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# An Overview of Complicated Differentiation and Integration

### Summary

I wish to shed some mild on advanced evaluation with out getting all of the technical particulars in the best way that are essential for the exact remedies that may be discovered in lots of wonderful commonplace textbooks.

Evaluation is about differentiation. Therefore, advanced differentiation might be my place to begin. It’s concurrently my end line as a result of its inverse, the advanced integration, is intently interwoven with advanced differentiation. By the dearth of particulars, I imply that I’ll typically assume a disc if a star-shaped area or a merely linked open set can be enough; or assume a differentiable operate if differentiability as much as finitely many factors would already be enough. Additionally, the typically essential strategies of gluing triangles for an integration path, or the epsilontic inside a area might be omitted.

The statements listed as theorems, nonetheless, might be exact. A few of them may typically enable a wider vary of validity, i.e. extra generality. Nonetheless, the reader will discover the essential concepts, definitions, methods, and theorems of the residue calculus; and if nothing else, see the place all of the ##pi##’s in integral formulation come from.

### Complicated Differentiation

A operate ##f:Urightarrow V## is differentiable at ##x_0## if there’s a linear approximation ##D_{x_0}f=(Df)_{x_0}##, such that (Karl T.W. Weierstraß)
start{equation*}
f(x_0+v)=f(x_{0})+(D_{x_0}f)cdot v+r(v)
finish{equation*}
the place the error operate ##r## has the property, that it converges quicker to zero than linear in any path ##v##, which suggests
\$\$lim_{v rightarrow 0}frac{r(v)}{v}=0.\$\$
If we rearrange this formulation, then
start{align*}
D_{x_0}f&= lim_{vto 0}D_{x_0}f = lim_{vto 0}left( dfrac{f(x_0+v)-f(x_0)}{v}-dfrac{r(v)}{v}proper)=lim_{vto 0} dfrac{f(x_0+v)-f(x_0)}{v}
finish{align*}
reveals the formulation that’s generally used to outline a spinoff. It additionally exhibits that
\$\$
D_{x_0}f=(Df)_{x_0}=left. dfrac{df}{dx}proper|_{x_0}=f'(x_0)
\$\$
If ##f## is a operate of a one-dimensional vary, then the path ##v## can solely be a one-dimensional vector and we assume ##v=1## once we write
\$\$f'(x_0)=f'(x_0)cdot 1= D_{x_0}fcdot v.\$\$
The great thing about Weierstraß’s formulation lies in the truth that it carries all of the essential properties of a spinoff:

• locality:
Weierstraß’s formulation requires validity in a area of ##x_0.##
• linear operator and Leibniz rule:
##D_{x_0}## is a derivation, i.e. ##D_{x_0}(alphacdot f+betacdot g)=alphacdot D_{x_0}(f) +betacdot D_{x_0}(g)## and ##D_{x_0}(fcdot g)=D_{x_0}(f)g(x_0)+f(x_0)D_{x_0}(g).##
• linearity:
##D_{x_0}f## is a linear operate, i.e. ##D_{x_0}f(alpha v+beta w)=alpha D_{x_0}f(v)+beta D_{x_0}f(w).##
• directionality, slope:
A spinoff ##(D_{x_0}f) v ## is directional, specifically the slope in path ##v.##
• approximation:
##r(v)=o(v).##

By permitting all elements ##x_0,f,v,r,D,D_{x_0},D_{x_0}f, D_{*}f## in Weierstraß’s formulation to be variables of an operator in its most basic that means, we instantly open the door to fiber bundles, world and native sections, differential types and tangent- in addition to cotangent bundles. Additionally word that we didn’t distinguish between actual and sophisticated capabilities, but. This generality shall be our place to begin.

