Brilliant Junior Integration Contest - Season 1 - Rajdeep Dhingra

Hi Brilliant! Just like what Aditya Kumar and Anastasiya Romanova conducted earlier, this year we would also like to conduct an integration contest for juniors.

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.

Eligibility:- People should fulfill either of the 2 following

17 years or below
Level 4 or below in Calculus

Eligible people 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.
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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.
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, digamma function,Harmonic numbers, trigonometric integral, Wallis' integral,

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

**[Post your problem here]**

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Problem 18:

Prove that: $\int_{0}^{1} \left(\frac{1}{\log x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} = -\frac{1}{2}+\frac{1}{2} \ln (2 \pi)-\frac{1}{2} \gamma.$

The Problem has been solved by Ishan Singh

Aditya Kumar - 3 years, 7 months ago

Lemma 1 : $\displaystyle \sum_{r=1}^{n} H_{r} = (n+1)H_{n} - n$

Proof : $\displaystyle \sum_{r=1}^{n} H_{r} = \int_{0}^{1} \sum_{r=1}^{n} \dfrac{x^r-1}{x-1} \mathrm{d}x$

$\displaystyle = \int_{0}^{1} \dfrac{x^{n+1} - x - nx + n}{(x-1)^2} \mathrm{d}x$

$\displaystyle = \int_{0}^{1} \dfrac{x^{n+1} - x}{(x-1)^2} \mathrm{d}x - \int_{0}^{1} \dfrac{n}{x-1} \mathrm{d}x$

$\displaystyle = \text{A} + \text{B}$

Using Integration By Parts on the first integral $\text{A}$ , we have,

$\displaystyle \text{A} = -\left[\dfrac{x^{n+1}-x}{x-1}\right]_{0}^{1} + \int_{0}^{1}\dfrac{(n+1)x^n - 1}{x-1} \mathrm{d}x$

$\displaystyle = -n + \int_{0}^{1}\dfrac{(n+1)x^n - 1}{x-1} \mathrm{d}x$

$\displaystyle \implies \text{A} +\text{B} = -n + (n+1) \cdot \int_{0}^{1}\dfrac{x^n - 1}{x-1} \mathrm{d}x$

$\displaystyle = (n+1)H_{n} - n$

Lemma 2 : $\displaystyle \int_{0}^{1} \dfrac{x^n - 1}{\ln x} \mathrm{d}x = \ln (n +1)$

Proof : $\displaystyle \int_{0}^{1} \dfrac{x^n - 1}{\ln x} \ \mathrm{d}x = \int_{0}^{1} \int_{0}^{n} x^y \ \mathrm{d}y \ \mathrm{d}x$

$\displaystyle = \int_{0}^{n} \int_{0}^{1} x^y \ \mathrm{d}x \ \mathrm{d}y$

$\displaystyle = \int_{0}^{n} \dfrac{1}{y+1} \ \mathrm{d}y$

$\displaystyle = \ln (n+1)$

Lemma 3 : $\displaystyle \int_{0}^{1} \left(\dfrac{1}{\ln (x)} + \dfrac{1}{1-x}\right) \mathrm{d}x = \gamma$

Proof : $\displaystyle \int_{0}^{1} \left( \dfrac{1}{1-x} + \dfrac{1}{\ln (x)}\right) \mathrm{d}x = \lim_{n \to \infty} \int_0^1\left(\dfrac{1-t^n}{1-t}-\frac{t^{n-1}-1}{\ln t} \mathrm{d}t \right)$

$\displaystyle = \lim _{n \to \infty} (H_{n} - \ln n)$

$\displaystyle = \gamma$

Now,

$\displaystyle \text{I} = \int_{0}^{1} \left(\dfrac{1}{\ln x} + \dfrac{1}{1-x} - \dfrac{1}{2}\right)\dfrac{\mathrm{d}x}{x-1}$

$\displaystyle = \lim_{n \to \infty} \sum_{r=1}^{n} \int_{0}^{1} \left(\dfrac{x^{r-1}}{\ln x} + \dfrac{x^{r-1}}{1-x} - \dfrac{x^{r-1}}{2} \right)$

Using Lemma 2 and Lemma 3 , we have,

$\displaystyle \text{I} = \lim_{n \to \infty} \sum_{r=1}^{n} \left(\gamma + \ln r - H_{r-1} -\dfrac{1}{2r} \right)$

$\displaystyle = \lim_{n \to \infty} \left( n\gamma + \ln (n!) - \sum_{r=1}^{n} H_{r-1} - \dfrac{H_{n}}{2} \right)$

