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# The Artwork of Integration | Physics Boards

### Summary

My faculty instructor used to say

“All people can differentiate, however it takes an artist to combine.”

The mathematical motive behind this phrase is, that differentiation is the calculation of a restrict
\$\$
f'(x)=lim_{vto 0} g(v)
\$\$
for which we’ve got many guidelines and theorems at hand. And if nothing else helps, we nonetheless can draw ##f(x)## and a tangent line. Geometric integration, nevertheless, is proscribed to rudimentary examples and even easy integrals such because the finite quantity of Gabriel’s horn with its infinite floor are exhausting to visualise. We can’t fill in a gallon of paint, however it takes infinitely many gallons to color it?!

Whereas the integrals of Gabriel’s horn
\$\$
textual content{quantity }=piint_1^infty dfrac{1}{x^2},dx; textual content{ and } ;textual content{floor }=2piint_1^infty dfrac{1}{x},sqrt{1+dfrac{1}{x^4}},dx
\$\$
can probably be solved by Riemann sums and thus with restrict calculations, too, a operate like
\$\$
\$\$
which is simple to distinguish
\$\$
dfrac{d}{dx}left(dfrac{log x}{a+1-x}-dfrac{log x}{a+x}proper)=
dfrac{1}{x(a+1-x)}+dfrac{log x}{(a+1-x)^2}-dfrac{1}{x(a+x)}-dfrac{log x}{(a+x)^2}
\$\$
can’t be built-in straightforwardly. Earlier than I symbolize the main strategies of integration which would be the content material of this text, allow us to take a look at a bit piece of artwork.

We need to discover the anti-derivative
\$\$
int_0^1 left(dfrac{log x}{a+1-x}-dfrac{log x}{a+x}proper),dx
\$\$
Outline ##F, : ,]0,1]longrightarrow mathbb{C}## by ##F(x):=dfrac{xlog x}{a+x}-log(a+x).##
Then
start{align*}
F'(x)&=dfrac{(a+x)(1+log x)-xlog x}{(a+x)^2}-dfrac{1}{a+x}=dfrac{alog x}{(a+x)^2}
finish{align*}
and
\$\$
aint_0^1 dfrac{log x}{(a+x)^2},dx=F(1)-F(0^+)=-log(a+1)-(-log a)=log a -log (a+1).
\$\$
Subsequent, we outline ##G, : ,]0,1]longrightarrow mathbb{C}## by ##G(x):=dfrac{xlog x}{a+1-x}+log(a+1-x).##
Then
start{align*}
G'(x)&=dfrac{(a+1-x)(1+log x)+xlog x}{(a+1-x)^2}-dfrac{1}{a+1-x}=dfrac{(a+1)log x}{(a+1-x)^2}
finish{align*}
and
\$\$
(a+1)int_0^1dfrac{log x}{(a+1-x)^2},dx=G(1)-G(0^+)=log a-log(a+1).
\$\$
Lastly, these integrals can be utilized to calculate
start{align*}
I(a)&=int_0^1 left(dfrac{log x}{a+1-x}-dfrac{log x}{a+x}proper),dx =int_0^1 int left(dfrac{log x}{(a+x)^2}-dfrac{log x}{(a+1-x)^2}proper),da,dx
&=int left(left(dfrac{1}{a}-dfrac{1}{a+1}proper)cdotleft(log a -log (a+1)proper)proper),da =dfrac{1}{2}left(log a -log (a+1)proper)^2+C.
finish{align*}
From ##displaystyle{lim_{a to infty}I(a)=0}## we get ##C=0## and so
\$\$
int_0^1 left(dfrac{log x}{a+1-x}-dfrac{log x}{a+x}proper),dx=dfrac{1}{2}left(log dfrac{a}{a+1}proper)^2.
\$\$

### The Basic Theorem of Calculus

If ##f:[a,b]rightarrow mathbb{R}## is a real-valued steady operate, then there exists a differentiable operate, the anti-derivative, ##F:[a,b]rightarrow mathbb{R}## such that
\$\$
F(x)=int_c^tf(t),dt textual content{ for }cin [a,b]textual content{ and }F'(x)=f(x)
\$\$
Furthermore \$\$int_a^b f(x),dx = left[F(x)right]_a^b=F(b)-F(a)\$\$

### Imply Worth Theorem of Integration

\$\$
int_a^b f(x)g(x),dx =f(xi) int_a^b g(x),dx
\$\$
for a steady operate ##f:[a,b]rightarrow mathbb{R}## and an integrable operate ##g## that has no signal adjustments, i.e. ##g(x) geq 0## or ##g(x)leq 0## on ##[a,b];,,xiin [a,b].##
\$\$
int_a^b f(x)g(x),dx = f(a)int_a^xi g(x),dx +f(b) int_xi^b g(x),dx
\$\$
if ##f(x)## is monotone, ##g(x)## steady.

These theorems are extra essential for estimations of integral values than truly computing their worth. Nonetheless, they’ve their attraction.

Is there at all times a place on a clean flooring for an oblong desk with 4 equal legs, such that the desk doesn’t wobble?

