Brilliant Integration Contest - Season 2 (Part 2)

Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 2(Part 1)

The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows:

  • I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  • You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  • Only make substantial comment that will contribute to the discussion.

  • Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  • If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  • The scope of questions is only computation of integrals either definite or indefinite integrals.

  • You are NOT allowed to post a multiple integrals problem.

  • It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

  • Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest

  • You are also NOT allowed to post a solution using a contour integration or residue method.

The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, trigonometric integral, Wallis' integral, Bessel function, contour integration and Ramanujan's Master Theorem (including Mellin Transform).

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

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**SOLUTION OF PROBLEM xxx (number of problem) :**

**[Post your solution here]**

**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":


UPDATE:

  • Tanishq Varshney has been banned from this contest indefinitely.

  • Contour integration is allowed in the contest.

Note by Aditya Kumar
4 years, 2 months ago

No vote yet
1 vote

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Problem 50:

Evaluate 0ln(1+x)x34(1+x)dx\large \int_{0}^{\infty} \dfrac{\ln(1+x)}{\sqrt[4]{x^3}(1+x)}\, dx

Surya Prakash - 4 years, 2 months ago

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Alternative method:

ln(1+x)1+x=n=1Hn(x)n\dfrac{\ln(1+x)}{1+x} = - \sum_{n=1}^{\infty} H_{n}(-x)^{n}

Now use RMT. That's it. ¨\ddot \smile

Surya Prakash - 4 years, 2 months ago

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Put 11+x=y\dfrac{1}{1+x}=y to get :

I=01y(1/4)(1y)3/4ln(y)dy \displaystyle I = -\int _{ 0 }^{ 1 }{ { y }^{ (-1/4) }{ (1-y) }^{ -3/4 }\ln { (y) } dy }

Use the definition of beta function we have :

01ym1(1y)n1dy=Γ(m)Γ(n)Γ(m+n) \displaystyle \int _{ 0 }^{ 1 }{ { y }^{ m-1 }{ (1-y) }^{ n-1 }dy } = \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) }

Differentiating it with respect to mm we have :

Γ(m)Γ(n)Γ(m+n)(ψ(m)ψ(m+n))=01ym1(1y)n1ln(y)dy \displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (\psi (m)-\psi (m+n)) = \int _{ 0 }^{ 1 }{ { y }^{ m-1 }{ (1-y) }^{ n-1 }\ln { (y) } dy }

Put m=34,n=14\displaystyle m=\dfrac{3}{4} , n=\dfrac{1}{4} to get :

I=Γ(3/4)Γ(1/4)Γ(1)(ψ(1)ψ(3/4)) \displaystyle I = \frac { \Gamma (3/4)\Gamma (1/4) }{ \Gamma (1) } (\psi (1)-\psi (3/4))

Using reflection formula for the gamma function we have :

Γ(1/4)Γ(11/4)=πsin(π4)=2π \displaystyle \Gamma(1/4)\Gamma(1-1/4) = \dfrac{\pi}{\sin(\dfrac{\pi}{4})}=\sqrt{2}\pi

We will use that ψ(x+1)+γ=011tx1tdt \displaystyle \psi(x+1) + \gamma = \int _{ 0 }^{ 1 }{ \frac { 1-{ t }^{ x } }{ 1-t } dt }

ψ(x+1)ψ(y+1)=01tytx1tdt \displaystyle \Rightarrow \psi(x+1)-\psi(y+1) = \int _{ 0 }^{ 1 }{ \frac { { t }^{ y }-{ t }^{ x } }{ 1-t } dt }

ψ(1)ψ(3/4)=01t1/411tdt=J \displaystyle \Rightarrow \psi(1)-\psi(3/4) = \int _{ 0 }^{ 1 }{ \frac { { t }^{ -1/4 }-1 }{ 1-t } dt }=J

For evaluting JJ we will put t=x4t={x}^{4} , to get :

J=4011x1x4x2dx\displaystyle J = 4\int _{ 0 }^{ 1 }{ \frac { 1-x }{ 1-{ x }^{ 4 } } { x }^{ 2 }dx }

J=401x2(1+x2)(1+x)dx \displaystyle J = 4\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ (1+{ x }^{ 2 })(1+x) } dx }

