{left( { â€“ 1} proper)}^{q â€“ 1}}}}{{{q^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 + prodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

Dirichlet Lambda Perform: ##lambda left( n proper): = sumlimits_{q = 0}^inftyÂ  {frac{1}{{{{left( {2q + 1} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{q = 1}^n {lambda _q^2} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

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Workouts

1) Confirm the remaining unit hypercube integral identities from the above abstract.

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Solutions to Workouts â€“ Unit Hypercube Integrals

1) Confirm the remaining unit hypercube integral identities from the above abstract.

a) The Lerch Transcendent: ##Phi left( {z,n,y} proper): = sumlimits_{q = 0}^inftyÂ  {frac{{{z^q}}}{{{{left( {q + y} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ zprodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {lambda _k^{y â€“ 1}d{lambda _k}} } } } }##

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\$\$start{eqnarray*}Phi ( z,n,y ) &=& int_0^1 int_0^1Â  cdots int_0^1 left( 1 â€“ zprodlimits_{q = 1}^n lambda _q proper) ^{ â€“ 1}prodlimits_{ok = 1}^n lambda _k^{y â€“ 1}d{lambda _k} &=& limlimits_{Ntoinfty}iint {mathopÂ  cdots limits_{C_N^n}} int sumlimits_{ok=0}^infty z^kprodlimits_{q = 1}^n lambda _q^{ok+y-1}d{lambda _q}Â  finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} Phi left( {z,n,y} proper)Â  &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty z^ok {(n-1)^{sumlimits_{p = 1}^n {tfrac{{{ok+y}}}{{{2N}}}} }}{{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{k+y}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{k+y}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{ok+y}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{ok+y}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} underbrace{{(n-1)^{sumlimits_{p = 1}^n {tfrac{{{ok+y}}}{{{2N}}}} }}}_{=1}sumlimits_{ok=0}^infty z^ok {{ {{tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{ok+y}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{ok+y}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(ok+y)n}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(ok+y)n}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty z^okÂ  tfrac{1}{(2N)^n} cdottfrac{(2N)^n}{(ok+y)^n} underbrace{Gamma ^nleft(Â  Â 1+ tfrac{ok+y}{2N} proper) mathord{left/{vphantom {tfrac{1}{(2N)^n}cdottfrac{(2N)^n}{(ok+y)^n}Gamma ^nleft( tfrac{ok+y}{2N} proper)Â  } Gamma left( 1 +tfrac{(ok+y)n}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3} } Â &=&Â  sumlimits_{ok=0}^inftyÂ  Â tfrac{z^ok}{(ok+y)^n} finish{eqnarray*}\$\$

b) Legendre Chi Perform: ##chi _n ( z ) = z int_0^1 int_0^1Â  cdots int_0^1 left( 1 â€“ z^2prodlimits_{q = 1}^n lambda _q^2 proper) ^{ â€“ 1} prodlimits_{ok = 1}^n d{lambda _k} ##

\$\$start{eqnarray*}chi _n ( z ) &=& zint_0^1 int_0^1Â  cdots int_0^1 left( 1 â€“ z^2prodlimits_{q = 1}^n lambda _q^2 proper) ^{ â€“ 1}prodlimits_{ok = 1}^n d{lambda _k} &=& zlimlimits_{Ntoinfty}iint {mathopÂ  cdots limits_{C_N^n} int sumlimits_{ok=0}^infty z^{2k}prodlimits_{q = 1}^n lambda _q^{2k}d{lambda _q} } &=& zlimlimits_{Ntoinfty}sumlimits_{ok=0}^infty z^{2k}iint {mathopÂ  cdots limits_{C_N^n} int prodlimits_{q = 1}^n lambda _q^{2k}d{lambda _q} } finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} chi _n left( z proper)Â  &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty z^{2k+1} {(n-1)^{sumlimits_{p = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} }}{{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{2k+1}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{2k+1}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} underbrace{{(n-1)^{sumlimits_{p = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} }}}_{=1}sumlimits_{ok=0}^infty z^{2k+1}{{ {{tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{2k+1}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{2k+1}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(2k+1)n}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(2k+1)n}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty z^{2k+1}Â  tfrac{1}{(2N)^n} cdottfrac{(2N)^n}{(2k+1)^n} underbrace{Gamma ^nleft(Â  Â 1+ tfrac{2k+1}{2N} proper) mathord{left/{vphantom {tfrac{1}{(2N)^n}cdottfrac{(2N)^n}{(2k+1)^n}Gamma ^nleft( tfrac{2k+1}{2N} proper)Â  } Gamma left( 1 +tfrac{(2k+1)n}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3 } } Â &=&Â  sumlimits_{ok=0}^inftyÂ  Â tfrac{z^{2k+1}}{(2k+1)^n} finish{eqnarray*}\$\$