We take into account the heights ##h## of its legs at a sure level ##x## on the ground measured by its angle to a set level (radial coordinates). The desk doesn’t wobble, if the sums of two reverse heights are equal, i.e. if
\$\$
f(x)=h(x)+h(x+pi)-h(x+d)-h(x+d+pi)=0
\$\$
for an angle ##din (0,pi).## We assume that our desk wobbles, say ##f(x_0)>0,.## By the periodicity of ##h(x)## (rotation of the desk) we get for ##H:=int_0^{2pi}h(x),dx## that ##H=int_0^{2pi}h(x+c),dx## for all ##c in mathbb{R},.## Therefore
\$\$
int_0^{2pi}f(x),dx=H+H-H-H=0=f(xi)cdot int_0^{2pi}dx=2pi f(xi)
\$\$
So we’ve got a degree (or moderately: an angle) ##xi## such that ##f(xi) = 0,## the desk doesn’t wobble on this place.

### Fubini’s Theorem

\$\$
int_{Xtimes Y}f(x,y),d(x,y)=int_Y int_Xf(x,y),dx,dy=int_Xint_Y f(x,y),dy,dx
\$\$
for steady capabilities ##f, : ,Xtimes Y rightarrow mathbb{R}.## It is likely one of the most used theorems each time an space or a quantity needs to be computed. Nonetheless, we want steady capabilities! An instance the place the concept can’t be utilized:
\$\$
underbrace{int_0^1int_0^1 dfrac{x-y}{(x+y)^3},dx,dy}_{=-1/2} =int_0^1int_0^1 dfrac{y-x}{(y+x)^3},dy,dx=-underbrace{int_0^1int_0^1dfrac{x-y}{(x+y)^3},dy,dx}_{=+1/2}
\$\$

### Partial Fraction Decomposition

We virtually used partial fraction decomposition within the earlier instance when the issue ##dfrac{1}{a}-dfrac{1}{a+1}=dfrac{1}{a(a+1)}## within the integrand occurred. Equations like this are meant by partial fraction decomposition. Whereas ##dfrac{1}{a(a+1)}## appears troublesome to combine, the integrals of ##dfrac{1}{a}## and ##dfrac{1}{a+1}## are easy logarithms. Assume we’ve got polynomials ##p_k(x)## and an integrand
\$\$
q_0(x)prod_{okay=1}^m p_k^{-1}(x)=dfrac{q_0(x)}{prod_{okay=1}^m p_k(x)}stackrel{!}{=}sum_{okay=1}^m dfrac{q_k(x)}{p_k(x)}=sum_{okay=1}^m dfrac{q_k(x)prod_{ineq okay}p_i(x)}{prod_{okay=1}^m p_k(x)}.
\$\$
Then all we want are appropriate polynomials ##q_k(x)## such that
\$\$
q_0(x)=sum_{okay=1}^m q_k(x)prod_{ineq okay}p_i(x)
\$\$
holds and every of the ##m## many phrases is simple to combine. The bounds of this system are apparent: the levels of the polynomials on the appropriate must be small. Partial fraction decomposition is certainly often used if the levels of ##q_k(x)## are quadratic at most, the much less the higher.

Instance: Combine ##displaystyle{int_2^{10} dfrac{x^2-x+4}{x^4-1},dx.}## This implies we’ve got to resolve
\$\$
x^2-x+4=q_1(x)underbrace{(x+1)(x-1)}_{x^2-1}+q_2(x)underbrace{(x^2+1)(x+1)}_{x^3+x^2+x+1}+q_3(x)underbrace{(x^2+1)(x-1)}_{x^3-x^2+x-1}
\$\$
Our unknowns are the polynomials ##q_k(x).## We set ##q_k=a_kx+b_k## and get by evaluating the coefficients
start{align*}
x^4, &: ,0= a_2+a_3
x^3, &: ,0= a_1+a_2-a_3+b_2+b_3
x^2, &: ,1= a_2+a_3+b_1+b_2-b_3
x^1, &: ,-1= -a_1+a_2-a_3+b_2+b_3
x^0, &: ,4= -b_1+b_2-b_3
finish{align*}
One answer is subsequently ##q_1(x)=dfrac{x-3}{2}, , ,q_2(x)=x, , ,q_3(x)=-dfrac{2x+5}{2}## and
start{align*}
int_2^{10} &dfrac{x^2-x+4}{x^4-1},dx =dfrac{1}{2}int_2^{10} dfrac{x}{x^2+1},dx – dfrac{3}{2}int_2^{10} dfrac{1}{x^2+1},dx
&phantom{=}+int_2^{10}dfrac{x}{x-1},dx -int_2^{10}dfrac{x}{x+1},dx-dfrac{5}{2}int_2^{10}dfrac{1}{x+1},dx
&=dfrac{1}{4}left[phantom{dfrac{}{}}log(x^2+1)right]_2^{10}-dfrac{3}{2}left[phantom{dfrac{}{}}operatorname{arc tan}(x)right]_2^{10}
&phantom{=}+left[phantom{dfrac{}{}}x+log( x-1)right]_2^{10}-left[phantom{dfrac{}{}}x-log(x+1)right]_2^{10}-dfrac{5}{2}left[phantom{dfrac{}{}}log(x+1)right]_2^{10}
&=-dfrac{3}{2}(operatorname{arctan}(10)-operatorname{arctan}(2))+dfrac{1}{4}logleft(dfrac{101cdot 9^4cdot 3^6}{5cdot 11^6}proper)approx 0,45375ldots
finish{align*}

