Brilliant Integration Contest - Season 2 (Part-1)

Hi Brilliant! Just like what Anastasiya Romanova conducted last year, this year I would also like to conduct an integration contest.

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":

View Part 2

Note by Aditya Kumar
4 years, 3 months ago

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Problem 25

Prove that

0π2ln(sinx)ln(cosx)tanxdx=18ζ(3).\large{\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\ln(\sin x) \ln(\cos x)}{\tan x} dx=\frac{1}{8} \zeta (3)}.

This problem has been solved by Julian Poon.

Tanishq Varshney - 4 years, 2 months ago

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Alternative approach could be (i know its too late :P) could be considering Beta function (Γ(a)Γ(b)Γ(a+b))(\dfrac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)}) in trignometric form and differentiating it with respect to a and b and then putting a = 0 and b = 1.

Harsh Shrivastava - 4 years, 1 month ago

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Great! That's the standard approach! Can you post the full solution?

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh Γ(a)Γ(b)Γ(a+b)=20π/2sin2a1xcos2b1xdx\displaystyle \frac { \Gamma (a)\Gamma (b) }{ \Gamma (a+b) } = 2\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2a-1 }{ x } \cos ^{ 2b-1 }{ x } dx }

Differentiating both sides first wrt a and then wrt b, we get

Γ(a)Γ(b)Γ(a+b)(((ψ(a)ψ(a+b))(ψ(b)ψ(a+b))ψ(a+b))=80π/2log(sin(x))log(cos(x))sin2a1xcos2b1xdx\displaystyle \frac { \Gamma (a)\Gamma (b) }{ \Gamma (a+b) } (((\psi (a)-\psi (a+b))(\psi (b)-\psi (a+b))-\psi '(a+b)) \displaystyle =8\int _{ 0 }^{ \pi /2 }{ log(sin(x))log(cos(x))\sin ^{ 2a-1 }{ x } \cos ^{ 2b-1 }{ x } dx }

Putting a = 0 and b = 1,

ψ(1)=80π2ln(sinx)ln(cosx)tanxdx-\psi ' (1) = 8\int^{\frac{\pi}{2}}_{0}\frac{\ln(\sin x) \ln(\cos x)}{\tan x} dx

Harsh Shrivastava - 4 years, 1 month ago

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@Harsh Shrivastava Even I got the same. Now your job is to prove it equal to 1/8 zeta(3).

Aditya Kumar - 4 years, 1 month ago

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@Aditya Kumar I don't have any idea how to proceed from here.

Harsh Shrivastava - 4 years, 1 month ago

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@Harsh Shrivastava Hint: What is the definition of a digamma function?

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh @Harsh Shrivastava Want to join our discussion group where we talk bout integrals and series?

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh Well I first need to clear my basics of series, so maybe after March or so, I'll join the group, BTW if you guys discuss from basics, then I'll surely join right now.

Harsh Shrivastava - 4 years, 1 month ago

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@Harsh Shrivastava Well, you can ask simple questions as well, we're happy to explain to you... Still want to join?

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh Yeah sure! I'll join right after my school , I have to go to school now, Thanks!

Harsh Shrivastava - 4 years, 1 month ago

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@Harsh Shrivastava Go here

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh When i enter my email address, it says already invited, what to do now? I am not able to access further.

Harsh Shrivastava - 4 years, 1 month ago

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@Harsh Shrivastava How's the registration coming along?

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh I checked my mail but nothin:'s there regarding registration, please help!

Harsh Shrivastava - 4 years, 1 month ago

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@Harsh Shrivastava Step 1: GO here and type your email.
Step 2: GO check your email which you have typed in.
Step 3: Click that email link (for verification...). Done!

If none of these seem to work, register another email address and repeat steps 1 to 3.

Pi Han Goh - 4 years, 1 month ago

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@Harsh Shrivastava Check your email.