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c) Polygamma Perform: ##{psi _n}left( z proper): = sumlimits_{q = 0}^inftyÂ  {frac{{{{left( { â€“ 1} proper)}^{n + 1}}n!}}{{{{left( {z + q} proper)}^{n + 1}}}} = {{left( { â€“ 1} proper)}^{n + 1}}n!int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{q = 1}^{n + 1} {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^{n + 1} {lambda _k^{z â€“ 1}d{lambda _k}} } } } }##

\$\$start{eqnarray*}psi _n ( z ) &=& (-1)^{n+1}n! int_0^1 int_0^1Â  cdots int_0^1 left( 1 â€“ prodlimits_{q = 1}^{n+1} lambda _q proper) ^{ â€“ 1}prodlimits_{ok = 1}^{n+1}lambda_k^{z-1} d{lambda _k} &=& (-1)^{n+1}n! limlimits_{Ntoinfty}sumlimits_{ok=0}^inftyiint {mathopÂ  cdots limits_{C_N^{n+1}} intÂ  prodlimits_{q = 1}^{n+1} lambda _q^{z+k-1}d{lambda _q} } finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} psi _n left( z proper)Â  &=& (-1)^{n+1}n!limlimits_{Ntoinfty} sumlimits_{q=0}^inftyÂ  {n^{sumlimits_{p = 1}^{n+1} {tfrac{{{z+q}}}{{{2N}}}} }}{{prodlimits_{ok = 1}^{n+1} {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{z+q}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{ok = 1}^{n+1} {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{z+q}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{lambda = 1}^{n+1} {tfrac{{{z+q}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{lambda = 1}^n {tfrac{{{z+q}}}{{{2N}}}} } proper)}} &=& (-1)^{n+1}n!limlimits_{Ntoinfty} underbrace{{n^{sumlimits_{p = 1}^{n+1} {tfrac{{{z+q}}}{{{2N}}}} }}}_{=n^0}sumlimits_{q=0}^infty tfrac{{(2N)^{n+1}}}{(z+q)^{n+1}}cdot{{ {{tfrac{{1}}{{{(2N)^{n+1}}}}Gamma ^{n+1}left( {1+tfrac{{{z+q}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{(2N)^{n+1}}}{(z+q)^{n+1}}cdottfrac{{1}}{{{(2N)^{n+1}}}}Gamma ^{n+1}left( {1+tfrac{{{z+q}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(2k+1)(n+1)}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(z+q)(n+1)}}}{{{2N}}}} } proper)}} &=& (-1)^{n+1}n!limlimits_{Ntoinfty} sumlimits_{q=0}^inftyÂ  tfrac{1}{(z+q)^{n+1}} underbrace{Gamma ^{n+1}left(Â  Â 1 + tfrac{z+q}{2N} proper) mathord{left/{vphantom {tfrac{1}{(z+q)^{n+1}}Gamma ^{n+1}left( 1+tfrac{z+q}{2N} proper)Â  } Gamma left( 1 +tfrac{(z+q)(n+1)}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3 } } Â &=&Â  sumlimits_{q=0}^inftyÂ  Â tfrac{(-1)^{n+1}n!}{(z+q)^{n+1}} finish{eqnarray*}\$\$