### Integration by Components – The Leibniz Rule

Integration by components is one other manner to have a look at the Leibniz rule.
start{align*}
(u(x)cdot v(x))’&=u(x)’cdot v(x)+u(x)cdot v(x)’
int (u(x)cdot v(x))’, dx&=int u(x)’cdot v(x),dx+int u(x)cdot v(x)’,dx
int_a^b u(x)’cdot v(x),dx &=left[u(x)v(x)right]_a^b -int_a^b u(x)cdot v(x)’,dx
finish{align*}
It’s a risk to shift the issue of integration of a product from one issue to the opposite, e.g.
start{align*}
int xlog x,dx&=dfrac{x^2}{2}log x-int dfrac{x^2}{2}cdot dfrac{1}{x},dx=dfrac{x^2}{4}left(-1+2log x proper)
finish{align*}
Integration by components is very helpful for trigonometric capabilities the place we will use the periodicity of sine and cosine beneath differentiation, e.g.
start{align*}
int sin xcos x,dx&=-cos^2 x-int cos xsin x,dx
int sin xcos x,dx&=-dfrac{1}{2}cos^2 x=dfrac{1}{2}(-1+sin^2 x)
int cos^2 x ,dx&=x-int sin^2x,dx=x+cos xsin x-int cos^2x,dx
&=dfrac{1}{2}(x+cos xsin x)
finish{align*}
part{Integration by Discount Formulation}
Integration by components usually gives recursions that result in so-called discount formulation since they successively lower a amount, often an exponent. Because of this an integral
\$\$
I_n=int f(n,x),dx
\$\$
could be expressed as a linear mixture of integrals ##I_k## with ##okay<n.## E.g.
start{align*}
underbrace{int x^n e^{alpha x},dx}_{=I_n}&= left[x^{n}alpha^{-1}e^{alpha x}right]-nalpha^{-1}underbrace{int x^{n-1} e^{alpha x},dx}_{=I_{n-1}}
finish{align*}
or
start{align*}
int cos^nx,dx&=left[sin xcos^{n-1}xright]+(n-1)int sin^2 xcos^{n-2}x,dx
&=left[sin xcos^{n-1}xright]+(n-1)int cos^{n-2}x,dx-(n-1)int cos^n,dx
finish{align*}
which ends up in the discount method
\$\$
int cos^nx,dx=dfrac{1}{n}left[sin xcos^{n-1}xright]+dfrac{n-1}{n}intcos^{n-2}x,dx
\$\$
Discount formulation could be sophisticated, and so they can have multiple pure quantity as parameters, cp.[1]. The discount method for the sine is
\$\$
int sin^nx,dx=-dfrac{1}{n}left[sin^{n-1} xcos xright]+dfrac{n-1}{n}intsin^{n-2}x,dx
\$\$

### Lobachevski’s Formulation

Let ##f(x)## be a real-valued steady operate that’s ##pi## periodic, i.e. ##f(x+pi)=f(x)=f(pi-x)## for all ##xgeq 0,## e.g. ##f(x)equiv 1.## Then
start{align*}
int_{0}^{infty }{dfrac{sin^{2}x}{x^{2}}}f(x),dx &=int_{0}^{infty }{dfrac{sin x}{x}}f(x),dx
=int_{0}^{pi /2}f(x),dx[6pt]
int_{0}^{infty }{dfrac{sin^{4}x}{x^{4}}}f(x),dx&
=int_{0}^{pi /2}f(t),dt-{dfrac{2}{3}}int_{0}^{pi /2}sin^{2}tf(t),dt
finish{align*}
\$\$
int_{0}^{infty }{dfrac{sin x}{x}},dx=int_{0}^{infty }{dfrac{sin^{2}x}{x^{2}}},dx={dfrac{pi }{2}}
; textual content{ and } ; int_{0}^{infty }{dfrac{sin^{4}x}{x^{4}}},dx={dfrac{pi }{3}}.
\$\$

### Substitutions

#### The Chain Rule

Substitutions are most likely probably the most utilized approach. Barely an integral that doesn’t use it, or how I prefer to put it:

Do away with what disturbs probably the most!

Substitution is technically a change of the mixing variable, a change of coordinates, the chain rule!
start{align*}
int_a^b f(g(x)),dx stackrel{y=g(x)}{=}int_{g(a)}^{g(b)} f(y)cdot left(dfrac{dg(x)}{dx}proper)^{-1},dy
finish{align*}
Notice that the mixing limits change, too! After all, not any coordinate transformation is useful, and what appears as if we sophisticated issues can truly simplify the integral. We will take away the disturbing operate ##g(x)## each time both ##g'(x)## or ##f(x)/g'(x)## are particularly easy, e.g.
start{align*}
int_0^2 xcos (x^2+1),dxstackrel{;y=x^2+1}{=}&int_1^5 xcos ydfrac{dy}{2x}=dfrac{1}{2}int_1^5cos y,dy =dfrac{1}{2}(sin (5)-sin (1))
int_0^1sqrt{1-x^2},dxstackrel{y=arcsin x}{=}&int_{0}^{pi/2}cos^2 y,dy=left[dfrac{1}{2}(x+cos x sin x)right]_0^{pi/2}=dfrac{pi}{4}
finish{align*}
start{align*}
int_{-infty}^{infty} f(x),dfrac{tan x}{x},dx =& sum_{kin mathbb{Z}} int_{(k-1/2) pi }^{(okay+1/2)pi}f(x),dfrac{tan x}{x},dx
stackrel{y=x-kpi}{=} sum_{kin mathbb{Z}} int_{-pi /2 }^{pi /2 } f(y) ,dfrac{tan y}{y+kpi},dy
=&int_{-pi /2 }^{pi /2 }f(y)underbrace{sum_{kin mathbb{Z}} dfrac{1}{y+kpi}}_{=cot y} ,tan y, dy
stackrel{z=y+pi /2}{=} int_{0}^{pi }f(z),dz
finish{align*}