Pi Han Goh - 4 years, 1 month ago

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Substitute tsin(x)t \to \sin(x). The integral becomes:

1201lntln(1t2)(1t)dt\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \ln { t } \ln { \left( 1-t^{ 2 } \right) } \left( \frac { 1 }{ t } \right) } dt

Substitute in the Taylor series for ln(1x)\ln (1-x):

1201lntn=1t2n1ndt=12n=11n01ln(t)t2n1dt-\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \ln { t } \sum _{ n=1 }^{ \infty } \frac { t^{ 2n-1 } }{ n } } dt=-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \int _{ 0 }^{ 1 }{ \ln { (t) } t^{ 2n-1 } } dt

Finally, integrate by parts to obtain

12n=11n14n2=18n=11n3-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \frac { -1 }{ 4{ n }^{ 2 } } =\frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ { n }^{ 3 } }

Julian Poon - 4 years, 2 months ago

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Problem 24:

Prove that

0π/21sin8x+cos8xdx=1022π. \large \int_0^{\pi /2} \dfrac1{\sin^8 x + \cos^8 x } \, dx = \dfrac{\sqrt{10-\sqrt2}}{2} \pi .

This problem has been solved by Tanishq Varshney.

Pi Han Goh - 4 years, 2 months ago

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Solution to Problem 24\text{Solution to Problem 24}

0xm11+xdx=πsinmπ0<m<1\large{\displaystyle \int ^{\infty}_{0} \frac{x^{m-1}}{1+x}dx=\frac{\pi}{\sin m \pi} \quad \quad 0<m<1}

Refer to problem 18 for this step.

Now in the integral tanxt\tan x \rightarrow t

0(1+t2)31+t8dt\large{\displaystyle \int ^{\infty}_{0} \frac{(1+t^2)^3}{1+t^{8}}dt}

t8z\large{t^8 \rightarrow z}

18(0z7811+z+3z5811+z+3z3811+z+z1811+zdz)\large{\frac{1}{8} \left(\displaystyle \int ^{\infty}_{0} \frac{z^{\frac{7}{8}-1}}{1+z}+ \frac{3z^{\frac{5}{8}-1}}{1+z}+ \frac{3z^{\frac{3}{8}-1}}{1+z}+ \frac{z^{\frac{1}{8}-1}}{1+z} dz \right)}

π4(1sin(π8)+3cos(π8))\large{\rightarrow \frac{\pi}{4} \left(\frac{1}{\sin \left(\frac{\pi}{8} \right)}+\frac{3}{\cos \left(\frac{\pi}{8} \right)} \right)}

=1022π\large{=\frac{\sqrt{10-\sqrt{2}}}{2} \pi}

Tanishq Varshney - 4 years, 2 months ago

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Problem 23:

Evaluate:01log(1+x)log(1x3)dx\int _{ 0 }^{ 1 }{ \log\left( 1+x \right) \log\left( 1-{ x }^{ 3 } \right) dx }

Due to time constraint, Aditya Kumar decided to post a solution himself.

Aditya Kumar - 4 years, 2 months ago

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Solution to Problem 23:

Substitute: x1x1+xx\longrightarrow \frac { 1-x }{ 1+x }

I=01ln(1+x)ln(1x3)dx=201ln(2x(3+x2)(1+x)3)ln(21+x)dx(1+x)2I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}

On separating the integrands,

ln(2x(3+x2)(1+x)3)ln(21+x)=ln22ln2lnx+ln2ln(3+x2)4ln2ln(1+x)lnxln(1+x)+3ln2(1+x)ln(1+x)ln(3+x2)\ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2)

1st integral:

\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12

2nd integral:

01lnx(1+x)2dx=n1(1)nn=ln2\int_0^1 \frac{\ln x}{(1+x)^2}dx=\sum_{n\geq1} \frac{(-1)^n}{n}=-\ln2

3rd integral:

\int_0^1 \frac{\ln(1+x)}{(1+x)^2}dx=-\frac{\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac{dx}{(1+x)^2}dx=\frac12-\frac{\ln2}{2}

4th integral:

\int_0^1 \frac{\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2x}{(1+x)(3+x^2)}dx \\=-\ln2+\ln3+2\operatorname{Re} \int_0^1 \frac{dx}{(1+x)(x+i\sqrt{3})}=-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}

5th integral:

\int_0^1 \frac{\ln x \ln(1+x)}{(1+x)^2}dx=-\frac{\ln x\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac1{1+x}\left(\frac{\ln(1+x)}{x}+\frac{\ln x}{1+x}\right)dx \\=-\ln2+\int_0^1 \frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{\pi^2}{12}-\frac12\ln^22-\ln2

6th integral:

\int_0^1 \frac{\ln^2(1+x)}{(1+x)^2}dx=-\frac{\ln^2(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2\ln(1+x)}{(1+x)^2}dx \\=1-\ln2-\frac12\ln^22

7th integral:

\int_0^1 \frac{\ln(1+x)\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(1+x)\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac1{1+x}\left(\frac{2x\ln(1+x)}{3+x^2}+\frac{\ln(3+x^2)}{1+x}\right)dx \\=-\ln^22-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}+2J

where,

J=01xln(1+x)(1+x)(3+x2)dx=Re01ln(1+x)(1+x)(x+i3)dx=18ln22Re1i3101ln(1+x)x+i3dxJ=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx

Now on using the formula:

01ln(1+x)x+adx=ln2lna+1a1+Li2(21a)Li2(11a)\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right)

Putting a=i3a=i\sqrt{3} then gives, after taking the real part, and assuming the principle value of the logarithm,

J=18ln22+π312ln2+14ReLi2(eiπ/3)34ImLi2(eiπ/3)14ReLi2(12eiπ/3)+34ImLi2(12eiπ/3)\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})

Using:

ReLi2(eiπ/3)=π236\displaystyle \,\, \operatorname{Re}\text{Li}_2(e^{i\pi/3})=\frac{\pi^2}{36},

ImLi2(eiπ/3)=123ψ1(13)π233\displaystyle \,\, \operatorname{Im}\text{Li}_2(e^{i\pi/3})=\frac1{2\sqrt{3}}\psi_1\left(\frac13\right)-\frac{\pi^2}{3\sqrt{3}},

Li2(12eiπ/3)=Li2(i3)12ln2(34i34)\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right),

and

ReLi2(i3)=n=1(1)n(2n)23n=14Li2(13)\displaystyle \,\, \operatorname{Re}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)^2\, 3^n}=\frac14\text{Li}_2\left(-\frac13\right)

we have ReLi2(12eiπ/3)=π27218ln2(34)14Li2(13)\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)

and ImLi2(12eiπ/3)=π12ln(34)ImLi2(i3).\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi}{12}\ln\left(\frac34\right)-\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).

Finally, putting everything together gives: I=ln2218ln23+2ln2ln332ln36ln2+6π43(2+ln3)37π272+12ψ1(13)14Li2(13)+3Li2(i3)I=\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )

......

This was not original. I found it very awesome, so I thought to share it here.

Aditya Kumar - 4 years, 2 months ago

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I don't know what's wrong with the rendering of the Latex, since it sometimes render and sometimes it doesn't. So I'm going to post in pictures the one's that didn't render. I think it's better to move this discussion to another note.

1st Integral:

3rd Integral:

4th Integral:

5th Integral:

6th Integral:

7th Integral:

Julian Poon - 4 years, 2 months ago

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Problem 22:

Evaluate the integral

0πxsin(x)(cos2(x)+3)2dx.\large \int_{0}^{\pi} \dfrac{x \sin (x)}{(\cos^2 (x) + 3)^2} dx .

This problem has been solved by Aditya Kumar.

