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Brilliant Junior Integration Contest - Season 1

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.

  • 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.

  • 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,

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":

Note by Rajdeep Dhingra
6 months, 1 week ago

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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 · 6 months, 1 week ago

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@Aditya Kumar 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 · 6 months, 1 week ago

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@Ishan Singh Nice. I used Binet's formula. Aditya Kumar · 6 months, 1 week ago

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@Ishan Singh Next problem please. Harsh Shrivastava · 6 months, 1 week ago

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@Harsh Shrivastava @Aditya Kumar May post another question if he wants. Ishan Singh · 6 months, 1 week ago

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@Ishan Singh 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 Hummus A · 6 months, 1 week ago

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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 · 6 months, 1 week ago

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@Rajdeep Dhingra Solution to Problem 1: \[{ I }_{ 1 }=\int _{ -\infty }^{ \infty }{ { \left( x+1 \right) }^{ 2 }{ e }^{ { -\left( x+1 \right) }^{ 2 } } } dx\]

Substitute: \({ \left( x+1 \right) }^{ 2 }=t\)

Therefore, we get: \[{ I }_{ 1 }=\frac { 1 }{ 2 }\int _{ -\infty }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt\]

This is symmetric about \(x=0\). Therefore, we can write it as: \[{ I }_{ 1 }=\int _{ 0 }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt\]

\[{ I }_{ 1 }=\Gamma \left( \frac { 3 }{ 2 } \right) =\frac { 1 }{ 2 } \Gamma \left( \frac { 1 }{ 2 } \right) =\frac { \sqrt { \pi } }{ 2 } \]

Similarly, \({ I }_{ 2 }=\sqrt { \pi } \)

Hence, \[\boxed{\frac{{ I }_{ 1 }}{{ I }_{ 2 }}=\frac{1}{2}}\]

My age is 17. Aditya Kumar · 6 months, 1 week ago

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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. Aditya Kumar · 6 months, 1 week ago

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@Aditya Kumar 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 \(I = A-B=\dfrac{5\pi^{2}}{96}\) Harsh Shrivastava · 6 months, 1 week ago

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@Aditya Kumar Generalization Ishan Singh · 6 months, 1 week ago

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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 Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra 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: \[\boxed{ \displaystyle \int_{0}^{\pi/2} {\arcsin{(c \cos{(x)})} \ dx} = \frac{c}{1^2} + \frac{c^3}{3^2} + \frac{c^5}{5^2} + \cdots \text{ (Upto Infinity)}}\] Aditya Kumar · 6 months, 1 week ago

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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. Vighnesh Shenoy · 6 months, 1 week ago

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@Vighnesh Shenoy \[I=\int_0^\infty x^3 \sum_{n=1}^\infty (-e^{-\alpha x})^n dx\] \[I= \sum_{n=1}^\infty (-1)^n \int_0^\infty x^3 e^{-\alpha n x}dx=\sum_{n=1}^\infty (-1)^n \dfrac{6}{(\alpha n)^4}\] \[I=\dfrac{6}{\alpha^4} (1-2^{-3}) \zeta(4)=\dfrac{1}{64}\] the result follows. Aareyan Manzoor · 6 months, 1 week ago

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@Vighnesh Shenoy Is the answer \(\huge \displaystyle \alpha = \pi {\left(\frac{56}{15}\right)}^{\frac14}\) ? Rajdeep Dhingra · 6 months, 1 week ago

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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. Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra 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

Now,

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

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

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

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

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,

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

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

\(\displaystyle = \boxed{\dfrac{\pi ^2}{4} - \dfrac{\pi}{2}}\) Ishan Singh · 6 months, 1 week ago

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@Rajdeep Dhingra What was your intended solution? Samuel Jones · 6 months, 1 week ago

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@Samuel Jones More or less the same. Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra @Ishan Singh @Rajdeep Dhingra Do any of you know another method to solve this question? Samuel Jones · 6 months, 1 week ago

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@Samuel Jones Ishan wrote that he has come up with several ways so I guess has many. @Ishan Singh Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra My other methods include using Dilogarithm and derivative of Beta function. Ishan Singh · 6 months, 1 week ago