#### Weierstraß Substitution

The Weierstraß substitution or tangent half-angle substitution is a method that transforms trigonometric capabilities into rational polynomial capabilities. We begin with
\$\$
tanleft(dfrac{x}{2}proper) = t
\$\$
and get
\$\$
sin x=dfrac{2t}{1+t^2}; , ;cos x=dfrac{1-t^2}{1+t^2}; , ;
dx=dfrac{2}{1+t^2},dt.
\$\$
These substitutions are particularly helpful in spherical geometry. The next examples might illustrate the precept.
start{align*}
int_0^{2pi}dfrac{dx}{2+cos x}&=int_{-infty }^infty dfrac{2}{3+t^2},dtstackrel{t=usqrt{3}}{=}dfrac{2}{sqrt{3}}int_{-infty }^infty dfrac{1}{1+u^2},du=dfrac{2pi}{sqrt{3}}
finish{align*}
start{align*}
int_0^{pi} dfrac{sin(varphi)}{3cos^2(varphi)+2cos(varphi)+3},dvarphi &= int_0^infty dfrac{t}{t^4+2}~dt stackrel{u=frac{1}{2}t^2}{=}
= frac{1}{2sqrt{2}}int_0^infty dfrac{du}{1+u^2}
&= frac{1}{2sqrt{2}}left[phantom{dfrac{}{}}operatorname{arctan} u;right]_0^infty=frac{pi}{4sqrt{2}}
finish{align*}

#### Euler Substitutions

Euler substitutions are a way to sort out integrands which might be rational expressions of sq. roots of a quadratic polynomial
\$\$
int Rleft(x,sqrt{ax^2+bx+c},proper),dx.
\$\$
Allow us to first take a look at an instance. With the intention to combine
\$\$int_1^5 dfrac{1}{sqrt{x^2+3x-4}},dx\$\$
we take a look at the zeros of the quadratic polynomial ##sqrt{x^2+3x-4}=sqrt{(x+4)(x-1)}.## We get from the substitution ##sqrt{(x+4)(x-1)}=(x+4)t##
\$\$x=dfrac{1+4t^2}{1-t^2}; , ;sqrt{x^2+3x-4}=dfrac{5t}{1-t^2}; , ;dx = dfrac{10t}{(1-t^2)^2},dt\$\$
start{align*}
int_1^5 frac{dx}{sqrt{x^2+3x-4}}&=int_{0}^{2/3}dfrac{2}{1-t^2},dt=int_{0}^{2/3}dfrac{1}{1+t},dt+int_{0}^{2/3}dfrac{1}{1-t},dt[6pt]
&=left[log(1+t)-log(1-t)right]_0^{2/3}=log(5)
finish{align*}
This was the third of Euler’s substitutions. All in all we’ve got

1. Euler’s First Substitution.
\$\$
sqrt{ax^2+bx+c}=pm sqrt{a}cdot x+t, , ,x=dfrac{c-t^2}{pm 2tsqrt{a}-b}; , ;a>0
\$\$
2. Euler’s Second Substitution.
\$\$
sqrt{ax^2+bx+c}=xtpm sqrt{c}, , ,x=dfrac{pm 2tsqrt{c}-b}{a-t^2}; , ;c>0
\$\$
3. Euler’s Third Substitution.
\$\$
sqrt{ax^2+bx+c}=sqrt{a(x-alpha)(x-beta)}=(x-alpha)t, , ,x=dfrac{abeta-alpha t^2}{a-t^2}
\$\$