Using Lemma 1,

$\displaystyle \text{I} = \lim_{n \to \infty} \left( n\gamma + \ln (n!) -n H_{n-1} + n - 1 -\dfrac{H_{n}}{2} \right)$

Using Stirling's Approximation, we have,

$\displaystyle \text{I} = \lim_{n \to \infty} \left( n\gamma + \dfrac{1}{2} \ln (2\pi) - n +\dfrac{1}{2} \ln (n) + n\ln (n) -n H_{n-1} + n - 1 -\dfrac{H_{n}}{2} \right)$

Simplifying using $\displaystyle \lim_{n \to \infty} (H_{n} - \ln n) = \gamma$ , we have,

$\displaystyle \text{I} = \boxed{\dfrac{1}{2} \ln (2\pi) -\dfrac{1}{2} \gamma - \dfrac{1}{2}}$

Ishan Singh - 3 years, 7 months ago

wow,that was probably why i gave up halfway through lol

i'm sorry i didn't attend the integration contest until now,there was a problem with my wifi and it got fixed

Hamza A - 3 years, 7 months ago

Next problem please.

Harsh Shrivastava - 3 years, 7 months ago

Nice. I used Binet's formula.

Let us start with an easy Problem.

Problem 1

Find the value of the following $\Large \displaystyle \dfrac{\displaystyle \int_{-\infty}^{\infty}{{(x+1)}^2 e^{-{(x+1)}^2}dx}}{\displaystyle \int_{-\infty}^{\infty}{e^{-x^2}dx}}$

This problem has been solved by Aditya Kumar.

Rajdeep Dhingra - 3 years, 7 months ago

Problem 2:

Prove that: $\int _{ 0 }^{ 1 }{ \frac { \arctan { \left( \sqrt { { x }^{ 2 }+2 } \right) } }{ \sqrt { { x }^{ 2 }+2 } \left( { x }^{ 2 }+1 \right) } \ dx } =\frac { 5{ \pi }^{ 2 } }{ 96 }$

This problem has been solved by Harsh Shrivastava.

Solution to Problem 2

$I = \int _{ 0 }^{ 1 }{ \frac { \arctan { \left( \sqrt { { x }^{ 2 }+2 } \right) } }{ \sqrt { { x }^{ 2 }+2 } \left( { x }^{ 2 }+1 \right) } \ dx } =\frac { 5{ \pi }^{ 2 } }{ 96 }$

Lets split $I$ as $A \text{ and } B$ by using the property that $\arctan(x) = \pi /2 -\arctan(1/x)$

Thus our integral becomes $I = A-B$ ,

where $A = \displaystyle \pi /2\int_{0}^{1}\dfrac{dx}{(1+x^{2})(\sqrt{2+x^{2}})}$

and $B= \displaystyle \int_{0}^{1}\dfrac{\arctan(\frac{1}{\sqrt{x^{2} + 2}})}{(1+x^{2})(\sqrt{2+x^{2}})}$

Evaluating A,

Let's evaluate A, without the limits,

$\displaystyle \pi /2\int\dfrac{dx}{(1+x^{2})(\sqrt{2+x^{2}})} = \pi /2\arctan \dfrac{x}{\sqrt{x^{2} + 2}}$

Which on substituting limits gives A = $\pi ^{2}/12$

Evaluating B,

Using $1/a \arctan(1/a) = \int_{0}^{a} \frac{dy}{y^{2}+a^{2}}$ ,

Let $a = \sqrt{x^2+2}$ .

We get $B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(2+x^{2} + y^{2})}$

$B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(1+y^{2})} - \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(2+x^{2} + y^{2})}$

$2B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(1+y^{2})}$

$2B = \displaystyle (\int_{0}^{1}\dfrac{dx}{1+x^{2}})^{2}$

Thus $B = \pi^{2} / 32$

Hence $I = A-B=\dfrac{5\pi^{2}}{96}$

Generalization

Problem 14:

If $\displaystyle \int_{0}^{\infty} \dfrac{x^{3}}{1+e^{\alpha x}}dx = \dfrac{1}{64}$
Find $\alpha$ given that , $\alpha$ is a positive constant and $\zeta(4) = \dfrac{\pi^{4}}{90}$

Problem has been solved by Rajdeep Dhingra and Aareyan Manzoor at the same instant. Aareyan has been given credits due to his solution.