Surya Prakash - 4 years, 2 months ago

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I=0πxsinx(cos2x+3)2dxI=π20πsinx(cos2x+3)2dxcosx=tI=π2111(t2+3)2dtt=3tanxI=π336π6π6(cos2x+1)dxI=π318{34+π6}I=\int _{ 0 }^{ \pi }{ \frac { xsinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ I=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ \frac { sinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ cosx=t\quad \\ I=\frac { \pi }{ 2 } \int _{ -1 }^{ 1 }{ \frac { 1 }{ { \left( { t }^{ 2 }+3 \right) }^{ 2 } } dt } \\ t=\sqrt { 3 } tanx\\ I=\frac { \pi \sqrt { 3 } }{ 36 } \int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \left( cos2x+1 \right) dx } \\ \therefore I=\frac { \pi \sqrt { 3 } }{ 18 } \left\{ \frac { \sqrt { 3 } }{ 4 } +\frac { \pi }{ 6 } \right\}

In the first step, I used: abf(x)=abf(a+bx)\int _{ a }^{ b }{ f(x) } =\int _{ a }^{ b }{ f(a+b-x) }

Aditya Kumar - 4 years, 2 months ago

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Problem 21:

For b<1|b|<1, find the value of the integral below in terms of bb.

π/2π/2ln(1+bsinx)sinxdx \int_{-\pi /2}^{\pi /2} \dfrac{ \ln(1+ b \sin x)}{\sin x} \, dx

This problem has been solved by Surya Prakesh.

Gautam Sharma - 4 years, 2 months ago

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f(b)=π/2π/2ln(1+bsin(x))sin(x)dxf(b)=π/2π/211+bsin(x)dxf(b) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1+ b \sin (x))}{\sin(x)} dx \\ f'(b) = \int_{-\pi/2}^{\pi/2}\dfrac{1}{1+ b\sin(x)} dx

Using Weierstrass Substitution i.e. t=tan(x/2)    sin(x)=2t1+t2t = \tan (x/2) \implies \sin(x) = \dfrac{2t}{1+t^2}.

f(b)=211dtt2+2bt+1=211dt(t+b)2+(1b2)2=21b2[arctan(t+b1b2)11=21b2[arctan(1+b1b2)arctan(1+b1b2)]=21b2[arctan(1+b1b2)+arctan(1b1b2)]=21b2[arctan(1+b1b)+arctan(1b1+b)]=π1b2\begin{aligned} f'(b) &= 2\int_{-1}^{1} \dfrac{dt}{t^2 + 2bt + 1}\\ &= 2\int_{-1}^{1} \dfrac{dt}{(t+b)^2 + (\sqrt{1-b^2})^2}\\ &= \dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{t+b}{\sqrt{1-b^2}} \right) \Bigr|_{-1}^{1}\\ &=\dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{1+b}{\sqrt{1-b^2}} \right) - \arctan\left( \dfrac{-1+b}{\sqrt{1-b^2}} \right)\Bigr] \\ &=\dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{1+b}{\sqrt{1-b^2}} \right) + \arctan\left( \dfrac{1-b}{\sqrt{1-b^2}} \right)\Bigr] \\&= \dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left(\sqrt{\dfrac{1+b}{1-b}} \right) + \arctan \left(\sqrt{\dfrac{1-b}{1+b}} \right)\Bigr] \\&= \dfrac{\pi}{\sqrt{1-b^2}} \end{aligned}

I used the fact that arctan(x)+arctan(1/x)=π/2\arctan(x) + \arctan(1/x) = \pi/2.

So,

f(b)=π1b2f(b)=π1b2dbf(b)=πarcsin(b)+cf'(b) = \dfrac{\pi}{\sqrt{1-b^2}} \\ f(b) = \int \dfrac{\pi}{\sqrt{1-b^2}} db \\ f(b) = \pi \arcsin(b) +c

But f(b)=π/2π/2ln(1+bsin(x))sin(x)dx    f(0)=π/2π/2ln(1)sin(x)dx=0f(b) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1+ b \sin (x))}{\sin(x)} dx \implies f(0) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1)}{\sin(x)} dx =0

So, c=0c=0. Therefore, f(b)=πarcsin(b)f(b) = \pi \arcsin(b).