#### Transformation Theorem

The chain rule works equally for capabilities in multiple variable. Let ##Usubseteq mathbb{R}^n## be an open set and ##varphi , : ,Urightarrow mathbb{R}^n## a diffeomorphism, and ##f## an actual valued operate outlined on ##varphi(U).## Then
\$\$
int_{varphi(U)} f(y),dy=int_U f(varphi (x))cdot |det underbrace{(Dvarphi(x) )}_{textual content{Jacobi matrix}}|,dx.
\$\$
That is particularly attention-grabbing if ##fequiv 1,## in order that the method turns into
\$\$
operatorname{vol}(varphi(U))=int_U |det (Dvarphi(x))|,dx = |det (Dvarphi(x))|cdotoperatorname{vol}(U),
\$\$
or in any circumstances the place the determinant equals one, e.g. for orthogonal transformations ##varphi ,## like rotations. For instance, we calculate the realm under the Gaussian bell curve.
start{align*}
int_{-infty }^infty dfrac{1}{sigma sqrt{2pi}} e^{-frac{1}{2}left(frac{x-mu}{sigma}proper)^2},dx&stackrel{t=frac{x-mu}{sigma sqrt{2}}}{=}
dfrac{1}{sqrt{pi}} int_{-infty }^infty e^{-t^2},dtstackrel{!}{=}1
finish{align*}
As a substitute of instantly calculating the integral, we are going to present that
\$\$
left(int_{-infty }^infty e^{-t^2},dtright)^2=int_{-infty }^infty e^{-x^2},dx cdot int_{-infty }^infty e^{-y^2},dy=int_{-infty }^inftyint_{-infty }^infty e^{-x^2-y^2},dx,dy=pi
\$\$
and make use of the truth that ##f(x,y)=e^{-x^2-y^2}=e^{-r^2}## is rotation-symmetric, i.e. we are going to make use of polar coordinates. Let ##U=mathbb{R}^+occasions (0,2pi)## and ##varphi(r,alpha)=(rcos alpha,r sin alpha).##
start{align*}
det start{pmatrix}dfrac{partial varphi_1 }{partial r }&dfrac{partial varphi_1 }{partial alpha }[12pt] dfrac{partial varphi_2 }{partial r }&dfrac{partial varphi_2 }{partial alpha }finish{pmatrix}=det start{pmatrix}cos alpha &-rsin alpha sin alpha &rcos alphaend{pmatrix}=r
finish{align*}
and eventually
start{align*}
int_{-infty }^inftyint_{-infty }^infty &e^{-x^2-y^2},dx,dy=int_{varphi(U)}e^{-x^2-y^2},dx,dy
&=int_U e^{-r^2cos^2alpha -r^2sin^2alpha }cdot |det(Dvarphi)(r,alpha)|,dr,dalpha
&=int_0^{2pi}int_0^infty re^{-r^2},dr,dalphastackrel{t=-r^2}{=}int_0^{2pi}dfrac{1}{2}int_{-infty }^0 e^{t},dt ,dalpha=int_0^{2pi}dfrac{1}{2},dalpha =pi
finish{align*}

### Translation Invariance

\$\$
int_{mathbb{R}^n} f(x),dx = int_{mathbb{R}^n} f(x-c),dx
\$\$
is nearly self-explaining since we combine over your complete area. Nonetheless, there are much less apparent circumstances
\$\$
int_{-infty}^{infty} fleft(x-dfrac{b}{x}proper),dx=int_{-infty}^{infty}f(x),dx, textual content{ for any } ,b>0
\$\$
textbf{Proof:} From
\$\$int_{-infty}^{infty}fleft(x-dfrac{b}{x}proper),dx
stackrel{u=-b/x}{=}int_{-infty}^{infty} fleft(u-dfrac{b}{u}proper)dfrac{b}{u^2},du\$\$
we get
start{align*}
int_{-infty}^{infty} fleft(x-dfrac{b}{x}proper),dx &+int_{-infty}^{infty} fleft(x-dfrac{b}{x}proper),dx=int_{-infty}^{infty} fleft(x-dfrac{b}{x}proper)left(1+dfrac{b}{x^2}proper),dx
&= int_{-infty}^{0} fleft(x-dfrac{b}{x}proper)left(1+dfrac{b}{x^2}proper),dx +int_{0}^{infty} fleft(x-dfrac{b}{x}proper)left(1+dfrac{b}{x^2}proper),dx
&stackrel{v=x-(b/x)}{=} int_{-infty}^{infty} f(v),dv +int_{-infty}^{infty} f(v),dv=2int_{-infty}^{infty} f(v),dv
finish{align*}
which proves the assertion.

### Symmetries

Symmetries are the core of arithmetic and physics on the whole. So it doesn’t shock that they play an important function in integration, too. The precept is a simplification:
\$\$
int_{-c}^c |x|,dx= 2cdot int_0^c x,dx = c^2
\$\$

A bit extra refined instance for additive symmetry is ##displaystyle{int_0^{pi/2}}log (sin(x)),dx.##

We use the symmetry between the sine and cosine capabilities.
start{align*}
int_0^{pi/2}log (sin x),dx&=-int_{pi/2}^0log (cos x),dx=int_0^{pi/2}log (cos x),dx
finish{align*}
start{align*}
2int_0^{pi/2}log (sin x),dx&=int_0^{pi/2}log (sin x),dx+int_0^{pi/2}log (cos x ,dx
&=int_0^{pi/2}log(sin x cos x)=int_0^{pi/2}logleft(dfrac{sin 2x}{2}proper),dx
&stackrel{u=2x}{=}-dfrac{pi}{2}log 2+dfrac{1}{2}int_0^{pi}log(sin u),du
&=-dfrac{pi}{2}log 2+int_0^{pi/2}log(sin u),du
finish{align*}
We thus have ##displaystyle{int_0^{pi/2}log (sin x),dx=-dfrac{pi}{2}log 2.}##

We will after all calculate symmetric areas. That simplifies the mixing boundaries.

The world of an astroid is enclosed by ##(x,y)=(cos^3 t,sin^3 t)## for ##0leq t leq 2pi.##
\$\$
4int_0^1 y,dxstackrel{x=cos^3(t)}{=}-12int_{pi/2}^0 sin^4 tcdot cos^2 t ,dt = dfrac{3}{8}pi
\$\$

#### Multiplicative Symmetry

A bit much less well-known are probably the multiplicative symmetries. We virtually bumped into one within the instance within the translation paragraph earlier than it lastly turned out to be an additive symmetry. Let’s illustrate with some examples what we imply, e.g. with a multiplicative asymmetry.
\$\$
int_0^pilog(1-2alpha cos(x)+alpha^2),dx=2pilog |alpha|; , ;|alpha|geq 1
\$\$
however how can we all know? We begin with a worth ##|alpha|leq 1## and ##0<x<pi.## Then by Gradshteyn, Ryzhik 1.514 [6]
\$\$
log(1-2alpha cos(x)+alpha^2)= -2sum_{okay=1}^{infty}dfrac{cos(kx)}{okay},alpha^okay
\$\$
and
##displaystyle{I(alpha)=int_0^pi log (1-2alpha cos(x)+alpha^2),dx = -2sum_{okay=1}^{infty} dfrac{alpha^okay}{okay} int_{0}^{pi} cos(kx),dx =0}.##