J=201dx1+x+012x1+x2dx201dx1+x2\displaystyle J =2\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+x } } +\int _{ 0 }^{ 1 }{ \frac { 2x }{ 1+{ x }^{ 2 } } dx } -2\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } }

J=3ln(2)π2 \displaystyle J = 3\ln { (2) }-\dfrac{\pi}{2}

I=32πln(2)π22 \displaystyle \Rightarrow I = 3\sqrt { 2 } \pi \ln { (2) } -\frac { { \pi }^{ 2 } }{ \sqrt { 2 } }

Ronak Agarwal - 4 years, 2 months ago

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Problem 49:

Evaluate 01ln(Γ(x))dx. \large \displaystyle \int _{ 0 }^{ 1 }{ \ln { (\Gamma (x)) } dx. }

This problem has been solved by Surya Prakesh.

Ronak Agarwal - 4 years, 2 months ago

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I=01ln(Γ(x))dx=01ln(Γ(1t))dtI = \int_{0}^{1} \ln(\Gamma(x)) dx = \int_{0}^{1} \ln(\Gamma(1-t)) dt

I used the substitution x+t=1x+t=1. But from Euler's reflection formula for Gamma function i.e.

Γ(x)Γ(1x)=πcsc(πx)    ln(Γ(1x))+ln(Γ(x))=ln(π)+ln(csc(πx))\Gamma(x) \Gamma(1-x)= \pi \csc(\pi x) \implies \ln(\Gamma(1-x)) + \ln(\Gamma(x)) = \ln(\pi) + \ln(\csc(\pi x))

I=01ln(Γ(1x))dxI=01ln(Γ(x))dx+01ln(π)dx+01ln(csc(πx))dxI=I+ln(π)01ln(sin(πx))dx2I=ln(π)1π0πln(sin(x))dx2I=ln(π)2π0π/2ln(sin(x))dx2I=ln(π)2π(π2ln(2))2I=ln(π)+ln(2)I=12ln(2π)\begin{aligned} I &= \int_{0}^{1} \ln(\Gamma(1-x)) dx\\ I &= -\int_{0}^{1} \ln(\Gamma(x)) dx + \int_{0}^{1} \ln(\pi) dx+ \int_{0}^{1} \ln(\csc(\pi x))dx \\ I &= -I + \ln(\pi) - \int_{0}^{1} \ln(\sin(\pi x))dx \\ 2I &= \ln(\pi) - \dfrac{1}{\pi} \int_{0}^{\pi} \ln(\sin(x))dx \\ 2I &= \ln(\pi) - \dfrac{2}{\pi} \int_{0}^{\pi / 2} \ln(\sin(x))dx \\ 2I &= \ln(\pi) - \dfrac{2}{\pi} \left(-\dfrac{\pi}{2} \ln(2) \right) \\ 2I &= \ln(\pi) + \ln(2) \\ I &= \dfrac{1}{2} \ln(2 \pi) \end{aligned}


In the above solution I used the result that 0π/2ln(sin(x))dx=π2ln(2)\int_{0}^{\pi/2} \ln(\sin(x))dx = -\dfrac{\pi}{2} \ln(2) . This is proved here as follows:

J=0π/2ln(sin(x))dx=0π/2ln(sin(π/2x))dx=0π/2ln(cos(x))dx2J=0π/2ln(sin(x))dx+0π/2ln(cos(x))dx2J=0π/2ln(sin(x)cos(x))dx2J=0π/2ln(sin(2x))dx0π/2ln(2)dx2J=120πln(sin(x))dxπ2ln(2)2J=12(20π/2ln(sin(x))dx)π2ln(2)2J=Jπ2ln(2)J=π2ln(2)\begin{aligned} J &=\int_{0}^{\pi/2} \ln(\sin(x))dx = \int_{0}^{\pi/2} \ln(\sin(\pi/2 - x))dx = \int_{0}^{\pi/2} \ln(\cos(x))dx \\ 2J &= \int_{0}^{\pi/2} \ln(\sin(x))dx + \int_{0}^{\pi/2} \ln(\cos(x))dx\\ 2J &= \int_{0}^{\pi/2} \ln(\sin(x) \cos(x))dx \\ 2J &= \int_{0}^{\pi/2} \ln(\sin(2x))dx - \int_{0}^{\pi/2} \ln(2)dx \\ 2J &= \dfrac{1}{2} \int_{0}^{\pi} \ln(\sin(x))dx - \dfrac{\pi}{2} \ln(2) \\ 2J &= \dfrac{1}{2}\left( 2\int_{0}^{\pi/2} \ln(\sin(x))dx \right) - \dfrac{\pi}{2} \ln(2) \\ 2J &= J - \dfrac{\pi}{2} \ln(2) \\ J &=- \dfrac{\pi}{2} \ln(2) \end{aligned}