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d) Polylogarithm of Order n: ##{textual content{L}}{{textual content{i}}_n}left( z proper): = sumlimits_{q = 1}^inftyÂ  {frac{{{z^q}}}{{{q^n}}} = zint_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ zprodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

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\$\$start{eqnarray*}{textual content{L}}{{textual content{i}}_n} ( z )Â  &=& zint_0^1 int_0^1Â  cdots int_0^1 left( 1 â€“ zprodlimits_{q = 1}^n lambda _q proper) ^{ â€“ 1}prodlimits_{ok = 1}^n d{lambda _k} &=& zlimlimits_{Ntoinfty}iint {mathopÂ  cdots limits_{C_N^n} int sumlimits_{ok=0}^infty z^{ok}prodlimits_{q = 1}^n lambda _q^{ok}d{lambda _q} } &=& limlimits_{Ntoinfty}sumlimits_{ok=0}^infty z^{ok+1}iint {mathopÂ  cdots limits_{C_N^n} int prodlimits_{q = 1}^n lambda _q^{ok}d{lambda _q} } finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} {textual content{L}}{{textual content{i}}_n} ( z ) Â &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty z^{ok+1} {(n-1)^{sumlimits_{p = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} }}{{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{k+1}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{k+1}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} underbrace{{(n-1)^{sumlimits_{p = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} }}}_{=n^0}sumlimits_{ok=0}^infty z^{ok+1}{{ {{tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{ok+1}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{ok+1}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(ok+1)n}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(ok+1)n}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty z^{ok+1}Â  tfrac{1}{(2N)^n} cdottfrac{(2N)^n}{(ok+1)^n} underbrace{Gamma ^nleft(Â  Â 1+ tfrac{ok+1}{2N} proper) mathord{left/{vphantom {tfrac{1}{(2N)^n}cdottfrac{(2N)^n}{(ok+1)^n}Gamma ^nleft( tfrac{ok+1}{2N} proper)Â  } Gamma left( 1 +tfrac{(ok+1)n}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3 } } Â &=&Â  sumlimits_{ok=1}^inftyÂ  Â tfrac{z^ok}{ok^n} finish{eqnarray*}\$\$

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e) Hurwitz Zeta Perform: ##zeta left( {n,z} proper): = sumlimits_{q = 0}^inftyÂ  {frac{1}{{{{left( {q + z} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {lambda _k^{z â€“ 1}d{lambda _k}} } } } }##

\$\$start{eqnarray*}zeta left( {n,z} proper) &=& int_0^1 int_0^1Â  cdots int_0^1 left( 1 â€“ prodlimits_{q = 1}^n lambda _q proper) ^{ â€“ 1}prodlimits_{ok = 1}^n lambda_k^{z-1} d{lambda _k} &=& limlimits_{Ntoinfty}iint {mathopÂ  cdots limits_{C_N^n} int sumlimits_{ok=0}^infty prodlimits_{q = 1}^n lambda _q^{z+k-1}d{lambda _q} } &=& limlimits_{Ntoinfty}sumlimits_{ok=0}^infty iint {mathopÂ  cdots limits_{C_N^n} int prodlimits_{q = 1}^n lambda _q^{z+k-1}d{lambda _q} } finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} zeta left( {n,z} proper) Â &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty {(n-1)^{sumlimits_{p = 1}^n {tfrac{{{z+ok}}}{{{2N}}}} }}{{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{z+k}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{z+k}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{z+ok}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{z+ok}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} underbrace{{(n-1)^{sumlimits_{p = 1}^n {tfrac{{{z+ok}}}{{{2N}}}} }}}_{=n^0}sumlimits_{ok=0}^infty {{ {{tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{z+ok}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{z+ok}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(z+ok)n}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(z+ok)n}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^inftyÂ  tfrac{1}{(2N)^n} cdottfrac{(2N)^n}{(z+ok)^n} underbrace{Gamma ^nleft(Â  Â 1+ tfrac{z+ok}{2N} proper) mathord{left/{vphantom {tfrac{1}{(2N)^n}cdottfrac{(2N)^n}{(z+ok)^n}Gamma ^nleft( tfrac{z+ok}{2N} proper)Â  } Gamma left( 1 +tfrac{(z+ok)n}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3 } } Â &=&Â  sumlimits_{ok=1}^inftyÂ  Â tfrac{1}{(z+ok)^n} finish{eqnarray*}\$\$