Surya Prakash - 4 years, 2 months ago

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Problem 20:

Evaluate

0dx(1+x)3+1\large \int_{0}^{\infty} \dfrac{dx}{(1+x)^3 +1}

This problem has been solved by Gautam Sharma.

Surya Prakash - 4 years, 2 months ago

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I used complex analysis

Nahom Assefa - 1 year, 3 months ago

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Solution to problem 20

Let (1+x)=tI=dtt3+1(1+x)=t \\ I=\large \int \dfrac{dt}{t^3 +1}

By partial fractions

I=2t3(t2t+1)dt+13(t+1)dtI=12(t2t+1)dt2t16(t2t+1)dt+13(t+1)dtI=12((t12)2+34)dt2t16(t2t+1)dt+13(t+1)dtI=13tan12t1316ln(t2t+1)+13ln(t+1)+cI=13tan12x+13+16ln((x+2)2x2+x+1)+cI=\int \frac { 2-t }{ 3({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ t }^{ 2 }-t+1) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ (t-\frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } ) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2t-1 }{ \sqrt { 3 } } } -\frac { 1 }{ 6 } \ln({ t }^{ 2 }-t+1)+\frac { 1 }{ 3 } \ln(t+1)+c \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2x+1 }{ \sqrt { 3 } } } +\frac { 1 }{ 6 } \ln\left(\frac { { (x+2) }^{ 2 } }{ { x }^{ 2 }+x+1 } \right)+c

Putting limits and evaluating we get π33ln23\frac{\pi}{3\sqrt3}-\frac{\ln2}{3}

Gautam Sharma - 4 years, 2 months ago

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Problem 19:

Prove that

0π3ln2(sinxsin(π3+x))dx=581π3.\large{\displaystyle \int^{\frac{\pi}{3}}_{0} \ln^{2} \left( \frac{\sin x}{\sin \left(\frac{\pi}{3}+x \right)} \right) \, dx=\frac{5}{81} \pi^{3}}.

This problem has been solved by Surya Prakesh.

Tanishq Varshney - 4 years, 2 months ago

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Alternate Solution to problem 19 :

Lemma: 0xm1ln2(x)1+xndx=(πn)3(csc(mπn)cot2(mπn)+csc3(mπn)) \displaystyle \int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 }{ ln }^{ 2 }(x) }{ 1+{ x }^{ n } } dx } ={ \left(\frac { \pi }{ n } \right) }^{ 3 }\left(\csc { \left(\frac { m\pi }{ n } \right) } \cot ^{ 2 }{ \left(\frac { m\pi }{ n } \right) } +\csc ^{ 3 }{ \left(\frac { m\pi }{ n } \right) } \right)\quad

Proof : We begin with identity I=0dx1+xn=πncsc(πn)\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } } =\frac { \pi }{ n } \csc { (\frac { \pi }{ n } ) }

To prove this put y=11+xn \displaystyle y = \dfrac{1}{1+{x}^{n}} to get :

I=1n01y1/n(1y)1/n1dy\displaystyle I =\dfrac{1}{n} \int _{ 0 }^{ 1 }{ { y }^{ -1/n }{ (1-y) }^{ 1/n-1 }dy }

Use definition of beta function and euler reflection formula to get :

I=1n01y1/n(1y)1/n1dy=1nΓ(1/n)Γ(11/n)Γ(1)=πncsc(πn)\displaystyle I = \dfrac{1}{n}\int _{ 0 }^{ 1 }{ { y }^{ -1/n }{ (1-y) }^{ 1/n-1 }dy } =\dfrac{1}{n}\frac { \Gamma (1/n)\Gamma (1-1/n) }{ \Gamma (1) } =\frac { \pi }{ n } \csc { (\frac { \pi }{ n } ) }