This implies for ##|alpha|geq 1##
start{align*}
int_0^pi log (1-2alpha cos(x)+alpha^2),dx &= int_0^pi logleft(alpha^2 left(dfrac{1}{alpha^2}-dfrac{2}{alpha}cos(x)+1right)proper),dx
&=2pi log|alpha| + underbrace{Ileft(dfrac{1}{alpha}proper)}_{=0}= 2pi log|alpha|
finish{align*}
So we used a consequence about ##I(1/a)## to calculate ##I(a).## The subsequent instance makes use of the multiplicative symmetry of the mixing variable
start{align*}
int_{n^{-1}}^{n} fleft(x+dfrac{1}{x}proper),dfrac{log x}{x},dx &stackrel{y=1/x}{=} int_{n}^{n^{-1}}fleft(dfrac{1}{y}+yright),dfrac{log y}{y},dy
&=-int_{n^{-1}}^{n} fleft(x+dfrac{1}{x}proper),dfrac{log x}{x},dx
finish{align*}
which is simply doable if the mixing worth is zero. By comparable arguments
\$\$
int_{frac{1}{2}}^{3} dfrac{1}{sqrt{x^2+1}},dfrac{log(x)}{sqrt{x}},dx , + , int_{frac{1}{3}}^{2} dfrac{1}{sqrt{x^2+1}},dfrac{log(x)}{sqrt{x}},dx =0
\$\$
and
\$\$
int_{pi^{-1}}^pi dfrac{1}{x}sin^2 left( -x-dfrac{1}{x} proper) log x,dx=0
\$\$

### Including an Integral

It’s generally useful to rewrite an integrand with a further integral, e.g. for ##alphanotin mathbb{Z}##
start{align*}
int_0^infty dfrac{dx}{(1+x)x^alpha} &= int_0^infty x^{-alpha} int_0^infty e^{-(1+x)t},dt,dx=int_0^inftyint_0^infty x^{-alpha} e^{-t}e^{-xt},dt,dx
&stackrel{u=tx}{=}int_0^inftyint_0^infty left(dfrac{u}{t}proper)^{-alpha}e^{-u}e^{-t}t^{-1},du,dx
&=int_0^infty u^{-alpha}e^{-u},du, int_0^infty t^{alpha -1} e^{-t},dt=Gamma (1-alpha) Gamma(alpha)=dfrac{pi}{sin pi alpha}
finish{align*}
or
start{align*}
&int_0^infty frac{x^{m}}{x^{n} + 1}dx = int_0^infty x^{m} int_0^infty e^{-x^n y} e^{-y} ,dy, dx
&stackrel{z=x^ny}{=}dfrac{1}{n} int_0^infty e^{-y} left(frac{1}{y}proper)^frac{m+1}{n} ,dy int_0^infty e^{-z} z^frac{m-n+1}{n} ,dz= dfrac{Gammaleft(dfrac{n-m-1}{n}proper) Gammaleft(dfrac{m+1}{n}proper)}{n}
finish{align*}

### The Mass at Infinity

If ##f_k: Xrightarrow mathbb{R}## is a sequence of pointwise absolute convergent, integrable capabilities, then
\$\$
int_X left(sum_{okay=1}^infty f_k(x)proper),dx=sum_{okay=1}^infty left(int_X f_k(x),dxright)
\$\$
Instance: The Wallis Sequence.
start{align*}
int_0^{pi/2}dfrac{2sin x}{2-sin^2 x},dx&=int_0^{pi/2}dfrac{2sin x}{1+cos x},dx=left[-2arctan(cos x)right]_0^{pi/2}=dfrac{pi}{2}
&=int_0^{pi/2}dfrac{sin x}{1-frac{1}{2}sin^2 x},dx=int_0^{pi/2}sum_{okay=0}^infty 2^{-k}sin^{2k+1}x ,dx
&=sum_{okay=0}^inftyint_0^{pi/2}2^{-k}sin^{2k+1}x ,dx=sum_{okay=0}^inftydfrac{1cdot 2cdot 3cdots okay}{3cdot 5 cdot 7 cdots (2k+1)}
finish{align*}
Absolutely the convergence is essential as we will see within the following instance. Let ##f_k=I_{[k,k+1,]}-I_{[k+1,k+2]}## with the indicator operate ##I_{[k,k+1,]}(x)=1## if ##xin[k,k+1]## and ##I_{[k,k+1,]}(x)=0## in any other case. If we sum up these capabilities, we get a telescope impact. The bump at ##[0,1]## stays untouched whereas the bump (the mass) ##[k+1,k+2]## escapes to infinity: ##displaystyle{sum_{okay=0}^n}f_k=I_{[0,1,]}-I_{[n+1,n+2]}## and ##displaystyle{sum_{okay=1}^infty f_k(x)}=I_{[0,1,]}(x).## Therefore,
\$\$
1=int_mathbb{R}I_{[0,1,]}(x),dx=int_mathbb{R} left(sum_{okay=1}^infty f_k(x)proper),dxneq sum_{okay=1}^infty left(int_mathbb{R} f_k(x),dxright)=0.
\$\$