A Former Brilliant Member - 3 years, 7 months ago

$I=\int_0^\infty x^3 \sum_{n=1}^\infty (-e^{-\alpha x})^n dx$ $I= \sum_{n=1}^\infty (-1)^n \int_0^\infty x^3 e^{-\alpha n x}dx=\sum_{n=1}^\infty (-1)^n \dfrac{6}{(\alpha n)^4}$ $I=\dfrac{6}{\alpha^4} (1-2^{-3}) \zeta(4)=\dfrac{1}{64}$ the result follows.

Aareyan Manzoor - 3 years, 7 months ago

Is the answer $\huge \displaystyle \alpha = \pi {\left(\frac{56}{15}\right)}^{\frac14}$ ?

Problem 6:

Find a closed form of the indefinite integral $\int H_{x} dx$

Where $H_x$ denotes harmonic number

Solution to Problem 6:

I'll use the definition: ${ H }_{ x }=\psi \left( x+1 \right) +\gamma$

$\int { { H }_{ x }dx } =\int { \left( \psi \left( x+1 \right) +\gamma \right) dx }$

We know that $\displaystyle \psi \left( x \right) =\frac { d\left( x! \right) }{ dx }$ . Hence, using this, we get: $\boxed{\displaystyle \int { { H }_{ x }dx } =ln\left( x! \right) +\gamma x+c}$

Problem 10

Let $S = \sin^2{(x)} + \dfrac{sin^4{(x)}}{3^2} + \dfrac{sin^6{(x)}}{5^2} + \cdots \text{ Upto Infinity }$

Prove that $\int_{0}^{\pi /2} {S \ dx} = \dfrac{{\pi}^2}{4} - \dfrac{\pi}2$

The Problem has been solved by Ishan Singh.

I have come up with several ways to prove it. Here is one of the more interesting methods.

Lemma: $\displaystyle f(a) = \dfrac{1}{1+a} + \sum_{r=1}^{\infty} \left[ \dfrac{1}{(2r+a+1)}\prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right] = \dfrac{\pi}{2^{a+1}} \dfrac{\Gamma (a+1)}{\Gamma^2\left(\dfrac{a+2}{2}\right)} \quad ; \quad a > -1$

Proof: I have proved it here

$\displaystyle \text{I} = \int_{0}^{\pi /2} {S \ \mathrm{d}x} = \sum_{r=1}^{\infty}\int_{0}^1 \dfrac{\sin ^{2r}}{(2r-1)^2} \mathrm{d}x$

$\displaystyle = \dfrac{1}{2} \sum_{r=1}^{\infty} \dfrac{\operatorname{B}\left(r+\frac{1}{2} , \frac{1}{2} \right)}{(2r-1)^2}$

$\displaystyle = \dfrac{1}{2} \sum_{r=1}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right) \Gamma \left(\frac{1}{2} \right)}{(2r-1)^2 \Gamma(r+1)}$

$\displaystyle = \dfrac{\pi}{2} \sum_{r=1}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right) }{(2r-1)^2 \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}$

Changing the index of summation and using $\Gamma(x+1) = x\Gamma(x)$ , we have,

$\displaystyle \text{I} = \dfrac{\pi}{4} \sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(r+1)(2r+1) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}$

$\displaystyle = \dfrac{\pi}{2} \left[\sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(2r+1) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)} - \sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(2r+2) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}\right]$

Using my solution here, we have,

$\displaystyle \text{I} = \dfrac{\pi}{2} \left[ 1+ \sum_{r=1}^{\infty} \dfrac{1}{(2r+1)} \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) - \dfrac{1}{2} - \sum_{r=1}^{\infty} \dfrac{1}{(2r+2)} \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right]$

$\displaystyle = \dfrac{\pi}{2} \left[f(0) - f(1) \right]$

$\displaystyle = \boxed{\dfrac{\pi ^2}{4} - \dfrac{\pi}{2}}$

What was your intended solution?

Samuel Jones - 3 years, 7 months ago

More or less the same.

@Ishan Singh @Rajdeep Dhingra Do any of you know another method to solve this question?

Ishan wrote that he has come up with several ways so I guess has many. @Ishan Singh

@Rajdeep Dhingra – My other methods include using Dilogarithm and derivative of Beta function.

Note to the Participants:

Please refrain from posting solution through images. It makes this note slow to load.
Up-vote good solutions and problems. I have see that people are not up-voting at all. This is a wrong sign in the contest. Up-voting encourages the users to improve.

Very True.I will change the note after it crosses 50 questions.

Maybe you should do it now, my browser has hanged twice loading this note!

Akshay Yadav - 3 years, 7 months ago

Change it after 20. 50 is too much.