Now consider the integral J(m)=0xm1dx1+xn \displaystyle J(m) = \int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 }dx }{ 1+{ x }^{ n } } }

Put xm=y{x}^{m}=y to get :

J(m)=1m0dy1+yn/m \displaystyle J(m) = \frac { 1 }{ m } \int _{ 0 }^{ \infty }{ \frac { dy }{ 1+{ y }^{ n/m } } }

Use the identity to get :

J(m)=πncsc(mπn) \displaystyle J(m) = \frac { \pi }{ n } \csc { (\frac { m\pi }{ n } ) }

Differentiate it two times with respect to m m to prove the lemma :

My initial steps are the same as surya prakash and I directly get till :

I=3201(1+y)ln2(y)1+y3dy \displaystyle I = \frac { \sqrt { 3 } }{ 2 } \int _{ 0 }^{ 1 }{ \frac { (1+y){ ln }^{ 2 }(y) }{ 1+{ y }^{ 3 } } dy }

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

I=321(1+x)ln2(x)1+x3dx \displaystyle I = \frac { \sqrt { 3 } }{ 2 } \int _{ 1 }^{ \infty }{ \frac { (1+x){ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx }

Adding these two forms we have :

I=340(1+x)ln2(x)1+x3dx \displaystyle I = \frac { \sqrt { 3 } }{ 4 } \int _{ 0 }^{ \infty }{ \frac { (1+x){ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx }

I=34(0ln2(x)1+x3dx+0xln2(x)1+x3dx) \displaystyle I = \frac { \sqrt { 3 } }{ 4 } (\int _{ 0 }^{ \infty }{ \frac { { ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } +\int _{ 0 }^{ \infty }{ \frac { x{ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } )

Use the lemma to get :

I=34((π3)3(23.13+833)+(π3)3(23.13+833))=5π381 \displaystyle I = \frac { \sqrt { 3 } }{ 4 } ({ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } )+{ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } ))=\frac { 5{ \pi }^{ 3 } }{ 81 }

Ronak Agarwal - 4 years, 2 months ago

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Lemma: 01xmlnn(x)dx=(1)nn!(m+1)n+1\int_{0}^{1} x^{m} \ln ^{n} (x) dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}}

Let y=sinxsin(x+π/3)    tanx=3y2y    dx=321y2y+1dyy=\dfrac{\sin x}{\sin \left(x+\pi/3 \right)} \implies \tan x = \dfrac{\sqrt{3}y}{2-y} \implies dx = \dfrac{\sqrt{3}}{2} \dfrac{1}{y^2 - y +1}dy .

So,

0π/3ln2(sinxsin(π3+x))dx=3201ln2(y)y2y+1dy=3201(1+y)ln2(y)1+y3dy\begin{aligned} \int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx &= \dfrac{\sqrt{3}}{2} \int_{0}^{1} \dfrac{\ln ^2 (y)}{y^2 - y +1} dy \\ &= \dfrac{\sqrt{3}}{2} \int_{0}^{1} \dfrac{(1+y)\ln ^2 (y)}{1+y^3} dy \end{aligned}

Since, 0<y<10 < y < 1, we can use the expansion 11+y3=1y3+y6+=k=0(1)ky3k\dfrac{1}{1+y^3} = 1-y^3 + y^6 + \ldots = \sum_{k=0}^{\infty} (-1)^k y^{3k}.