Surya Prakash - 4 years, 2 months ago

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Problem 48:

Evaluate 0π/2csc(x)tanxsinx+cosxdx.\int _{ 0 }^{ \pi /2 }{ \frac { \csc ( x ) \sqrt { \tan x } }{ \sin x+\cos x } \, dx . }

This problem has been solved by Surya Prakesh.

Julian Poon - 4 years, 2 months ago

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Take the substitution t=tanxt= \tan x .

0π/2cscxtanxsinx+cosxdx=20π/2tanx(1+tanx)(sin2x)dx=20t(1+t)(2t)dt=0t(1+t)(t)dt\begin{aligned} \int_{0}^{\pi/2} \dfrac{ \csc x \sqrt{\tan x}}{\sin x + \cos x} dx &= 2\int_{0}^{\pi/2} \dfrac{\sqrt{\tan x}}{(1+\tan x ) (\sin 2x)} dx \\ &= 2\int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(2t)} dt \\ &= \int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(t)} dt \end{aligned}

Take t=tan2(y)t = tan^2 (y).

0t(1+t)(t)dt=0π/2tanysec2(y)tan2(y)(2tan(y)sec2(y)dy)=20π/2dy=π.\int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(t)} dt = \int_{0}^{\pi /2} \dfrac{\tan y}{\sec^2 (y) \tan^2 (y)} (2 \tan(y) \sec^2 (y) dy) = 2 \int_{0}^{\pi/2}dy = \pi.

Surya Prakash - 4 years, 2 months ago

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Problem 47:

Evaluate 0ex2ln(x)dx. \displaystyle \large \int _{ 0 }^{ \infty }{ { e }^{ -{ x }^{ 2 } }\ln { (x) } \, dx. }

This problem has been solved by Julian Poon.

Ronak Agarwal - 4 years, 2 months ago

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Lemma 0xaex2dx=Γ(a+12)2\int _{ 0 }^{ \infty }{ x^{ a }\cdot e^{ -x^{ 2 } }dx } =\frac { \Gamma \left( \frac { a+1 }{ 2 } \right) }{ 2 }

Giving the substitution x2=tx^2=t, the integral becomes:

0xaex2dx=120ta12etdt=Γ(a+12)2\int _{ 0 }^{ \infty }{ x^{ a }\cdot e^{ -x^{ 2 } }dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ t^{ \frac { a-1 }{ 2 } }\cdot e^{ -t }dt } =\frac { \Gamma \left( \frac { a+1 }{ 2 } \right) }{ 2 }

So the required integral is

=14[Γ(12)ψ(12)]=14π(γ+ln(4))=\frac { 1 }{ 4 } \left[ \Gamma \left( \frac { 1 }{ 2 } \right) \psi \left( \frac { 1 }{ 2 } \right) \right] =-\frac { 1 }{ 4 } \sqrt { \pi } (\gamma +\ln{(4)})


To show that ψ(1/2)=γln(4)\psi(1/2)=-\gamma-\ln(4), I'll use ψ(a)=Ha1γ\psi(a)=H_{a-1}-\gamma

ψ(1/2)=H1/2γ\psi(1/2)=H_{-1/2} - \gamma

From the definition of harmonic numbers:

H1/2=H1/22H_{-1/2}=H_{1/2} - 2

H1/2=011x1xdx=201t(1t1t2)dtH_{1/2}=\int _{ 0 }^{ 1 }{ \frac { 1-\sqrt { x } }{ 1-x } dx } =2\int _{ 0 }^{ 1 }{ t\left( \frac { 1-t }{ 1-t^{ 2 } } \right) dt }