Each advanced linear operate ##varphi : mathbb{C}rightarrow mathbb{C}## is an actual linear operate, too, and might be written as
start{align*}
varphi(x+iy)&=underbrace{start{pmatrix}varphi_{11}&varphi_{12}varphi_{21}&varphi_{22}finish{pmatrix}}_{in mathbb{M}(2,mathbb{R})}(x+iy)=(varphi_{11}(x)+varphi_{12}(y))+i(varphi_{21}(x)+varphi_{22}(y))
finish{align*}
##mathbb{C}##-linearity signifies that
\$\$
(a+ib)varphi(x+iy)=varphi((ax-by)+i(bx+ay))
\$\$
and thus
start{align*}
(a+ib)&cdot (varphi_{11}(x)+varphi_{12}(y))+i(varphi_{21}(x)+varphi_{22}(y))
&=(varphi_{11}(ax)+varphi_{12}(ay)-varphi_{21}(bx)-varphi_{22}(by))
&phantom{=}+i(varphi_{11}(bx)+varphi_{12}(by)+varphi_{21}(ax)+varphi_{22}(ay))
&=(varphi_{11}(ax)-varphi_{11}(by)+varphi_{12}(bx)+varphi_{12}(ay))
&phantom{=}+i(varphi_{21}(ax)-varphi_{21}(by)+varphi_{22}(bx)+varphi_{22}(ay))
finish{align*}
start{align*}
varphi_{11}=varphi_{22}, &textual content{ and } ,varphi_{21}+varphi_{12}=0quad (*)
finish{align*}
We all know that ##D_{z_0}## is a ##mathbb{C}##-linear operator, and ##D_{z_0}f## is a ##mathbb{C}##-linear operate. If we write ##f=u+iv## then
\$\$
D_{z_0}f=D_{z_0}u+iD_{z_0}v=left( left.dfrac{partial u}{partial x}proper|_{z_0}dx+left.dfrac{partial u}{partial y}proper|_{z_0}dyright)+ileft(left.dfrac{partial v}{partial x}proper|_{z_0}dx+left.dfrac{partial v}{partial y}proper|_{z_0}dyright)
\$\$
and our situation ##(*)## of ##mathbb{C}##-linearity turns into
\$\$
left. dfrac{partial u}{partial x }proper|_{z_0} = left. dfrac{partial v}{partial y}proper|_{z_0}textual content{ and }left. dfrac{partial u}{partial y}proper|_{z_0}+left. dfrac{partial v}{partial x}proper|_{z_0}=0,,
\$\$
the Cauchy-Riemann equations. Which means that advanced differentiability is actual differentiability plus the Cauchy-Riemann equations. That is the primary distinction why advanced evaluation is greater than ##mathbb{R}^2## evaluation, advanced linearity.

### Holomorphic Capabilities

We see by induction that skew-symmetric matrices with a continuing diagonal (spiral symmetry) preserve their construction if we multiply them by themselves
start{align*}
start{pmatrix}a&b-b&aend{pmatrix}^{n+1}&=start{pmatrix}c&d-d&cend{pmatrix}cdot start{pmatrix}a&b-b&aend{pmatrix}=start{pmatrix}ca-db&cb+da-da-cb&-db+caend{pmatrix}
finish{align*}
Which means the Cauchy-Riemann property is invariant beneath exponentiation
\$\$
start{pmatrix}dfrac{partial u}{partial x}&dfrac{partial u}{partial y}[16pt] dfrac{partial v}{partial x}&dfrac{partial v}{partial y}finish{pmatrix}^n=start{pmatrix}dfrac{partial u}{partial x}&dfrac{partial u}{partial y}[16pt] -dfrac{partial u}{partial y}&dfrac{partial u}{partial x}finish{pmatrix}^n=
start{pmatrix}
pleft(dfrac{partial u}{partial x}, , ,dfrac{partial u}{partial y}proper) & qleft(dfrac{partial u}{partial x}, , ,dfrac{partial u}{partial y}proper) [16pt]
-qleft(dfrac{partial u}{partial x}, , ,dfrac{partial u}{partial y}proper) & pleft(dfrac{partial u}{partial x}, , ,dfrac{partial u}{partial y}proper)
finish{pmatrix}
\$\$
for some actual polynomials ##p(X,Y,n),q(X,Y,n)in mathbb{R}[X,Y].##

Definition: A posh operate ##f:Urightarrow V## is holomorphic whether it is advanced differentiable in a non-empty, open, and linked set ##U##, i.e. in a area. ##f## is known as meromorphic whether it is holomorphic as much as a set of remoted factors, its poles. If ##f## is holomorphic on the complete advanced airplane, then it’s referred to as an whole operate. An actual, or advanced operate is analytic at some extent ##z_0in U##, if there’s a energy sequence that converges to the operate worth in an open neighborhood of ##z_0##
\$\$
f(z)=sum_{n=0}^infty a_n(z-z_0)^n.
\$\$

Theorem: The next statements for ##f=u+iv, : ,Urightarrow V## are equal:

1. ##f## is advanced differentiable in a area ##U,## ##fin C^1(U).##
2. ##f## is unfair typically advanced differentiable in a area ##U,## ##fin C^infty (U).##
3. ##f## is holomorphic on ##U,## ##fin mathcal{O}(U).##
4. The true and imaginary elements of ##f## are repeatedly actual differentiable and fulfill the Cauchy-Riemann partial differential equations (CR).
5. ##f## is analytic on ##U.##
6. ##f## is actual differentiable and ##{displaystyle {tfrac{partial f}{partial {bar {z}}}}:={tfrac{1}{2}} left({tfrac{partial }{partial x}} + i {tfrac{partial }{partial y}}proper)}(f) =0.##

### Complicated Line Integrals

We’ve seen that the definition of advanced differentiation doesn’t require a change of actual differentiation, simply the reminder that the sector ##mathbb{C}## which replaces the sector ##mathbb{R}## is probably not confused with the actual vector house ##mathbb{R}^2.## We’re used to imagining advanced numbers as factors within the advanced airplane, however the advanced numbers will not be an actual airplane, they’re a one-dimensional advanced vector house over themselves.

The inverse operation is integration. And actual integration is the calculation of an oriented quantity, an space (quantity of the world beneath the operate graph) within the case of a operate ##g:mathbb{R}rightarrow mathbb{R}.## Oriented signifies that we distinguish the areas above the ##x##-axis from the areas under the ##x##-axis, in addition to the combination path from ##x=a## to ##x=b## from the combination path from ##x=b## to ##x=a.## If the operate itself is a straight, say ##g(x)=r,## then the world is that of a single rectangle
\$\$
int_a^b g(x),dx=int_a^b r,dx=[rx]_a^b=rb-ra=rcdot (b-a).
\$\$
If we proceed as above and solely substitute the actual numbers with advanced numbers, then we get
start{align*}
int_{a+ic}^{b+id} g(x),dx&=int_{a+ic}^{b+id} (r+is),dz=[(r+is)z]_{a+ic}^{b+id}
&=(r(b-a)-s(d-c))+i(r(d-c)+s(b-a))
&=int_a^b r,dz +int_{ic}^{id}(is),dz+ int_{ic}^{id}r,dz+int_{a}^b (is),dz ,.
finish{align*}
Solely the variety of distinguishable orientations will increase as a result of ##i cdot i = -1## creates one other destructive space and imaginary volumes are allowed. This implies now we have to analyze the combination path extra rigorously than simply from left to proper or from proper to left as we did in actual integration if we wish to perceive a fancy quantity constructed by advanced lengths. Fortuitously, the integral itself cares about all orientations so long as we don’t make an indication error, and so long as we aren’t all in favour of an precise geometric quantity, that needed to be optimistic and actual.

\$\$
int_a^b f(t),dt := int_a^b u(t),dt +iint_a^b v(t),dt
\$\$
for (piecewise) steady capabilities ##f=u+iv : mathbb{R}supset [a,b] rightarrow mathbb{C}.##

We additionally use this for the definition of the advanced line integral ##displaystyle{int_{z_1}^{z_2}f(z) ,dz}## the place we combine alongside a (actual) parameterized path ##gamma: [a,b] longrightarrow Gsubseteq mathbb{C}## from ##gamma (a)=z_1in G,gamma (b)=z_2in G## for a fancy operate ##f: Glongrightarrow mathbb{C}## outlined on a area ##Gsubseteq mathbb{C}##
start{align*}
int_gamma f(z),dz &=…textual content{ (substitution }z=gamma(t), , ,dz=gamma’,dt)
…&=int_a^b (fcirc gamma )(t)cdot gamma’,dt =int_a^b f(gamma (t))cdot gamma'(t),dt
finish{align*}
The advanced line integral will depend on a priori on the combination path! This may simply be proven for ##f(z)=bar{z}.## Nevertheless, if ##f## is holomorphic, then its line integrals are path unbiased.