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Problem 6:

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

Where \(H_x\) denotes harmonic number

This problem has been solved by Aditya Kumar. Aareyan Manzoor · 6 months, 1 week ago

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@Aareyan Manzoor 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: \[\boxed{\displaystyle \int { { H }_{ x }dx } =ln\left( x! \right) +\gamma x+c}\] Aditya Kumar · 6 months, 1 week ago

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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 \(x=\frac { 1 }{ u^{ n } } \) Hummus A · 6 months ago

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@Hummus A Let \(\displaystyle \text{I} = \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx\)

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

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

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

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

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

\(\displaystyle = \dfrac{n}{n-1} - \zeta(n) \) Samuel Jones · 5 months, 4 weeks ago

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@Hummus A Does the Curly brackets mean fractional part ? Rajdeep Dhingra · 6 months ago

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@Rajdeep Dhingra It is pointless to continue the contest. It hasn't garnered attention like the previous original integration contests. Samuel Jones · 5 months, 4 weeks ago

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@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. Rajdeep Dhingra · 5 months, 4 weeks ago

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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.

Aditya Kumar · 6 months, 1 week ago

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@Aditya Kumar Very True.I will change the note after it crosses 50 questions. Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra Change it after 20. 50 is too much. Aditya Kumar · 6 months, 1 week ago

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@Rajdeep Dhingra Maybe you should do it now, my browser has hanged twice loading this note! Akshay Yadav · 6 months, 1 week ago

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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. Aareyan Manzoor · 6 months, 1 week ago

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@Aareyan Manzoor Solution to Problem 9

Page 1

Page 1

Page 2

Page 2

Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra The second page is not at all clear. Please LaTeX your solution. Aditya Kumar · 6 months, 1 week ago

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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 \(t=\frac{x^k}{y}\) Hummus A · 5 months, 4 weeks ago

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@Hummus A 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: \[\displaystyle\int _{0}^{1}{\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } =\frac { 2k+1 }{ (k+1)^{ 2 } } -\frac { \gamma }{ k+1 }} \] Aditya Kumar · 5 months, 3 weeks ago

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Problem 22 :

Prove That

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

This problem has been solved by Aditya Kumar. Samuel Jones · 6 months ago

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@Samuel Jones 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, \[I=\int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } =\frac{\pi}{2}\] Aditya Kumar · 6 months ago

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@Aditya Kumar @Rajdeep Dhingra post the next problem. I can't come up with one of this level. Aditya Kumar · 6 months ago

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@Aditya Kumar I don't have anymore problems. Any one can post the Next Question. @Samuel Jones@Harsh Shrivastava@Aareyan Manzoor Rajdeep Dhingra · 6 months ago

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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. Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra 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 \(f(1) = \boxed{2\pi}\). Aareyan Manzoor · 6 months, 1 week ago

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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. Harsh Shrivastava · 6 months, 1 week ago

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@Harsh Shrivastava Solution to Problem 3

Let us start with the integral \[I(a) = \int_{0}^{1}{ \dfrac{\ln{(1+ax)}}{x\sqrt{1-x^2}} \ dx}\]. Differentiate with respect to a. \[I'(a) = \int_{0}^{1}{ \dfrac1{(1+ax)\sqrt{1-x^2}} \ dx}\] This can easily be solved by substituting \(1 + ax = \dfrac1t\).We now get \[I'(a) = \dfrac{\arccos{(a)}}{\sqrt{1-a^2}}\] Now we find I(a). \[I(a) = \int {\dfrac{\arccos{(a)}}{\sqrt{1-a^2}} \ da} = -\dfrac12\arccos^2{(x)} + C\] Now we put a = 1. We get \[I(1) = C\] Now we put a = -1. We get \[I(-1) = -\dfrac{{\pi}^2}2 + C\] We get \[I(1) - I(-1) = \dfrac{{\pi}^2}2\]

Q.E.D Rajdeep Dhingra · 6 months, 1 week ago

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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 \(x+y=t\) Hummus A · 5 months, 2 weeks ago

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Problem 26:

Prove that:

\[\int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx = -\frac{\gamma}{2}\] Aditya Kumar · 5 months, 3 weeks ago