I used the substitution t=xt=\sqrt{x} above

01t(1t1t2)dt=01t1+tdt=01111+tdt=1ln(2)\int _{ 0 }^{ 1 }{ t\left( \frac { 1-t }{ 1-t^{ 2 } } \right) dt } =\int _{ 0 }^{ 1 }{ \frac{t}{1+t}dt } =\int _{ 0 }^{ 1 }{ 1-\frac{1}{1+t} dt}=1-\ln(2)

Hence

H1/2=H1/22=2(1ln(2))2=ln(4)H_{-1/2}=H_{1/2} - 2=2\left(1-\ln \left(2\right)\right)-2=-\ln(4)

ψ(1/2)=ln(4)γ\psi(1/2)=-\ln(4) - \gamma

Julian Poon - 4 years, 2 months ago

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Problem 46:

Evaluate

limn0π/4n(cos2nxsin2nx)tan2x dx.\large \lim_{n \to \infty} \int_{0}^{{\pi} / {4}} n \left(\cos^{2n} x - \sin^{2n} x \right) \tan 2x \ \mathrm{d}x.

Ishan Singh - 4 years, 2 months ago

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Take f(n)=0π4n(cos2nθsin2nθ)tan(2θ)dθ \displaystyle f(n) = \int _{ 0 }^{ \frac { \pi }{ 4 } }{ n(\cos ^{ 2n }{ \theta } -\sin ^{ 2n }{ \theta } )\tan { (2\theta ) } d\theta }

Take cos(2θ)=xcos(2\theta)=x , to get :

f(n)=n2n+101(1+x)n(1x)nxdx\displaystyle f(n) = \frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ \frac { { (1+x) }^{ n }-{ (1-x) }^{ n } }{ x } dx }

We have to find limnf(n)\displaystyle \lim _{ n\rightarrow \infty }{ f(n) }

We will first find the functional equation :

n2n+101(1+x)n(1x)nxdx=n2n+101(1+x)n1(1x)n1xdx+n2n+101(1+x)n1+(1x)n1dx \displaystyle \frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ \frac { { (1+x) }^{ n }-{ (1-x) }^{ n } }{ x } dx } =\frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ \frac { { (1+x) }^{ n-1 }-{ (1-x) }^{ n-1 } }{ x } dx } +\frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ { (1+x) }^{ n-1 }+{ (1-x) }^{ n-1 }dx }

f(n)=nf(n1)2(n1)+12 \displaystyle f(n) = \frac { nf(n-1) }{ 2(n-1) } +\frac { 1 }{ 2 }

Let the limit be LL

limnf(n)=limnnf(n1)2(n1)+12 \displaystyle \lim _{ n\rightarrow \infty }{ f(n) } =\lim _{ n\rightarrow \infty }{ \frac { nf(n-1) }{ 2(n-1) } +\frac { 1 }{ 2 } }

Hence L=12L+12\displaystyle L = \dfrac{1}{2}L+\dfrac{1}{2}

L=1\Rightarrow \boxed{L=1}

Ronak Agarwal - 4 years, 2 months ago

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Problem 45:

Prove that 0π/2sin6(x)tan(x)sin(tanx)dx=5π96e. \large \int_0^{\pi /2} \sin^6 (x) \tan (x ) \sin(\tan x) \, dx = \dfrac{5\pi}{96e} .

This problem has been solved by Ishan Singh.

Pi Han Goh - 4 years, 2 months ago

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Substitute tanxx\displaystyle \tan x \mapsto x

    I=0x7sinx(1+x2)4dx\displaystyle \implies \text{I} = \int_{0}^{\infty} \dfrac{x^{7}\sin x}{(1+x^2)^4} \mathrm{d}x

From solution of Problem 28, we have,

\displaystyle \int_{0}^{\infty} \dfrac{x\sin mx}{(a^2+x^2)^3} \mathrm{d}x = \dfrac{\pi b e^{-a}}{16a^3} (1+am) \quad \quad \quad \quad \tag{*}

Differentiating ()(*) 66 times w.r.t. mm , 11 time w.r.t. aa, at a=m=1a=m=1 and simplifying, we get,

I=5π96e\displaystyle \text{I} = \boxed{\dfrac{5 \pi}{96e}}

Ishan Singh - 4 years, 2 months ago

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@Kazem Sepehrinia@Ronak Agarwal @Tanishq Varshney @Ishan Singh @Rajorshi Chaudhuri@Aman Rajput

Can you guys join Slack? I intend to form a discussion group for the series+integrals enthusiast for us.