Problem 17

Prove that if $c < 1$ , $\int_{0}^{\pi/2} {\arcsin{(c \cos{(x)})} \ dx} = \frac{c}{1^2} + \frac{c^3}{3^2} + \frac{c^5}{5^2} + \cdots \text{ (Upto Infinity)}$

The Problem has been solved by Aditya Kumar

Solution to Problem 17:

Maclaurin series of $\arcsin { \left( x \right) }$ is: $\displaystyle \arcsin { \left( x \right) } =\sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) }{ \sqrt { \pi } \left( 2n+1 \right) n! } { x }^{ 2n+1 } }$

Here $x=c \cos(x)$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) }{ \sqrt { \pi } \left( 2n+1 \right) n! } { \left( c\cos { \left( x \right) } \right) }^{ 2n+1 } } dx }$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) { c }^{ 2n+1 } }{ \sqrt { \pi } \left( 2n+1 \right) n! } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( \cos { \left( x \right) } \right) }^{ 2n+1 }dx } }$

Therefore, by beta function we get:

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\sum _{ n=0 }^{ \infty }{ \left( \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) { c }^{ 2n+1 } }{ \sqrt { \pi } \left( 2n+1 \right) n! }\frac { 1 }{ 2 } \frac { \Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( n+1 \right) }{ \Gamma \left( n+\frac { 3 }{ 2 } \right) } \right) }$

Hence, on simplifying, we get: $\boxed{ \displaystyle \int_{0}^{\pi/2} {\arcsin{(c \cos{(x)})} \ dx} = \frac{c}{1^2} + \frac{c^3}{3^2} + \frac{c^5}{5^2} + \cdots \text{ (Upto Infinity)}}$

Problem 9

Prove $\int_0^{\pi/2} x\ln(\tan(x)) dx= \dfrac{7\zeta(3)}{8}$

The Problem has been solved by Rajdeep Dhingra.

Solution to Problem 9

Page 1

Page 2

The second page is not at all clear. Please LaTeX your solution.

Problem 23

Prove that for an integer $n\ge2$

$\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx=\frac { n }{ n-1 } -\zeta (n)$

hint:let $x=\frac { 1 }{ u^{ n } }$

Let $\displaystyle \text{I} = \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx$

Substitute $\displaystyle x = \frac{1}{t^n}$

$\displaystyle \implies \text{I} = n \int_{1}^{\infty} \dfrac{ \left \{t \right \} }{t^{n+1}} dt$

$\displaystyle = n \sum_{r=1}^{\infty} \int_{r}^{r+1} \dfrac{t-r}{t^{n+1}} dt$

$\displaystyle = n \sum_{r=1}^{\infty} \left[ \dfrac{t^{1-n}}{1-n} +r \dfrac{t^{-n}}{n} \right]_{r}^{r+1}$

$\displaystyle = -n \sum_{r=1}^{\infty} \left( \dfrac{1}{n(n+1) (r+1)^{n-1}} + \dfrac{1}{n(r+1)^n} \right)$

$\displaystyle = \dfrac{n}{n-1} - \zeta(n)$

Does the Curly brackets mean fractional part ?

It is pointless to continue the contest. It hasn't garnered attention like the previous original integration contests.

@Samuel Jones – No Problem. Still it is open. Please post the next question if you want or if don't want to post, then @Hummus a will post.

Problem 3:

Prove that $\displaystyle \int_{0}^{1} \dfrac{(\ln(1+x) - \ln(1-x))}{x(\sqrt{1-x^{2}})}dx=\frac{\pi^{2}}{2}$

This problem has been solved by Rajdeep Dhingra.

Problem 8

Prove that $\int_{0}^{2 \pi} { e^{\cos{(\theta)}} \cos{(\sin{(\theta)})} \ d{\theta}} = 2\pi$

Not Original

The Problem has been solved by Aareyan Manzoor.

solution to problem 8

Considering the function $f(t) = \int_{0}^{2\pi} e^{t \cos(\theta)} \cos(t\sin(\theta)) \, d\theta$ , we see that $\begin{array}{rcl} tf'(t) & = & \displaystyle t \int_0^{2\pi} e^{t\cos\theta}\big[\cos\theta \cos(t\sin\theta) - \sin\theta\sin(t\sin\theta)\big]\,d\theta \\ & = & \displaystyle \int_0^{2\pi} \frac{\partial}{\partial \theta}\big[e^{t\cos\theta} \sin(t\sin\theta)\big]\,d\theta \\ & = & \Big[ e^{t\cos\theta}\sin(t\sin\theta)\Big]_0^{2\pi} \; = \; 0\\ &&f'(t)=0\to f(t)=C \end{array}$ Since $f(0) = 2\pi$ , the desired integral is $f(1) = \boxed{2\pi}$ .