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@Aditya Kumar

Lemma : \[\displaystyle \int_{0}^{\infty} \dfrac{\cos x - e^x}{x} \mathrm{d}x = 0\]

Proof : \(\displaystyle \int_{0}^{\infty} \dfrac{\cos x - e^x}{x} \mathrm{d}x = \int_{0}^{\infty} \int_{0}^{\infty} \left(e^{-yx}\cos x - e^{-x(y+1)} \right) \mathrm{d}x \ \mathrm{d}y \)

\(\displaystyle = \int_{0}^{\infty} \left( \dfrac{y}{y^2 + 1} - \dfrac{1}{y+1} \right) \mathrm{d}y\)

\(\displaystyle = \left[\ln \left( \dfrac{\sqrt{y^2+1}}{y+1} \right)\right]_{y=0}^{y \to \infty} = 0\)

Proposition : \[\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \gamma \]

Proof : \(\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x \)

\(\displaystyle = \int_{0}^{\infty} \int_{0}^{1} (1-e^{-x})e^{-yx} \mathrm{d}x \ \mathrm{d}y - \int_{0}^{\infty} \int_{1}^{\infty} e^{-x(y+1)} \mathrm{d}x \ \mathrm{d}y \)

\(\displaystyle = \int_{0}^{\infty} \left(\dfrac{1-e^{-y}}{y} + \dfrac{e^{-(y+1)} - 1}{y+1} \right) \mathrm{d}y - \int_{0}^{\infty} \dfrac{e^{(y+1)}}{y+1} \mathrm{d}y \)

\(\displaystyle = -\int_{0}^{\infty} \left(e^{-y} - \dfrac{1}{y+1} \right)\dfrac{\mathrm{d}y}{y} \)

\(\displaystyle = -\int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x}\)

Using the Lemma,

\(\displaystyle = - \int_{0}^{\infty} \left(\cos x - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x} \)

\(\displaystyle = \gamma\) (Using my solution here (see Problem 39))

Now,

\(\displaystyle \text{I} = \int_{0}^{\infty} \dfrac{\cos x - e^{-x^2}}{x} \mathrm{d}x \)

\(\displaystyle = \int_{0}^{\infty} \dfrac{e^{-x} - e^{-x^2}}{x} \mathrm{d}x \quad (\text{Using Lemma}) \)

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

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

\(\displaystyle = -\gamma + \int_{0}^{1} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x^2}}{x} \mathrm{d}x \quad (\text{Using Proposition}) \)

Substitute \(x^2 \mapsto x\)

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

\( \displaystyle = -\gamma + \dfrac{\gamma}{2} \quad (\text{Using Proposition})\)

\( \displaystyle = \boxed{-\dfrac{\gamma}{2}}\) Ishan Singh · 5 months, 2 weeks ago

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@Ishan Singh Post the next problem. Aditya Kumar · 5 months, 2 weeks ago

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@Aditya Kumar I don't have one appropriate for this contest right now, someone else may post the next problem. Ishan Singh · 5 months, 2 weeks ago

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@Ishan Singh Nice one! There's a shorter method that uses rmt! Aditya Kumar · 5 months, 2 weeks ago

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@Aditya Kumar R.M.T. on which function? Ishan Singh · 5 months, 2 weeks ago

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@Ishan Singh RMT on both. You'll get \[\lim _{ s\rightarrow 0 }{ \left( \Gamma \left( s \right) \cos \left( \frac { \pi \, s }{ 2 } \right) -\frac { 1 }{ 2 } \, \Gamma \left( \frac { s }{ 2 } \right) \right) } =\frac { -\gamma }{ 2 } \] Aditya Kumar · 5 months, 2 weeks ago

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@Aditya Kumar Nice! Ishan Singh · 5 months, 2 weeks ago

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Problem 24:

Prove that

\[\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}\]

The Problem has been solved by Rajdeep Dhingra. Samuel Jones · 5 months, 4 weeks ago

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@Samuel Jones Solution to Problem 24 :