### Limits and Integrals

Mass at infinity is strictly an change of a restrict and an integral. So what about limits on the whole? Even when all capabilities ##f_k## of a convergent sequence are steady or integrable, the identical is just not essentially true for the restrict. Take a look at
\$\$
f_k(x)=start{circumstances}okay^2x &textual content{ if }0leq xleq 1/okay 1/x&textual content{ if }1/kleq xleq 1end{circumstances}
\$\$
Each operate is steady and subsequently integrable, however the restrict ##displaystyle{lim_{okay to infty}f_k(x)=dfrac{1}{x}}## is neither. Even when all ##f_k## and the restrict operate ##f## are integrable, their integrals don’t have to match, e.g.
\$\$
triangle_r(x)=start{circumstances}0 &textual content{ if }|x|geq r dfracx{r^2}&textual content{ if }|x|leq rend{circumstances}; , ;lim_{r to 0}triangle_r(x)=triangle_0(x):=start{circumstances}infty &textual content{ if }x=0&textual content{ if }xneq 0end{circumstances}
\$\$
We thus have
\$\$
lim_{r to 0}inttriangle_r(x),dx =lim_{r to 0} 1=1neq 0=int triangle_0,dx=intlim_{r to 0}triangle_r(x),dx
\$\$
Once more, the mass escapes to infinity. To stop this from occurring, we want an integrable operate as an higher sure.

Let ##f_0,f_1,f_2,ldots## be a sequence of actual integrable capabilities that converge pointwise to ##displaystyle{lim_{okay to infty}f_k(x)}=f(x).## If their is an actual integrable operate ##h(x)## such that ##|f_k|leq h## and ##int h(x),dx <infty ,## then ##f## is integrable and
\$\$
displaystyle{lim_{okay to infty}int f_k(x),dx=int lim_{okay to infty}f_k(x),dx=int f(x),dx}
\$\$
The concept behind the proof is straightforward. We get from ##|f_k|leq h## that ##|f|leq h## and so ##|f-f_k|leq|f|+|f_k|leq 2h.## By Fatou’s lemma, we’ve got
start{align*}
int 2h&=int liminfleft(2h-|f_k-f|proper)leq liminfintleft(2h-|f_k-f|proper)
&=liminfleft(int 2h-int |f_k-f|proper)=int 2h-limsupint|f_k-f|
finish{align*}
Since ##int h<infty ,## we conclude ##lim int|f_k-f|=0.##

Instance: The Stirling Method.
\$\$
lim_{n to infty}dfrac{n!}{sqrt{n}left(frac{n}{e}proper)^n}=sqrt{2pi}
\$\$
We use the ability collection of the logarithm for ##|s|<1##
\$\$
log(1+s)=s-dfrac{s^2}{2}+dfrac{s^3}{3}-dfrac{s^4}{4}pm ldots
\$\$
and get for ##s=t/sqrt{n}## with ##-sqrt{n}<t<sqrt{n}##
\$\$
nleft(logleft(1+dfrac{t}{sqrt{n}}proper)-dfrac{t}{sqrt{n}}proper)=-dfrac{t^2}{2}+dfrac{t^3}{sqrt{n}}-dfrac{t^4}{4n}pmldotsstackrel{nto infty }{longrightarrow }-dfrac{t^2}{2}
\$\$
\$\$
n!=Gamma(n+1)=int_0^infty x^ne^{-x},dx
\$\$
so we get
start{align*}
dfrac{n!}{sqrt{n}left(frac{n}{e}proper)^n}=&int_0^infty
expleft(nleft(logleft(dfrac{x}{n}proper)+1-dfrac{x}{n}proper)proper),dfrac{dx}{sqrt{n}}
stackrel{x=sqrt{n}t+n}{=}&int_{-sqrt{n}}^{infty }exp
left(nleft(logleft(1+dfrac{t}{sqrt{n}}proper)-dfrac{t}{sqrt{n}}proper)proper),dt
=&int_{-infty }^infty underbrace{exp
left(nleft(logleft(1+dfrac{t}{sqrt{n}}proper)-dfrac{t}{sqrt{n}}proper)proper)cdot I_{ [-sqrt{n},infty [ }}_{=f_n(t)},dt
end{align*}
We saw that the sequence ##f_n(t)## converges pointwise to ##e^{-t^2/2},## is integrable and bounded from above by an integrable function. Hence, we can switch limit and integration.
\$\$
lim_{n to infty}dfrac{n!}{sqrt{n}left(frac{n}{e}right)^n}=lim_{n to infty}int_mathbb{R}f_n(t),dt=int_mathbb{R}lim_{n to infty}f_n(t),dt=int_mathbb{R}e^{-t^2/2},dt = sqrt{2pi}
\$\$

### The Feynman Trick – Parameter Integrals

“Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.” (Richard Feynman, 1918-1988)