Problem 11

Prove that $\int_{0}^{\pi/2} { {\left( \dfrac{x}{\sin{(x)}} \right)}^2 \ dx} = \pi \log{(2)}$

The Problem has been solved Aareyan Manzoor.

solution to problem 11 $\int uv'=uv-\int u'v$ $I=\int_0^{\pi/2} x^2 \csc^2(x) dx$ $u=x^2\to u'=2x, v'=\csc^2 x\to v=-cot(x)$ $I=-x^2\cot(x)\mid_0^{\pi/2} +2\int_0^{\pi/2} x \cot(x) dx=2\int_0^{\pi/2} x \cot(x) dx$ $u=x\to u'=1, v'=\cot(x)\to v=-\ln(sin(x))$ $I=2x\ln(\sin(x))\mid_0^{\pi/2}+2\int_0^{\pi/2} \ln(\sin(x)) dx=\int_0^{\pi/2} \ln(\sin^2(x)) dx$ $B(a,1/2)=\int_0^{\pi/2} 2\sin^{2a-1}(x) dx$ $B'(a,1/2)=\int_0^{\pi/2} 2\ln(\sin^2(x))\sin^{2a-1}(x) dx$ $B'(a,1/2)=B(a,1/2)(\psi(a)-\psi(a+1/2))$ $a=1/2$ $\int_0^{\pi/2} 2\ln(\sin^2(x))dx=B(1/2,1/2)(\psi(1/2)-\psi(1))=\pi(-\gamma-(-\gamma-2\ln(2)))=2\pi\ln(2)$ $I=\int_0^{\pi/2} \ln(\sin^2(x)) dx=\boxed{\pi\ln(2)}$

Problem 22 :

Prove That

$\int_{0}^{\infty} \dfrac{\sin \left(\frac{x^2 + 1}{2x} \right) \cos \left(\frac{x^2 - 1}{2x} \right) }{x} \mathrm{d}x = \dfrac{\pi}{2}$

Solution to Problem 22:

Use the formula $2\sin { \left( A \right) } \cos { \left( B \right) } =\sin { \left( A+B \right) } -\sin { \left( A-B \right) }$

Therefore, we get: $I=\frac { 1 }{ 2 } \left\{ \int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } +\int _{ 0 }^{ \infty }{ \frac { \sin { \left( \frac { 1 }{ x } \right) } }{ x } dx } \right\}$

It is easy to see that both the integrals are same. Hence, $I=\int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } =\frac{\pi}{2}$

Problem 25

Let $k$ be a positive real number

then prove

$\displaystyle\int _{0}^{1}{\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } =\frac { 2k+1 }{ (k+1)^{ 2 } } -\frac { \gamma }{ k+1 }}$

Hint:let $t=\frac{x^k}{y}$

Solution to Problem 25:

$I=\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } }$

$I=\int _{ 0 }^{ 1 }{ \left( \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dy } \right) dx }$

By IBP, we get: $I={ \left( \frac { { x }^{ k+1 } }{ k+1 } \int _{ { x }^{ k } }^{ \infty }{ \frac { \left\{ t \right\} }{ { t }^{ 2 } } dt } \right) }_{ 0 }^{ 1 }+\frac { k }{ k+1 } \int _{ 0 }^{ 1 }{ { x }^{ k }dx }$

Hence on simplifying, we get: $\displaystyle\int _{0}^{1}{\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } =\frac { 2k+1 }{ (k+1)^{ 2 } } -\frac { \gamma }{ k+1 }}$

Problem 27

Evaluate

$\displaystyle\int _{0}^{1}{\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x+y } \right\} ^{ n }dxdy }}$

with $n\in\mathbb{N}$

Hint:let $x+y=t$

Problem 4

Prove $\int_{0}^{\infty} { \dfrac{\sqrt{x}}{(1+x)(2+x)(3+x)} \ dx} = \dfrac{\pi}2 \left(- 2\sqrt{2} - 1 + \sqrt{3}\right)$

Problem 5:

$\displaystyle \int_{0}^{1} \dfrac{\ln (1+x)}{x}$

This problem was first solved by Aaryen Manzoor using a disallowed method and then solved by Aditya Kumar using a legal method. Credit is still given to Aaryen.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$