Cauchy’s Integral Theorem: If ##f:Glongrightarrow mathbb{C}## is a steady advanced operate then ##f## has an anti-derivative on ##G## if and provided that ##displaystyle{oint_gamma f(z),dz=0}## for any closed integration path ##gamma## in ##G##, ##gamma(a)=z_0=gamma(b).##

Each is true for a star-shaped area ##G,## e.g. a convex area, and a holomorphic operate ##f.##

Instance: Take into account the border ##gamma(t)=z_0+re^{i t}, , ,0leq tleq 2pi## of the disc ##G=D_r(z_0)## round a central level ##z_0.## Then
start{align*}
oint_gamma dfrac{1}{z-z_0},dz&=int_0^{2pi}dfrac{1}{re^{it}} cdot ire^{it} ,dt=icdot int_0^{2pi}dt=2pi i
oint_gamma left(z-z_0right)^p,dz&stackrel{(pneq -1)}{=}int_0^{2pi} r^p e^{i p t} cdot ire^{it},dt=ir^{p+1}int_0^{2pi} e^{i(p+1)t},dt
&=ir^{p+1}left[dfrac{e^{i(p+1)t}}{i(p+1)}right]_0^{2pi}=dfrac{r^{p+1}}{p+1}(e^{2(p+1)pi i t}-e^0)=0,.
finish{align*}
Therefore, we get the essential formulation
\$\$
oint_gamma left(z-z_0right)^p,dz=start{instances}
2pi i & textual content{ if } p = -1
0 & textual content{ if } pneq -1 ,.
finish{instances}
\$\$

Corollary: If ##Dsubset mathbb{C}## is a disc, and ##zin mathbb{C} backslash partial D## then
\$\$
oint_{partial D} ,dfrac{dzeta}{zeta -z}=start{instances}
2pi i & textual content{ if } zin D
0 & textual content{ else } ,.
finish{instances}
\$\$

### Complicated Integration

If ##f:Grightarrow mathbb{C}## is holomorphic, then
\$\$
f(zeta)=f(z)+underbrace{left( (D_{z}f)+underbrace{dfrac{r(zeta-z)}{zeta-z}}_{stackrel{zeta to z}{longrightarrow };;0} proper)}_{=:Delta_z(zeta)}cdot (zeta-z)
\$\$
with an in every single place steady and moreover ##z## even holomorphic operate ##Delta_z.## Thus
start{align*}
0&=oint_{partial D}Delta(zeta),dzeta =oint_{partial D}dfrac{f(zeta)-f(z)}{zeta -z},dzeta
&=oint_{partial D}dfrac{f(zeta)}{zeta -z},dzeta – f(z) oint_{partial D}dfrac{1}{zeta -z},dzeta =oint_{partial D}dfrac{f(zeta)}{zeta -z},dzeta – f(z)cdot 2pi i
finish{align*}
and
\$\$
f(z)=dfrac{1}{2pi i}oint_{partial D}dfrac{f(zeta)}{zeta -z},dzeta
\$\$
We even have

Theorem: The next statements for ##f=u+iv, : ,Urightarrow V## are equal:

1. ##f## is holomorphic on ##U.##
2. ##f## is steady and its path integral over any closed, contractible path vanishes.
3. The operate values of ##f## on the inside of a disc ##Dsubseteq U## might be decided by the operate values on the border of this disc by
Cauchy’s integral formulation
\$\$ f(z)=dfrac{1}{2pi i} oint_{partial D} dfrac{f(zeta)}{zeta – z},dzeta,.\$\$

Allow us to now introduce the trick with the geometric sequence. ##f:Drightarrow mathbb{C}## be a steady operate an a compact set ##D, , , z_0notin D ## and ##R=operatorname{dist}(z_0,D).## Then
start{align*}
dfrac{1}{zeta – z}&=dfrac{1}{(zeta -z_0)-(z-z_0)}=dfrac{1}{zeta -z_0}cdot dfrac{1}{1-dfrac{z-z_0}{zeta-z_0}}=dfrac{1}{zeta -z_0}cdot sum_{n=0}^infty left(dfrac{z-z_0}{zeta-z_0}proper)^n
finish{align*}
for ##|z-z_0|<Rleq |zeta-z_0|## and ##zin D_R(z_0), , ,zeta in D.## Since ##f## is bounded on the compact set ##D,## say ##0<|f(zeta)|<C,## now we have
\$\$
left| dfrac{f(zeta)}{(zeta-z_0)^{n+1}}cdot(z-z_0)^n proper|leq dfrac{C}{R}cdot left(dfrac{z-z_0}{R}proper)^n
\$\$
and the sequence on the correct converges for any fastened ##zin D_R(z_0).## Thus
\$\$
dfrac{f(zeta)}{zeta – z}=dfrac{f(zeta)}{zeta -z_0}cdot sum_{n=0}^infty left(dfrac{z-z_0}{zeta-z_0}proper)^n=sum_{n=0}^infty dfrac{f(zeta)}{(zeta-z_0)^{n+1}}cdot (z-z_0)^n
\$\$
converges completely and uniformly for a hard and fast ##z## for ##zeta in D## by Weierstraß’s criterion. Lastly,
start{align*}
f(z)=dfrac{1}{2pi i}oint_gamma dfrac{f(zeta)}{zeta – z},dzeta= sum_{n=0}^infty underbrace{left(dfrac{1}{2pi i} oint_gamma dfrac{f(zeta)}{(zeta-z_0)^{n+1}},dzeta proper)}_{=:a_n}cdot (z-z_0)^n
finish{align*}
converges absolute and uniformly on the inside of ##D_R(z_0)## versus the in every single place holomorphic, and by building analytic operate
\$\$
f(z)=sum_{n=0}^infty a_n(z-z_0)^n,.
\$\$
If we apply this to the border of a disc ##gamma(t)=z_0+re^{it}## with ##0<r<R## and ##0leq tleq 2pi ,## we get