\$\$
dfrac{d}{dx}int f(x,y),dy stackrel{(*)}{=}int dfrac{partial }{partial x}f(x,y),dy
\$\$
Let us consider an example and define ##f(x,y)=xcdot |x|cdot e^{-x^2y^2}.## Therefore
begin{align*}
dfrac{d}{dx}int_mathbb{R} f(x,y),dy&=dfrac{d}{dx}int_mathbb{R}x |x| e^{-x^2y^2}stackrel{t=xy}{=}dfrac{d}{dx}xint_mathbb{R}e^{-t^2},dt=sqrt{pi}[6pt]
int_mathbb{R} dfrac{partial }{partial x}f(x,y),dy&=int_mathbb{R} dfrac{partial }{partial x}x |x| e^{-x^2y^2},dy=2int_mathbb{R}left(|x|e^{-x^2y^2}- |x|x^2y^2 e^{-x^2y^2}proper),dy[6pt]
&stackrel{t=xy, , ,xneq 0}{=}2 int_mathbb{R}e^{-t^2},dt -2 int_mathbb{R}t^2e^{-t^2},dt =2sqrt{pi}-2dfrac{sqrt{pi}}{2}=sqrt{pi}
finish{align*}

This exhibits us that differentiation beneath the integral works advantageous for ##xneq 0## whereas at ##x=0## we’ve got
\$\$
sqrt{pi}=dfrac{d}{dx}int xcdot |x|cdot e^{-x^2y^2},dy neqint dfrac{partial }{partial x}xcdot |x|cdot e^{-x^2y^2},dy=0
\$\$
The lacking situation is
\$\$
(*)qquad … textual content{ if }f(x,y)textual content{ is textbf{repeatedly differentiable} alongside the parameter }x.
\$\$

### Residues

Residue calculus deserves an article by itself, see [7]. So we are going to solely give a couple of examples right here. Residue calculus is the artwork of fixing actual integrals (and complicated integrals) through the use of complicated capabilities and numerous theorems of Augustin-Louis Cauchy.

Let ##f(z)## be an analytic operate ##f(z)## with a pole at ##z_0## and
\$\$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}\$\$ be the Laurent collection of ##f(z)## on a disc with radius ##rho## round ##z_0.## The coefficient at ##n=-1## known as the residue of ##f## at ##z_0,##
\$\$
operatorname{Res}_{z_0}(f)=a_{-1}=dfrac{1}{2pi i} oint_{partial D_rho(z_0)} f(zeta) ,dzeta.
\$\$

Mellin-Transformation.

\$\$
int_0^infty x^{lambda -1}R(x),dx=dfrac{pi}{sin(lambda pi)}sum_{z_k}operatorname{Res}_{z_k}left(fright); , ;lambda in mathbb{R}^+backslash mathbb{Z}
\$\$
the place ##R(x)=p(x)/q(x)## with ##p(x),q(x)in mathbb{R}[x],## ##q(x)neq 0 textual content{ for }xin mathbb{R}^+,## ##q(z_k)=0## for poles ##z_kin mathbb{C}## and the integral exists, i.e. ##lambda +deg p<deg q.##Notice that
\$\$
f(z),dx:={(-z)}^{lambda -1}R(z)=expleft((lambda -1)log(-z)proper)R(z)
\$\$
is holomorphic for ##zin mathbb{C}backslash mathbb{R}^+## if we selected the primary department of the complicated logarithm. We outline a closed curve
\$\$
alpha=start{circumstances}
alpha_1(t)=te^{ivarphi }&textual content{ if } tin left[1/r,rright]
alpha_2(t)=re^{it}&textual content{ if }t inleft[varphi ,2pi -varphi right]
alpha_3(t)=-te^{-ivarphi }&textual content{ if }tin left[-r,-1/rright]
alpha_4(t)=frac{1}{r}e^{i (2pi-t)}&textual content{ if }tin left[varphi ,2pi – varphi right]
finish{circumstances}
\$\$
such that the poles are all enclosed by ##alpha## for giant sufficient ##r## and sufficiently small ##varphi## and
\$\$
int_{alpha_1}f(z),dz+underbrace{int_{alpha_2}f(z),dz}_{stackrel{varphi to 0}{longrightarrow },0}+int_{alpha_3}f(z),dz+underbrace{int_{alpha_4}f(z),dz}_{stackrel{varphi to 0}{longrightarrow },0}=2pi i sum_{z_k}operatorname{Res}_{z_k}left(fright)
\$\$
We will present that
start{align*}
displaystyle{lim_{varphi to 0}}&int_{alpha_1}f(z),dz=-e^{-lambda pi i}int_{1/r}^r t^{lambda -1}R(t),dt
displaystyle{lim_{varphi to 0}}&int_{alpha_3}f(z),dz=e^{lambda pi i}int_{1/r}^r t^{lambda -1}R(t),dt
finish{align*}
and eventually get
\$\$
2pi i sum_{z_k}operatorname{Res}_{z_k}left(fright) =lim_{r to infty}int_alpha f(t),dt=2 i sin(lambda pi) int_{-infty }^infty t^{lambda -1}R(t),dt.
\$\$

The sinc Perform.

The sinc operate is outlined as ##operatorname{sinc}(x)=dfrac{sin x}{x}## for ##xneq 0## and ##operatorname{sinc(0)=1}.##
start{align*}
int_0^infty dfrac{sin x}{x},dx&=dfrac{1}{2}lim_{r to infty}int_{-r}^rdfrac{e^{iz}-e^{-iz}}{2i z},dz=operatorname{Res}_0left(dfrac{e^{iz}}{4 i z}proper)=dfrac{pi}{2}
finish{align*}

Guidelines for Residues.

We lastly embrace some common calculation guidelines for residue calculus that we took from [7].
\$\$
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}
\$\$