Do let me know if you do not know how to log in.

Pi Han Goh - 4 years, 2 months ago

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Hellolo sir, could I be part of this discussion? i'm also pretty interested in Integrals and series as well

Seidemann Frost - 3 years, 11 months ago

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Join Slack first.

Pi Han Goh - 3 years, 11 months ago

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@Pi Han Goh Click here

Pi Han Goh - 3 years, 11 months ago

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@Pi Han Goh Thanks!

Seidemann Frost - 3 years, 11 months ago

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@Pi Han Goh How do you do you that? It isn't very clear

Seidemann Frost - 3 years, 11 months ago

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I'm in.

Ronak Agarwal - 4 years, 2 months ago

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message me (3.14159han)

Pi Han Goh - 4 years, 2 months ago

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@Pi Han Goh After entering my email Id and clicking on get my invite, nothing is happening. Please help

Parasuram Ivln - 3 years, 8 months ago

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Hi participants. Contour integration is allowed from now onwards.

Aditya Kumar - 4 years, 2 months ago

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Problem 44:

Evaluate 0(ln(1+x)ln2(x)+π2)dxx2.\displaystyle \int _{ 0 }^{ \infty }{ \left( \dfrac{ \ln { (1+x) } }{ { \ln }^{ 2 }(x)+{ \pi }^{ 2 } }\right) \frac { dx }{ {x}^{2} } } .

Due to time constraint, the author decided to post their own solution.

Ronak Agarwal - 4 years, 2 months ago

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Lemma : I=0ekyy2+π2dy=(1)kπkπsin(x)xdx \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ -ky } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } =\frac{{ (-1) }^{ k }}{\pi} \int _{ k\pi }^{ \infty }{ \frac { sin(x) }{ x } dx } , for kk is a whole number.

Proof : We being with noting that :

1w0estsinwtdt=1s2+w2 \displaystyle \frac { 1 }{ w } \int _{ 0 }^{ \infty }{ { e }^{ -st }sinwtdt } =\frac { 1 }{ { s }^{ 2 }+{ w }^{ 2 } }

So we can rewrite the integral on the left side as :

I=1π0ekydy0etysinπtdt=1π0sinπt0e(k+t)ydydt \displaystyle I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ { e }^{ -ky }dy } \int _{ 0 }^{ \infty }{ { e }^{ -ty }sin\pi tdt } =\frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ sin\pi t\int _{ 0 }^{ \infty }{ { e }^{ -(k+t)y }dy } dt }

I=1π0sinπtt+kdt \displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ \frac { sin\pi t }{ t+k } dt }

I=1πksin(πtπk)tdt \displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ k }^{ \infty }{ \frac { sin(\pi t-\pi k) }{ t } dt }

I=(1)kksin(πt)tdt\displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k }^{ \infty }{ \frac { sin(\pi t) }{ t } dt }

I=(1)kkπsin(t)tdt \displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k\pi }^{ \infty }{ \frac { sin(t) }{ t } dt }

Hence proved

Solution :

We start with x=eyx={e}^{-y} to get out integral as :

I=eyln(1+ey)y2+π2dy \displaystyle I = \int _{ -\infty }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

Break in into two parts :

I=0eyln(1+ey)y2+π2dy+0eyln(1+ey)y2+π2dy \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ -\infty }^{ 0 }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

In the second part put y=xy=-x to get :

I=0eyln(1+ey)y2+π2dy+0exln(1+ex)x2+π2dy\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { e }^{ -x }\ln { (1+{ e }^{ x }) } }{ { x }^{ 2 }+{ \pi }^{ 2 } } dy }

Manipulating it we have :

I=0eyln(1+ey)y2+π2dy+0eyln(1+ey)y2+π2dy+0yeyy2+π2dy \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { e }^{ -y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

I=0(ey+ey)ln(1+ey)y2+π2dy+0yeyy2+π2dy\displaystyle \Rightarrow I= \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ y }+{ e }^{ -y })\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