Cauchy’s Growth Theorem: Let ##f:Grightarrow mathbb{C}## be a fancy, holomorphic operate on a area ##G## and ##z_0in G.## Let additional be ##R## the maximal radius of an open disc round ##z_0## that matches into ##G.## Then there’s a energy sequence
\$\$
f(z)=sum_{n=0}^infty a_n(z-z_0)^n,
\$\$
that converges for all ##0<r<R## on the disc ##D_r(z_0)## completely and uniformly versus ##f(z).## For each such ##r## is
\$\$
displaystyle{a_n=dfrac{1}{2pi i} oint_{partial D_r(z_0)} dfrac{f(zeta)}{(zeta-z_0)^{n+1}},dzeta}
\$\$
and ##f## is on ##G## arbitrary typically advanced differentiable.

### Residue Theorem

Allow us to begin with Laurent’s decomposition trick for a holomorphic operate ##f:mathcal{R}rightarrow mathbb{C}## on a hoop ##mathcal{R}={zin mathbb{C},|,0<r<|z|<R}.## There exists a decomposition
\$\$
f(z)=g(z)+hleft(dfrac{1}{z}proper)
\$\$
into holomorphic capabilities on discs
\$\$
g:D_R(0)rightarrow mathbb{C}; , ;h:D_{1/r}(0)rightarrow mathbb{C}
\$\$
The decomposition turns into distinctive if we require ##h(0)=0.## Since holomorphic capabilities are analytic, now we have an enlargement into energy sequence
\$\$
f(z)= sum_{n=0}^infty a_nz^n+sum_{n=0}^infty b_nleft(dfrac{1}{z}proper)^n=sum_{n=-infty }^infty c_nz^n
\$\$
For an enlargement at ##z_0## we obtain a Laurent sequence ##(r<rho<R)##
\$\$
f(z)=sum_{n=-infty }^infty a_n (z-z_0)^n; , ;a_n=displaystyle{dfrac{1}{2pi i} oint_{partial D_rho(z_0)} dfrac{f(zeta)}{(zeta-z_0)^{n+1}},dzeta}
\$\$
Notice that ##r=0,R=infty ,## and ##r=R## are doable settings. These radii might be calculated by the formulation of Cauchy-Hadamard
\$\$
r=limsup_{nto infty } sqrt[n]{|a_{-n}|}; , ;R=dfrac{1}{displaystyle{limsup_{nto infty } sqrt[n]{|a_{n}|}}}
\$\$
Definition: Let ##f:Grightarrow mathbb{C}## be a holomorphic operate that doesn’t utterly vanish on the area ##Gsubseteq mathbb{C}.## The coefficient at ##-1##
\$\$
operatorname{Res}_{z_0}(f)=a_{-1}=dfrac{1}{2pi i} oint_{partial D_rho(z_0)} f(zeta) ,dzeta
\$\$
is known as the residue of ##f## at ##z_0.## It’s the solely coefficient with out a ##z-z_0## time period within the Laurent sequence of ##f.##