Let \[I = \int_{-\infty}^{\infty} { e^{- x^2} \ dx}\] We can write it as \[I = \sqrt{I^2}\] \[ I = \sqrt{\int_{-\infty}^{\infty} { e^{- y^2} \ dy} \int_{-\infty}^{\infty} { e^{- x^2} \ dx}} \] \[I = \sqrt{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} { e^{- x^2 + y^2} \ dy \ dx }}\] Switching to Polar Coordinates of one integral. \[I = \sqrt{\int_{0}^{2\pi} { \int_{0}^{\infty} { e^{-r^2} r \ dr \ d{\theta}}}} \] The 'dr' one integral can be calculated by integration by parts. The other one is very easy. \[I = \sqrt{ \pi}\] Rajdeep Dhingra · 5 months, 4 weeks ago

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@Rajdeep Dhingra Nice! Please post the next problem. Samuel Jones · 5 months, 4 weeks ago

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Problem 21:

Prove That:

\[ \int_{0}^1\dfrac {1}{x^x} dx = \sum_{n=1}^{\infty} \dfrac {1}{n^n} \] Samuel Jones · 6 months ago

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@Harsh Shrivastava solution to problem 21 \[\int_0^1 x^{-x}dx=\int_0^1 e^{-x\ln(x)} dx= \int_0^1 \sum_{n=0}^\infty \dfrac{(-x\ln(x))^n}{n!} dx=\sum_{n=0}^\infty \dfrac{(-1)^n}{n!} \int_0^1 x^n\ln^n(x) dx\] Lemma :

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

Consider the integral,
\( \displaystyle \int_{0}^{1} x^{m}dx = \dfrac{1}{m+1} \)

Differentiate n times with respect to m,
\( \displaystyle \int_{0}^{1} x^{m}\ln^{n}x dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}} \)
put m=n to have our original integral \[\sum_{n=0}^\infty \dfrac{(-1)^n}{n!} \int_0^1 x^n\ln^n(x) dx=\sum_{n=0}^\infty \dfrac{1}{(n+1)^{n+1}}=\sum_{n=1}^\infty \dfrac{1}{n^n} \] Aareyan Manzoor · 6 months ago

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@Rajdeep Dhingra Since @Aareyan Manzoor hasn't posted a problem, can I post (according to the rules) ? Samuel Jones · 6 months ago

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@Samuel Jones Yup. Rajdeep Dhingra · 6 months ago

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@Rajdeep Dhingra Posted! Samuel Jones · 6 months ago

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@Aareyan Manzoor just a bit elaboration for your solution. Aareyan Manzoor · 6 months ago

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@Aareyan Manzoor You may post next problem, cause i am out of any good problem. Harsh Shrivastava · 6 months ago

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Problem 20 prove \[\int_0^1 \dfrac{\ln(x^2+x+1)}{x} dx=\dfrac{2\zeta(2)}{3}\] Aareyan Manzoor · 6 months ago

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@Aareyan Manzoor \(\displaystyle \int_{0}^1 \dfrac{\ln (x^2+x+1)}{x} dx = \int_{0}^1 \dfrac{\ln (1-x^3)}{x} dx - \int_{0}^1 \dfrac{\ln (1-x)}{x} dx\)

\(\displaystyle = -\sum_{n=1}^{\infty} \int_{0}^1\dfrac {x^{3n-1}}{n} dx + \sum_{n=1}^{\infty} \int_{0}^1 \dfrac {x^{n-1}}{n} dx \)

\(\displaystyle = \dfrac {2}{3} \zeta (2)\) Samuel Jones · 6 months ago

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@Samuel Jones well done, you are correct. Aareyan Manzoor · 6 months ago

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@Samuel Jones Awesome!