We write I=J+K I = J+K

J=0(ex+ex)ln(1+ex)x2+π2dx \displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x })\ln { (1+{ e }^{ -x }) } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx

Wrting ln(1+ex) \ln(1+{e}^{-x}) in it's taylor series we have :

J=0(ex+ex)x2+π2r=1(1)r1rerxdx\displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x }) }{ { x }^{ 2 }+{ \pi }^{ 2 } } \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } { e }^{ -rx } } } dx

We interchange the integral and sum :

J=r=1(1)r1r(0e(r1)xx2+π2dx+0e(r+1)xx2+π2dx) \displaystyle J = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } (\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r-1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx+\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r+1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx } )

Using the lemma we have :

J=1πr=11r((r1)πsin(x)xdx+(r+1)πsin(x)xdx) \displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ (r-1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } +\int _{ (r+1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } } dx) }

Now it is worthy noting that :

rπsin(x)xdx=1sin(rπx)xdx \displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx } , for all positive rr, what if it was applied for r=0r=0, then the left integral is π2\frac{\pi}{2}, while the write one is 0, so for r=0r=0, we can write it as :

rπsin(x)xdx=1sin(rπx)xdx+π2 \displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx } +\frac { \pi }{ 2 } , for r=0 r=0, having said that we proceed further :

J=1πr=11r(12sin(2(r1)πx)+sin(2(r+1)πx)xdx)+12 \displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { \sin { (2(r-1)\pi x) } +\sin { (2(r+1)\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }

J=2πr=11r(12cos(2πx)sin(2rπx)xdx)+12 \displaystyle \Rightarrow J = \frac { 2 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { cos(2\pi x)\sin { (2r\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }

Again changing sum and integral we have :

J=2π12cos(2πx)r=1sin(2rπx)rxdx+12 \displaystyle \Rightarrow J = \frac { 2 }{ \pi } \int _{ \frac { 1 }{ 2 } }^{ \infty }{ \cos { (2\pi x) } \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ rx } } dx } +\frac { 1 }{ 2 }

It is a well known result that :

r=1sin(2rπx)rπ=x+12x \displaystyle \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ r\pi } } =\left\lfloor x \right\rfloor +\frac { 1 }{ 2 }-x

Using this we have :

J=212cos(2πx)x(x+x+12)dx+12 \displaystyle J = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 }

Leaving this let's evaluate KK first :

Using the same techniques as used in proving the lemma, we can show that :

K=0xexx2+π2dx=1/2cos(2πx)xdx \displaystyle K = \int _{ 0 }^{ \infty }{ \frac { x{ e }^{ -x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } dx } =-\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx }

Now I=J+K=212cos(2πx)x(x+x+12)dx+121/2cos(2πx)xdx=12+21/2cos(2πx)x{x}dx \displaystyle I = J+K = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 } -\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } =\frac { 1 }{ 2 } +2\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }

I=12+21cos(2πx)x{x}dx+2121cos(2πx)dx=12+21cos(2πx)x{x}dx \displaystyle I = \frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx } +2\int _{ \frac { 1 }{ 2 } }^{ 1 }{ \cos { (2\pi x) } dx } =\frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }

Now we will be evaluating :

M=1cos(2πx)x{x}dx \displaystyle M = \int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }

It can be written as :

M=n=1nn+1cos(2πx)x(xn)dx \displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } (x-n)dx } }

M=n=1nnn+1cos(2πx)xdx\displaystyle \Rightarrow M = -\sum _{ n=1 }^{ \infty }{ n\int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } dx } }

MM can also be written as :

M=n=1ncos(2πx)xdx\displaystyle M =-\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } }

This manipulation can easily be justified with the help of properties of double summations :

Using integration by parts we have :

M=n=1nsin(2πx)2πx2dxn=1sin(2πn)2πn \displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } } -\sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi n) } }{ 2\pi n } }

M=n=1nsin(2πx)2πx2dx\displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } }

M=n=11sin(2πnx)2πnx2dx \displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ 1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } dx } }

Again changing sum and integral we have :

M=1n=1sin(2πnx)2πnx2dx \displaystyle \Rightarrow M = \quad \int _{ 1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } } dx }

Using the result we have used above we have :

M=12112x2{x}x2dx \displaystyle M = \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }

M=12112x2{x}x2dx \displaystyle \Rightarrow M = \quad \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }

For solving H=1{x}x2dx \displaystyle H = \int _{ 1 }^{ \infty }{ \frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }

write it as a sum :

H=n=1nn+11xnx2dx \displaystyle H = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { 1 }{ x } -\frac { n }{ { x }^{ 2 } } dx } }

H=n=1ln(n+1)ln(n)1n+1 \displaystyle \Rightarrow H = \sum _{ n=1 }^{ \infty }{ \ln { (n+1) } -\ln { (n) } -\frac { 1 }{ n+1 } }

H=limn1+ln(n+1)Hn+1=1γ \displaystyle H = \lim _{ n\rightarrow \infty }{ 1+\ln { (n+1)-{ H }_{ n+1 } } } =1-\gamma

Hence M=γ214\displaystyle M = \frac { \gamma }{ 2 } -\frac { 1 }{ 4 }

Finally we got :

I=γ \Large \boxed { I=\gamma }

Ronak Agarwal - 4 years, 2 months ago

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Problem 43:

Evaluate x3ex2(x2+1)2dx.\large \int \dfrac{x^3 e^{x^2} }{(x^2 + 1)^2} \, dx .

This problem has been solved by Ronak Agarwal.

Surya Prakash - 4 years, 2 months ago

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Put x2=y {x}^{2}=y to get the integral as :

I=12yey(y+1)2dy\displaystyle I = \frac { 1 }{ 2 } \int { \frac { y{ e }^{ y } }{ { (y+1) }^{ 2 } } dy }

It can be also written as :

I=12ey(y+1)ey(y+1)2dy=12d(eyy+1)dydy \displaystyle I = \frac { 1 }{ 2 } \int { \frac { { e }^{ y } }{ (y+1) } -\frac { { e }^{ y } }{ { (y+1) }^{ 2 } } dy } =\frac { 1 }{ 2 } \int { \frac { d(\frac { { e }^{ y } }{ y+1 } ) }{ dy } dy }

I=ey2(y+1)+C=ex22(x2+1)+C I = \dfrac { { e }^{ y } }{ 2(y+1) } +C=\dfrac { { e }^{ { x }^{ 2 } } }{ 2({ x }^{ 2 }+1) } +C

Ronak Agarwal - 4 years, 2 months ago

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Another way to solve is if we consider quotient rule in reverse its easy to guess our integral to be of the form ce^{x^2}/[x^2+1] it remains to solve for c and we're done

D S - 1 year, 9 months ago

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Problem 42:

Evaluate 0π2ln(2cos(x2))dx\displaystyle \large \int _{ 0 }^{ \frac{\pi}{2} }{ \ln { \left( 2\cos \left( \frac { x }{ 2 } \right) \right) } \,dx }

This problem has been solved by Surya Prakesh.

Julian Poon - 4 years, 2 months ago

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ln(2cos(x2))=ln(eix/2+eix/2)=ln(1+eix)ix2\ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) = \ln \left(e^{ix/2} + e^{-ix/2} \right) = \ln \left( 1 + e^{ix} \right) - i\dfrac{x}{2}

So,

0π/2ln(2cos(x2))dx=0π/2ln(1+eix)dxi0π/2x2dx=0π/2r=1(1)r+1eirxrdxiπ216\begin{aligned} \int_{0}^{\pi / 2} \ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) dx &= \int_{0}^{\pi / 2} \ln \left( 1 + e^{ix} \right) dx - i \int_{0}^{\pi / 2}\dfrac{x}{2} dx \\ &= \int_{0}^{\pi /2} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} e^{irx}}{r} dx - i \dfrac{\pi^2}{16} \end{aligned}

We shall evaluate the first integral

0π/2r=1(1)r+1eirxrdx=r=1(1)r+1r0π/2eirxdx=r=1(1)r+1r[eirxir0π/2=1ir=1(1)r+1(eirπ/21)r2\begin{aligned} \int_{0}^{\pi /2} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} e^{irx}}{r} dx &= \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1}}{r} \int_{0}^{\pi /2} e^{irx} dx \\ &= \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1}}{r} \left[ \dfrac{e^{irx}}{ir} \right|_{0}^{\pi/2}\\ &= \dfrac{1}{i} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} \end{aligned}