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f) Dirichlet Beta Perform: ##beta ( n ) : = sumlimits_{q = 0}^inftyÂ  {frac{{{{left( { â€“ 1} proper)}^q}}}{{{{left( {2q + 1} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 + prodlimits_{q = 1}^n {lambda _q^2} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

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\$\$start{eqnarray*}beta ( n ) &=& int_0^1 int_0^1Â  cdots int_0^1 left( 1 + prodlimits_{q = 1}^n lambda _q^2right) ^{ â€“ 1}prodlimits_{ok = 1}^n d{lambda _k} &=& limlimits_{Ntoinfty}iint {mathopÂ  cdots limits_{C_N^n} int sumlimits_{ok=0}^infty (-1)^kprodlimits_{q = 1}^n lambda _q^{2k}d{lambda _q} } &=& limlimits_{Ntoinfty}sumlimits_{ok=0}^infty (-1)^ok iint {mathopÂ  cdots limits_{C_N^n} int prodlimits_{q = 1}^n lambda _q^{2k}d{lambda _q} } finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} beta ( n ) &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty (-1)^ok {(n-1)^{sumlimits_{p = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} }}{{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{2k+1}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{2k+1}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} underbrace{{(n-1)^{sumlimits_{p = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} }}}_{=n^0}sumlimits_{ok=0}^infty (-1)^ok {{ {{tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{2k+1}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{2k+1}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(2k+1)n}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(2k+1)n}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty (-1)^ok tfrac{1}{(2N)^n} cdottfrac{(2N)^n}{(2k+1)^n} underbrace{Gamma ^nleft(Â  Â 1+ tfrac{2k+1}{2N} proper) mathord{left/{vphantom {tfrac{1}{(2N)^n}cdottfrac{(2N)^n}{(2k+1)^n}Gamma ^nleft( tfrac{2k+1}{2N} proper)Â  } Gamma left( 1 +tfrac{(2k+1)n}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3 } } Â &=&Â  sumlimits_{ok=0}^inftyÂ  Â tfrac{(-1)^ok}{(2k+1)^n} finish{eqnarray*}\$\$

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g) Dirichlet Eta Perform: ##eta left( n proper): = sumlimits_{q = 1}^inftyÂ  {frac{{{{left( { â€“ 1} proper)}^{q â€“ 1}}}}{{{q^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 + prodlimits_{q = 1}^n {{lambda _q}} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

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\$\$start{eqnarray*}eta ( n ) &=& int_0^1 int_0^1Â  cdots int_0^1 left( 1 + prodlimits_{q = 1}^n lambda _qright) ^{ â€“ 1}prodlimits_{ok = 1}^n d{lambda _k} &=& limlimits_{Ntoinfty}iint {mathopÂ  cdots limits_{C_N^n} int sumlimits_{ok=0}^infty (-1)^kprodlimits_{q = 1}^n lambda _q^{ok}d{lambda _q} } &=& limlimits_{Ntoinfty}sumlimits_{ok=0}^infty (-1)^ok iint {mathopÂ  cdots limits_{C_N^n} int prodlimits_{q = 1}^n lambda _q^{ok}d{lambda _q} } finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} eta ( n ) &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty (-1)^ok {(n-1)^{sumlimits_{p = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} }}{{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{k+1}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{k+1}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} underbrace{{(n-1)^{sumlimits_{p = 1}^n {tfrac{{{ok+1}}}{{{2N}}}} }}}_{=n^0}sumlimits_{ok=0}^infty (-1)^ok {{ {{tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{ok+1}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{ok+1}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(ok+1)n}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(ok+1)n}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^infty (-1)^ok tfrac{1}{(2N)^n} cdottfrac{(2N)^n}{(ok+1)^n} underbrace{Gamma ^nleft(Â  Â 1+ tfrac{ok+1}{2N} proper) mathord{left/{vphantom {tfrac{1}{(2N)^n}cdottfrac{(2N)^n}{(ok+1)^n}Gamma ^nleft( tfrac{ok+1}{2N} proper)Â  } Gamma left( 1 +tfrac{(ok+1)n}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3 } } Â &=&Â  sumlimits_{ok=1}^inftyÂ  Â tfrac{(-1)^{k-1}}{ok^n} finish{eqnarray*}\$\$

