# 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).

Format your post is as follows:

 1 2 3 4 5 6 7 **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

$\large{\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\ln(\sin x) \ln(\cos x)}{\tan x} dx=\frac{1}{8} \zeta (3)}.$

###### This problem has been solved by Julian Poon.

- 4 years, 2 months ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

$\displaystyle \frac { \Gamma (a)\Gamma (b) }{ \Gamma (a+b) } = 2\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2a-1 }{ x } \cos ^{ 2b-1 }{ x } dx }$

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

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

Putting a = 0 and b = 1,

$-\psi ' (1) = 8\int^{\frac{\pi}{2}}_{0}\frac{\ln(\sin x) \ln(\cos x)}{\tan x} dx$

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Hint: What is the definition of a digamma function?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Go here

- 4 years, 1 month ago

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

- 4 years, 1 month ago

How's the registration coming along?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

- 4 years, 1 month ago

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

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

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

$-\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \ln { t } \sum _{ n=1 }^{ \infty } \frac { t^{ 2n-1 } }{ n } } dt=-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \int _{ 0 }^{ 1 }{ \ln { (t) } t^{ 2n-1 } } dt$

Finally, integrate by parts to obtain

$-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \frac { -1 }{ 4{ n }^{ 2 } } =\frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ { n }^{ 3 } }$

- 4 years, 2 months ago

Problem 24:

Prove that

$\large \int_0^{\pi /2} \dfrac1{\sin^8 x + \cos^8 x } \, dx = \dfrac{\sqrt{10-\sqrt2}}{2} \pi .$

###### This problem has been solved by Tanishq Varshney.

- 4 years, 2 months ago

$\text{Solution to Problem 24}$

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

Refer to problem 18 for this step.

Now in the integral $\tan x \rightarrow t$

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

$\large{t^8 \rightarrow z}$

$\large{\frac{1}{8} \left(\displaystyle \int ^{\infty}_{0} \frac{z^{\frac{7}{8}-1}}{1+z}+ \frac{3z^{\frac{5}{8}-1}}{1+z}+ \frac{3z^{\frac{3}{8}-1}}{1+z}+ \frac{z^{\frac{1}{8}-1}}{1+z} dz \right)}$

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

$\large{=\frac{\sqrt{10-\sqrt{2}}}{2} \pi}$

- 4 years, 2 months ago

Problem 23:

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

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

- 4 years, 2 months ago

Solution to Problem 23:

Substitute: $x\longrightarrow \frac { 1-x }{ 1+x }$

$I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}$

On separating the integrands,

$\ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2)$

1st integral:

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

2nd integral:

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

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,

$J=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx$

Now on using the formula:

$\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right)$

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

$\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})$

Using:

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

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

$\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right)$,

and

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

we have $\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)$

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

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

$...$

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

- 4 years, 2 months ago

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:

- 4 years, 2 months ago

Problem 22:

Evaluate the integral

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

###### This problem has been solved by Aditya Kumar.

- 4 years, 2 months ago

$I=\int _{ 0 }^{ \pi }{ \frac { xsinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ I=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ \frac { sinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ cosx=t\quad \\ I=\frac { \pi }{ 2 } \int _{ -1 }^{ 1 }{ \frac { 1 }{ { \left( { t }^{ 2 }+3 \right) }^{ 2 } } dt } \\ t=\sqrt { 3 } tanx\\ I=\frac { \pi \sqrt { 3 } }{ 36 } \int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \left( cos2x+1 \right) dx } \\ \therefore I=\frac { \pi \sqrt { 3 } }{ 18 } \left\{ \frac { \sqrt { 3 } }{ 4 } +\frac { \pi }{ 6 } \right\}$

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

- 4 years, 2 months ago

Problem 21:

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

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

###### This problem has been solved by Surya Prakesh.

- 4 years, 2 months ago

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

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

\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 $\arctan(x) + \arctan(1/x) = \pi/2$.

So,

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

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

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

- 4 years, 2 months ago

Problem 20:

Evaluate

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

###### This problem has been solved by Gautam Sharma.

- 4 years, 2 months ago

I used complex analysis

- 1 year, 3 months ago

Solution to problem 20

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

By partial fractions

$I=\int \frac { 2-t }{ 3({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ t }^{ 2 }-t+1) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ (t-\frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } ) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2t-1 }{ \sqrt { 3 } } } -\frac { 1 }{ 6 } \ln({ t }^{ 2 }-t+1)+\frac { 1 }{ 3 } \ln(t+1)+c \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2x+1 }{ \sqrt { 3 } } } +\frac { 1 }{ 6 } \ln\left(\frac { { (x+2) }^{ 2 } }{ { x }^{ 2 }+x+1 } \right)+c$

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

- 4 years, 2 months ago

Problem 19:

Prove that

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

###### This problem has been solved by Surya Prakesh.

- 4 years, 2 months ago

Alternate Solution to problem 19 :

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

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

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

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

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

$\displaystyle I = \dfrac{1}{n}\int _{ 0 }^{ 1 }{ { y }^{ -1/n }{ (1-y) }^{ 1/n-1 }dy } =\dfrac{1}{n}\frac { \Gamma (1/n)\Gamma (1-1/n) }{ \Gamma (1) } =\frac { \pi }{ n } \csc { (\frac { \pi }{ n } ) }$

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

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

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

Use the identity to get :

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

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

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

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

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

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

Adding these two forms we have :

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

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

Use the lemma to get :

$\displaystyle I = \frac { \sqrt { 3 } }{ 4 } ({ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } )+{ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } ))=\frac { 5{ \pi }^{ 3 } }{ 81 }$

- 4 years, 2 months ago

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

Let $y=\dfrac{\sin x}{\sin \left(x+\pi/3 \right)} \implies \tan x = \dfrac{\sqrt{3}y}{2-y} \implies dx = \dfrac{\sqrt{3}}{2} \dfrac{1}{y^2 - y +1}dy$.

So,

\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, $0 < y < 1$, we can use the expansion $\dfrac{1}{1+y^3} = 1-y^3 + y^6 + \ldots = \sum_{k=0}^{\infty} (-1)^k y^{3k}$.

So, the integral becomes,

\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,

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

Here comes the tricky part,

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

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

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

$\dfrac{1}{q^3} - \dfrac{1}{(1+q)^3} + \dfrac{1}{(2+q)^3} - \ldots = \dfrac{1}{16} \left(\psi_{(2)} \left(\dfrac{q+1}{2} \right) - \psi_{(2)} \left(\dfrac{q}{2} \right) \right)$

So, finally

\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

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

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

Notify me if there are any typing mistakes.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

Problem 18:

Evaluate

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

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

- 4 years, 2 months ago

$\text{Solution to problem 18}$

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

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

The whole expression becomes

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

Now

$\large{\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } dx=\frac { 1 }{ a+1 } \\ \displaystyle \int _{ 0 }^{ 1 }{ \frac { { \partial }^{ s } }{ { \partial x }^{ s } } } { x }^{ a }\quad dx=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } { \ln { ^{ s } } { x } }dx=\frac { { \left( -1 \right) }^{ s }s! }{ { \left( a+1 \right) }^{ s+1 } } }$

we have

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

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

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

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

Now for the answer

Use of Dirichlet Beta function

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

- 4 years, 2 months ago

Problem 17:

Prove that

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

###### This problem has been solved by Surya Prakesh.

- 4 years, 2 months ago

\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}
\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}