A degree ##z_0## is known as a zero of order ##m## if there’s a holomorphic operate ##g:D_r(z_0)rightarrow mathbb{C}## such that
\$\$
f(z)=(z-z_0)^m g(z); , ;g(z_0)neq 0 .
\$\$
A degree ##z_0## is known as a pole (remoted singularity) of order ##m## if there’s a holomorphic operate ##g:D_r(z_0)rightarrow mathbb{C}## such that
\$\$
f(z)=(z-z_0)^{-m}g(z); , ;g(z_0)neq 0 .
\$\$
A pole is a zero of ##1/f.## The residue of a pole of order ##m## is
\$\$
operatorname{Res}_{z_0}(f)= dfrac{1}{(n-1)!} lim_{zto z_0}{dfrac{d^{n-1}}{dz^{n-1}}} left((z-z_0)^{n}f(z)proper)
\$\$
We’ve seen that surrounding a disc as soon as means ##;dfrac{1}{2pi i}displaystyle{oint_{partial D} ,dfrac{dzeta}{zeta -z}}=1.##

If we as a substitute have any closed curve ##gamma ## round ##z_0,## then we outline the winding variety of ##gamma ## round ##z_0## as
\$\$
operatorname{Ind}_gamma (z_0):=dfrac{1}{2pi i}oint_{gamma} ,dfrac{dzeta}{zeta -z}.
\$\$

Residue Theorem: Let ##f:Grightarrow mathbb{C}## be a holomorphic capabilities as much as finitely many remoted singularities ##z_1,ldots,z_q,## i.e. ##f## is meromorphic, and ##gamma ## a closed, piecewise easy curve within the area ##G,## then
\$\$
oint_gamma f(z),dz=2pi i sum_{ok=1}^q operatorname{Ind}_gamma (z_k) operatorname{Res}_{z_k}(f)
\$\$
Let’s name the set of zeros ##Z## and the set of poles ##P## of a meromorphic operate ##f,## and require that our integration path ##gamma ## doesn’t include any of them. We have already got seen that we will write ##f(z) =(z-z_0)^{m}g(z) = (z-z_0)^{operatorname{ord}_{z_0}(f)}## for any ##z_0in Zcup P## of order ##|m|.## Then
\$\$
dfrac{f'(z)}{f(z)}=dfrac{m}{z-z_0}+dfrac{g'(z)}{g(z)}.
\$\$
##dfrac{g'(z)}{g(z)}## is holomorphic at ##z_0## since ##g(z_0)neq 0.## The residue of ##f’/f## at ##z_0## equals due to this fact precisely the order ##m## of the zero or pole ##z_0## of ##f## and
\$\$
dfrac{1}{2pi i}oint_gamma dfrac{f'(z)}{f(z)},dz=sum_{z_0in Zcup P} operatorname{Ind}_gamma (z_0)cdot operatorname{Res}_{z_0}left(dfrac{f’}{f}proper) =sum_{z_0in Zcup P} operatorname{Ind}_gamma (z_0)cdot operatorname{ord}_{z_0}(f)
\$\$
the place
\$\$
operatorname{Res}_{z_0}left(dfrac{f’}{f}proper) = operatorname{ord}_{z_0}(f)=
start{instances}
ok & textual content{if } z_0 in Z textual content{ of order ok }
-k & textual content{if } z_0 in P textual content{ of order ok }
0 & textual content{ else }
finish{instances}
\$\$

### Properties of Residues

Cauchy’s integral theorem says that ##operatorname{Res}_{z_0}(f)=0## for a holomorphic operate ##f:Grightarrow mathbb{C}## and ##z_0in G.## Let’s see how residues might be calculated in different instances. Say ##z_m## is a zero of order ##m## and ##p_m## a pole of order ##m## of a meromorphic operate ##f:Grightarrow mathbb{C}.## Let additional be ##h:Grightarrow mathbb{C}## holomorphic at these factors, and ##g## one other meromorphic operate. Then ##operatorname{Res}_{z_0}## is ##mathbb{C}##-linear and