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h) Dirichlet Lambda Perform: ##lambda left( n proper): = sumlimits_{q = 0}^inftyÂ  {frac{1}{{{{left( {2q + 1} proper)}^n}}} = int_0^1 {int_0^1 { cdots int_0^1 {{{left( {1 â€“ prodlimits_{q = 1}^n {lambda _q^2} } proper)}^{ â€“ 1}}prodlimits_{ok = 1}^n {d{lambda _k}} } } } }##

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\$\$start{eqnarray*}lambda ( n ) &=& int_0^1 int_0^1Â  cdots int_0^1 left( 1 â€“ prodlimits_{q = 1}^n lambda _q^2right) ^{ â€“ 1}prodlimits_{ok = 1}^n d{lambda _k} &=& limlimits_{Ntoinfty}iint {mathopÂ  cdots limits_{C_N^n} int sumlimits_{ok=0}^infty prodlimits_{q = 1}^n lambda _q^{2k}d{lambda _q} } &=& limlimits_{Ntoinfty}sumlimits_{ok=0}^inftyÂ  iint {mathopÂ  cdots limits_{C_N^n} int prodlimits_{q = 1}^n lambda _q^{2k}d{lambda _q} } finish{eqnarray*}\$\$

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Then by Theorem 1.1, we’ve

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\$\$start{eqnarray*} lambda ( n ) &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^inftyÂ  {(n-1)^{sumlimits_{p = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} }}{{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{2k+1}}}{{{2N}}}} right)} right]} } mathord{left/{vphantom {{prodlimits_{q = 1}^n {left[ {tfrac{{1}}{{{2N}}}Gamma left( {tfrac{{{2k+1}}}{{{2N}}}} right)} right]} } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 + sumlimits_{ok = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} underbrace{{(n-1)^{sumlimits_{p = 1}^n {tfrac{{{2k+1}}}{{{2N}}}} }}}_{=n^0}sumlimits_{ok=0}^infty {{ {{tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{2k+1}}}{{{2N}}}} proper)}} } mathord{left/{vphantom {{ { {tfrac{{1}}{{{(2N)^n}}}Gamma ^nleft( {tfrac{{{2k+1}}}{{{2N}}}} proper)} } } {Gamma left( {1 +{tfrac{{{(2k+1)n}}}{{{2N}}}} } proper)}}} proper. } {Gamma left( {1 +Â  {tfrac{{{(2k+1)n}}}{{{2N}}}} } proper)}} &=& limlimits_{Ntoinfty} sumlimits_{ok=0}^inftyÂ  tfrac{1}{(2N)^n} cdottfrac{(2N)^n}{(2k+1)^n} underbrace{Gamma ^nleft(Â  Â 1+ tfrac{2k+1}{2N} proper) mathord{left/{vphantom {tfrac{1}{(2N)^n}cdottfrac{(2N)^n}{(2k+1)^n}Gamma ^nleft( tfrac{2k+1}{2N} proper)Â  } Gamma left( 1 +tfrac{(2k+1)n}{2N}Â  proper) } proper. }Â  }_{ =1 textual content{ Restrict 3.3 } } Â &=&Â  sumlimits_{ok=0}^inftyÂ  Â tfrac{1}{(2k+1)^n} finish{eqnarray*}\$\$

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