Now,

eirπ/21={i1r=4k+12r=4k+2i1r=4k+30r=4k+4 e^{ ir\pi /2 }-1=\quad \begin{cases} i-1 & \quad \quad r=4k+1 \\ -2 & \quad \quad r=4k+2 \\ -i-1 & \quad \quad r=4k+3 \\ 0 & \quad \quad r=4k+4 \end{cases}

r=1(1)r+1(eirπ/21)r2=k=0(1)4k+2(i1)(4k+1)2+k=0(1)4k+3(2)(4k+2)2+k=0(1)4k+4(i1)(4k+3)2+k=0(1)4k+5(0)(4k+4)2=k=0(i1)(4k+1)2+k=02(4k+2)2+k=0i1(4k+3)2+0=k=0(i1)(4k+1)2+k=02(4k+2)2+k=0i1(4k+3)2\begin{aligned}\sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+2}(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+3}(-2)}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+4}(-i-1)}{(4k+3)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+5}(0)}{(4k+4)^{2}} \\ &= \sum_{k=0}^{\infty} \dfrac{(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{2}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{-i-1}{(4k+3)^{2}} + 0 \\ &= \sum_{k=0}^{\infty} \dfrac{(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{2}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{-i-1}{(4k+3)^{2}} \end{aligned}

k=01(4k+1)2=116ζ(2,14)k=01(4k+2)2=116ζ(2,12)k=01(4k+3)2=116ζ(2,34)\sum_{k=0}^{\infty} \dfrac{1}{(4k+1)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{4} \right) \\ \sum_{k=0}^{\infty} \dfrac{1}{(4k+2)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{2} \right) \\ \sum_{k=0}^{\infty} \dfrac{1}{(4k+3)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{3}{4} \right)

So,

r=1(1)r+1(eirπ/21)r2=(i1)116ζ(2,14)+2116ζ(2,12)+(i1)116ζ(2,34)=116(i(ζ(2,14)ζ(2,34))(ζ(2,14)+ζ(2,34))+2ζ(2,12))\begin{aligned} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= (i-1) \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{4} \right) + 2 \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{2} \right) + (-i-1) \dfrac{1}{16} \zeta \left(2 , \dfrac{3}{4} \right)\\ &= \dfrac{1}{16} \left( i \left( \zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) \right) - \left( \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) \right) + 2 \zeta \left(2 , \dfrac{1}{2} \right) \right) \end{aligned}

1:
ζ(2,14)ζ(2,34)=16k=0(1(4k+1)21(4k+3)2)=16k=0(1)k(2k+1)2=16G\zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) = 16 \sum_{k=0}^{\infty} \left(\dfrac{1}{(4k+1)^2} - \dfrac{1}{(4k+3)^2}\right) = 16 \sum_{k=0}^{\infty} \dfrac{(-1)^{k}}{(2k+1)^2} = 16G

2:

ζ(2,14)+ζ(2,34)=16k=0(1(4k+1)2+1(4k+3)2)=16k=01(2k+1)2=16π28=2π2\begin{aligned} \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) &= 16 \sum_{k=0}^{\infty} \left(\dfrac{1}{(4k+1)^2} + \dfrac{1}{(4k+3)^2}\right) \\ &= 16 \sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2}\\ &=16 \dfrac{\pi^2}{8}\\ &= 2 \pi^2 \end{aligned}

3:

ζ(2,12)=4k=01(2k+1)2=4π28=π22\zeta \left(2 , \dfrac{1}{2} \right) = 4 \sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2} = 4 \dfrac{\pi^2}{8} = \dfrac{\pi^2}{2}

So,

r=1(1)r+1(eirπ/21)r2=116(i(ζ(2,14)ζ(2,34))(ζ(2,14)+ζ(2,34))+2ζ(2,12))=116(i(16G)(2π2)+2π22)=iGπ216\begin{aligned} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= \dfrac{1}{16} \left( i \left( \zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) \right) - \left( \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) \right) + 2 \zeta \left(2 , \dfrac{1}{2} \right) \right)\\ &= \dfrac{1}{16} \left( i (16G) - (2\pi^2) + 2 \dfrac{\pi^2}{2} \right)\\ &= iG - \dfrac{\pi^2}{16} \end{aligned}