\$\$
start{array}{ll}
operatorname{Res}_{z_0}(alpha f+beta g)=alphaoperatorname{Res}_{z_0}(f)+betaoperatorname{Res}_{z_0}(g) &(z_0in G , , ,alpha, beta in mathbb{C}) [16pt]
operatorname{Res}_{z_1}left(dfrac{h}{f}proper)=dfrac{h(z_1)}{f'(z_1)}&
operatorname{Res}_{z_1}left(dfrac{1}{f}proper)=dfrac{1}{f'(z_1)}[16pt]
operatorname{Res}_{z_m}left(hdfrac{f’}{f}proper)=h(z_m)cdot m&operatorname{Res}_{z_m}left(dfrac{f’}{f}proper)=m[16pt]
operatorname{Res}_{p_1}(hcdot f)=h(p_1)cdot operatorname{Res}_{p_1}(f)&operatorname{Res}_{p_1}(f)=displaystyle{lim_{z to p_1}((z-p_1)f(z))}[16pt]
operatorname{Res}_{p_m}(f)=dfrac{1}{(m-1)!}displaystyle{lim_{z to p_m}dfrac{partial^{m-1} }{partial z^{m-1}}left( (z-p_m)^m f(z) proper)}&operatorname{Res}_{p_m}left(dfrac{f’}{f}proper)=-m[16pt]
operatorname{Res}_{infty }(f)=operatorname{Res}_0left(-dfrac{1}{z^2}fleft(dfrac{1}{z}proper)proper)&operatorname{Res}_{p_m}left(hdfrac{f’}{f}proper)=-h(p_m)cdot m[16pt]
operatorname{Res}_{z_0}(h)=0 & operatorname{Res}_0left(dfrac{1}{z}proper)=1[16pt]
operatorname{Res}_1left(dfrac{z}{z^2-1}proper)=operatorname{Res}_{-1}left(dfrac{z}{z^2-1}proper)=dfrac{1}{2}&operatorname{Res}_0left(dfrac{e^z}{z^m}proper)=dfrac{1}{(m-1)!}
finish{array}
\$\$

### Actual Integration with Residues

Take into account an actual rational operate ##f:mathbb{R}rightarrow mathbb{R}## with ##deg(f)leq -2## that has no poles on the actual axis. We wish to combine ##f## alongside the actual axis, however take this as part of the advanced airplane. Our main integration path is a composite of the actual path ##gamma :[-r,r]rightarrow [-r,r]subseteq mathbb{C}, , ,gamma(t)=t## and the advanced path ##tilde{gamma }:[0,pi]rightarrow mathbb{C}, , ,tilde{gamma }(t)=re^{it}## within the higher half of the airplane the place ##mathfrak{Im}(z)geq 0.## The residue theorem says

\$\$
displaystyle{oint_{gamma oplus tilde{gamma }}f(z),dz =int_gamma f(z),dz +int_{tilde{gamma }}f(z),dz= 2pi isum_{ok=1}^qoperatorname{Res}_{z_k}(f). }
\$\$
and all poles of ##f## lie inside our closed integration path so long as the radius ##r## is massive sufficient. Furthermore, for any normed, i.e. the main coefficient equals ##1##, advanced polynomial ##P(z)## of diploma ##n## there’s an ##Rin mathbb{R}^+## such that for all ##zin mathbb{C}## with ##|z|geq R##
\$\$
dfrac{1}{2}|z|^nleq |P(z)|leq dfrac{3}{2}|z|^n
\$\$
which might be confirmed with the triangle inequalities. We get for our rational operate that there’s a optimistic actual fixed ##cin mathbb{R}^+## such that (##L##= size)
\$\$
displaystyle{left|int_{tilde{gamma }}f(z),dzright|leq L(gamma )cdot max_=r|f(z)|leq pi r cdot cr^{-2}=pidfrac{c}{r};stackrel{rto infty }{longrightarrow };0}
\$\$
Therefore
start{align*}
int_{-infty }^infty f(x),dx&= displaystyle{lim_{r to infty} int_{-r}^rf(x),dx}=displaystyle{lim_{r to infty} int_gamma f(z),dz +0}
&=displaystyle{lim_{r to infty} int_gamma f(z),dz + lim_{r to infty}int_{tilde{gamma }}f(z),dz }= 2pi isum_{ok=1}^qoperatorname{Res}_{z_k}(f).
finish{align*}