So, the integral becomes,

01(1+y)ln2(y)1+y3dy=01(1+y)ln2(y)k=0(1)ky3kdy=k=0(1)k[01y3kln2(y)dy+01y3k+1ln2(y)dy]=2k=0(1)k(3k+1)3+2k=0(1)k(3k+2)3\begin{aligned} \int_{0}^{1} \dfrac{(1+y)\ln ^2 (y)}{1+y^3} dy &= \int_{0}^{1} (1+y)\ln ^2 (y) \sum_{k=0}^{\infty} (-1)^k y^{3k} dy \\ &= \sum_{k=0}^{\infty} (-1)^k \Bigr[ \int_{0}^{1} y^{3k} \ln^2 (y) dy + \int_{0}^{1} y^{3k+1} \ln^2 (y) dy \Bigr] \\ &= 2 \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + 2 \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \end{aligned}

So,

0π/3ln2(sinxsin(π3+x))dx=3[k=0(1)k(3k+1)3+k=0(1)k(3k+2)3]\int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx = \sqrt{3} \Bigr[ \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \Bigr]

Here comes the tricky part,

Consider the series (0<q<1)(0< q< 1)

1q31(1+q)3+1(2+q)3\dfrac{1}{q^3} - \dfrac{1}{(1+q)^3} + \dfrac{1}{(2+q)^3} - \ldots

It is easy to prove that above series is equal to 18(ζ(3,q2)ζ(3,q+12))\dfrac{1}{8} \left( \zeta \left(3, \dfrac{q}{2} \right) - \zeta \left(3, \dfrac{q+1}{2} \right) \right), where ζ(s,q)\zeta(s,q) is Hurwitz Zeta Function.

Using the relation between Polygamma function and Hurwitz Zeta Function, We get

1q31(1+q)3+1(2+q)3=116(ψ(2)(q+12)ψ(2)(q2))\dfrac{1}{q^3} - \dfrac{1}{(1+q)^3} + \dfrac{1}{(2+q)^3} - \ldots = \dfrac{1}{16} \left(\psi_{(2)} \left(\dfrac{q+1}{2} \right) - \psi_{(2)} \left(\dfrac{q}{2} \right) \right)

So, finally

k=0(1)k(3k+1)3+k=0(1)k(3k+2)3=127(k=0(1)k(k+1/3)3+k=0(1)k(k+2/3)3)=127116(ψ(2)(16)ψ(2)(23)+ψ(2)(13)ψ(2)(56))=53π3243\begin{aligned} \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} &= \dfrac{1}{27} \left( \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1/3)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+2/3)^3} \right) \\ &= \dfrac{1}{27} \dfrac{1}{16} \left(\psi_{(2)} \left(\dfrac{1}{6}\right)-\psi_{(2)}\left(\dfrac{2}{3}\right)+\psi_{(2)}\left(\dfrac{1}{3}\right)-\psi_{(2)}\left(\dfrac{5}{6}\right) \right) = \dfrac{5 \sqrt{3} \pi^3}{243} \end{aligned}

Above final calculation is done using Euler's Reflection Formula for polygamma function

So, finally

0π/3ln2(sinxsin(π3+x))dx=3[k=0(1)k(3k+1)3+k=0(1)k(3k+2)3]=353π3243=5π381 \int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx = \sqrt{3} \Bigr[ \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \Bigr] = \sqrt{3} \dfrac{5 \sqrt{3}\pi^3}{243} = \dfrac{5 \pi^3}{81}

Sorry for missing the calculation part, As it became too lengthy to type here.

Notify me if there are any typing mistakes.

Surya Prakash - 4 years, 2 months ago

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Thanks for everyone. Because of this competition and you guys I learnt a lot of integrations skills and develop my integration skills so fast. I thank everyone. @Pi Han Goh @Aditya Kumar @Tanishq Varshney @Sudeep Salgia @Abhishek Bakshi

Surya Prakash - 4 years, 2 months ago

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Problem 18:

Evaluate

0log2(t)1+t2dt\large \int_{0}^{\infty} \dfrac{\log^2 (t)}{1+t^2} \, dt

This problem has been solved by Tanishq Varshney (first) and Sudeep Salgia (second) almost at the same time.

Surya Prakash - 4 years, 2 months ago

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Solution to problem 18\text{Solution to problem 18}

01ln2(t)1+t2dt+1ln2(t)1+t2dt\large{\displaystyle \int^{1}_{0} \frac{\ln^{2} (t)}{1+t^2} dt+\displaystyle \int^{\infty}_{1} \frac{\ln^{2} (t)}{1+t^2} dt}

In the latter integral t1tt\rightarrow \frac{1}{t}

The whole expression becomes

201ln2(t)1+t2dt\large{2 \displaystyle \int^{1}_{0} \frac{\ln^{2} (t)}{1+t^2} dt}

Now

01xadx=1a+101sxsxadx=01xalnsxdx=(1)ss!(a+1)s+1\large{\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } dx=\frac { 1 }{ a+1 } \\ \displaystyle \int _{ 0 }^{ 1 }{ \frac { { \partial }^{ s } }{ { \partial x }^{ s } } } { x }^{ a }\quad dx=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } { \ln { ^{ s } } { x } }dx=\frac { { \left( -1 \right) }^{ s }s! }{ { \left( a+1 \right) }^{ s+1 } } }

we have

11+t2=r=0(1)rt2r\large{\frac{1}{1+t^2}=\displaystyle \sum^{\infty}_{r=0} (-1)^{r} t^{2r}}

201r=0ln2(t)(1)rt2rdt\large{\longrightarrow 2 \displaystyle \int^{1}_{0} \displaystyle \sum^{\infty}_{r=0} \ln^{2} (t) (-1)^{r} t^{2r} dt}

2r=0(1)r01t2rln2(t)dt\large{\longrightarrow 2 \displaystyle \sum^{\infty}_{r=0} (-1)^{r} \displaystyle \int^{1}_{0} t^{2r} \ln^{2} (t) dt}

4r=0(1)r(2r+1)3\large{4 \displaystyle \sum^{\infty}_{r=0} \frac{(-1)^{r}}{(2r+1)^{3}}}

Now for the answer

Use of Dirichlet Beta function

which gives 4β(3)=π38\large{4 \beta(3)=\frac{\pi^{3}}{8}}

Tanishq Varshney - 4 years, 2 months ago

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Problem 17:

Prove that

(x2x4x3+x2x+1dx)÷(xx4x3+x2x+1dx)=3+52.\large \left(\int_{-\infty}^\infty \dfrac{x^2}{x^4 - x^3+x^2-x+1} \, dx \right) \div \left(\int_{-\infty}^\infty \dfrac{x}{x^4 - x^3+x^2-x+1} \, dx \right) = \dfrac{3+\sqrt5}2.

This problem has been solved by Surya Prakesh.

Pi Han Goh - 4 years, 2 months ago

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x2x4x3+x2x+1dx=x3+x2x5+1dx=0x3+x2x5+1dx+0x3+x2x5+1dx=0x2x31x5dx+0x3+x2x5+1dx=20x2x81x10dx\begin{aligned} \int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx &=\int_{-\infty}^{\infty} \dfrac{x^3 + x^2}{x^5+1}dx \\ &= \int_{-\infty}^{0} \dfrac{x^3 + x^2}{x^5+1}dx + \int_{0}^{\infty} \dfrac{x^3 + x^2}{x^5+1} dx \\ &= \int_{0}^{\infty} \dfrac{x^2 - x^3}{1-x^5}dx + \int_{0}^{\infty} \dfrac{x^3 + x^2}{x^5+1} dx \\ &= 2 \int_{0}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx \end{aligned}

0x2x81x10dx=01x2x81x10dx+1x2x81x10dx=011+x2x6x81x10dx\begin{aligned} \int_{0}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx &= \int_{0}^{1} \dfrac{x^2 - x^8}{1-x^{10}} dx + \int_{1}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx \\ &= \int_{0}^{1} \dfrac{1+ x^2 - x^6 - x^8}{1-x^{10}}dx \end{aligned}