Instance: We wish to discover ##displaystyle{int_{-infty }^infty dfrac{dx}{(1+x^2)^n}}## for a optimistic integer ##nin mathbb{N}## which equals ##2pi i operatorname{Res}_{i}(f)## for the meromorphic operate \$\$f(x)=dfrac{1}{(z^2+1)^n}=dfrac{1}{(z+i)^n(z-i)^n}\$\$ that has ##z=i## as solely pole of order ##n## within the higher half of the advanced airplane.
start{align*}
operatorname{Res}_{i}left(dfrac{1}{(1+z^2)^n}proper)&=dfrac{1}{(n-1)!}displaystyle{;lim_{z to i}};dfrac{partial^{n-1}}{partial z^{n-1}}left((z-i)^ndfrac{1}{(1+z^2)^n}proper)
&=dfrac{1}{(n-1)!}displaystyle{;lim_{z to i}};dfrac{partial^{n-1}}{partial z^{n-1}}(z+i)^{-n}
&=dfrac{1}{(n-1)!}displaystyle{;lim_{z to i}
(-n)(-n-1)cdots (-2n+2)(z+i)^{-2n+1}}
&=dfrac{1}{(n-1)!}(-1)^{n-1}dfrac{(2n-2)!}{(n-1)!}(2i)^{-2n+1}=-dfrac{i}{2^{-2n+1}}binom{2n-2}{n-1}
finish{align*}
and
\$\$
displaystyle{int_{-infty }^infty dfrac{dx}{(1+x^2)^n}= dfrac{pi}{2^{2n-2}}}binom{2n-2}{n-1}.
\$\$
It may be confirmed by comparable strategies as above that (cp. Fourier remodel)
\$\$
int_{-infty }^infty f(x)e^{ix},dx=2pi i sum_{ok=1}^qoperatorname{Res}_{z_k}left(f(z)e^{iz}proper)
\$\$
for an actual rational operate ##f## with ##deg(f)leq -2## and poles ##z_1,ldots,z_q## on the strict higher half of the advanced airplane, i.e. ##mathfrak{Im}(z_k)>0.##

Instance:

start{align*}
int_{-infty }^infty dfrac{cos x}{1+x^2},dx&=mathfrak{Re}left(int_{-infty }^infty dfrac{frac{1}{2}(e^{ix}+e^{-ix})}{1+x^2},dx proper)=mathfrak{Re}left(int_{-infty }^infty dfrac{e^{ix}}{1+x^2},dxright)
&=mathfrak{Re}left(2pi i operatorname{Res}_{i}left(dfrac{e^{iz}}{1+z^2}proper)proper)=mathfrak{Re}left(2pi i displaystyle{;lim_{z to i}}(z-i)dfrac{e^{iz}}{1+z^2} proper)
&=mathfrak{Re}left(2pi i cdot dfrac{e^{-1}}{(i+i)}proper)=dfrac{pi}{e}
finish{align*}
These strategies are used to show the marginally extra basic

Jordan’s Lemma: Let ##f(z)=e^{ialpha z}g(z):mathbb{C}longrightarrow mathbb{C}## be a complex-valued, steady operate with ##alphain mathbb{R}^+## and ##gamma :[0,pi]rightarrow mathbb{C}, , ,gamma(t)=re^{it}, , ,rin mathbb{R}^+.## Then
\$\$
left| int_{gamma } f(z) , dz proper| =left| int_{gamma } e^{ialpha z}g(z) , dz proper|le frac{pi}{alpha} M_r quad textual content{the place} quad M_r := max_{t in [0,pi]} left| g left(r e^{i t}proper) proper|.
\$\$
We get particularly for capabilities ##g## with a uniform convergence ##displaystyle{lim_ to inftyg(z)=0}## for ##zin {zin mathbb,mathfrak{Im}(z)>0}## that
\$\$
displaystyle{lim_{r to infty}int_gamma f(z),dz =lim_{r to infty}int_gamma e^{ialpha z}g(z),dz=0}
\$\$
That is additionally true for ##alpha=0## if ##displaystyle{lim_ to inftyzcdot g(z)=0}.##

Some extra examples that may be calculated by these strategies are

start{align*}
int_{-infty }^infty dfrac{cos (alpha x)}{1+x^2},dx=dfrac{pi}{e^alpha}; &, ; int_{-infty }^infty dfrac{xsin (alpha x)}{1+x^2},dx=dfrac{pi}{e^alpha} [6pt]
int_0^infty dfrac{1}{1+x^3},dx = dfrac{2pi}{3sqrt{3}}; &, ;int_0^infty dfrac{1}{4+x^4},dx = dfrac{pi}{8}[6pt]
int_{-infty }^infty dfrac{1}{e^x+e^{-x}},dx =dfrac{pi }{2}; &, ;int_{-infty }^infty dfrac{x}{e^x-e^{-x}},dx =dfrac{pi^2 }{4}
finish{align*}