Post the next problem. Harsh Shrivastava · 6 months ago

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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. Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra solution to problem 11 \[\int uv'=uv-\int u'v\] \[I=\int_0^{\pi/2} x^2 \csc^2(x) dx\] \[u=x^2\to u'=2x, v'=\csc^2 x\to v=-cot(x)\] \[I=-x^2\cot(x)\mid_0^{\pi/2} +2\int_0^{\pi/2} x \cot(x) dx=2\int_0^{\pi/2} x \cot(x) dx\] \[u=x\to u'=1, v'=\cot(x)\to v=-\ln(sin(x))\] \[I=2x\ln(\sin(x))\mid_0^{\pi/2}+2\int_0^{\pi/2} \ln(\sin(x)) dx=\int_0^{\pi/2} \ln(\sin^2(x)) dx\] \[B(a,1/2)=\int_0^{\pi/2} 2\sin^{2a-1}(x) dx\] \[B'(a,1/2)=\int_0^{\pi/2} 2\ln(\sin^2(x))\sin^{2a-1}(x) dx\] \[B'(a,1/2)=B(a,1/2)(\psi(a)-\psi(a+1/2))\] \[a=1/2\] \[\int_0^{\pi/2} 2\ln(\sin^2(x))dx=B(1/2,1/2)(\psi(1/2)-\psi(1))=\pi(-\gamma-(-\gamma-2\ln(2)))=2\pi\ln(2)\] \[I=\int_0^{\pi/2} \ln(\sin^2(x)) dx=\boxed{\pi\ln(2)}\] Aareyan Manzoor · 6 months, 1 week ago

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Problem 5:

Evaluate

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

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. Harsh Shrivastava · 6 months, 1 week ago

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@Harsh Shrivastava Solution to Problem 5:

\[I=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( 1+x \right) } }{ x } dx } \]

\[I=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( -x \right) } }{ 1-x } dx } \]

\[I=-\int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) } }{ 1+x } dx } \]

Now, I'll use power series of \(\frac{1}{x+1}\), I'll write the integral as:

\[I=-\sum _{ n=0 }^{ \infty }{ { \left( -1 \right) }^{ n }\int _{ 0 }^{ 1 }{ { x }^{ n }\ln { x } dx } } \]

\[I=\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } } \]

\[\boxed{\therefore I=\frac { \zeta \left( 2 \right) }{ 2 } =\frac { { \pi }^{ 2 } }{ 12 } }\] Aditya Kumar · 6 months, 1 week ago

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@Harsh Shrivastava solution to problem 5

replace x with -x to have \[\int_0^{-1} \dfrac{\ln(1-x)}{x} dx\] \[\sum_{n≥1} \dfrac{x^n}{n}=\ln(1-x)\] dividing by x and integrating from 0 to a \[\int_0^{a} \dfrac{\ln(1-x)}{x} dx=\sum_{n≥1} \dfrac{a^n}{n^2}\] We have \[\int_0^{-1} \dfrac{\ln(1-x)}{x} dx= \sum_{n≥1}\dfrac{(-1)^n}{n^2}=\sum_{n≥1} \dfrac{(-1)^n}{n^2}=\sum_{n≥1} \dfrac{1}{n^2}-2\sum_{n≥1} \dfrac{1}{(2n)^2}=\\\dfrac{1}{2}\sum_{n≥1} \dfrac{1}{n^2}=\boxed{\dfrac{\pi^2}{12}}\] Aareyan Manzoor · 6 months, 1 week ago

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@Aareyan Manzoor Yeah there is a hell easy method.

:) Harsh Shrivastava · 6 months, 1 week ago

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@Harsh Shrivastava \(\text{Alternate solution to Problem 5:}\)Using MacLaurin expansion of \(\ln(1+x)\):\[\ln(1+x)=\sum_{r=1}^{\infty} (-1)^{r+1}\dfrac{x^r}{r}\]\[\Longrightarrow \dfrac{\ln(1+x)}{x}=\sum_{r=1}^{r=\infty}(-1)^{r+1}\dfrac{x^{r-1}}{r}\]\[\Longrightarrow I=\int_{0}^{1}(\sum_{r=1}^{r=\infty}(-1)^{r+1}\dfrac{x^{r-1}}{r})dx\]\[\Longrightarrow I=\sum_{r=1}^{\infty}(-1)^{r+1}\dfrac{\color{red}{1}}{r^2}=\dfrac{{\pi}^2}{12}\]. Adarsh Kumar · 6 months, 1 week ago

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Problem 4

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

This problem has been solved by Harsh Shrivastava. Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra

Harsh Shrivastava · 6 months, 1 week ago

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Problem 19

Prove that \[\int_0^1 \dfrac{\ln^2(1-x)}{x}dx=2\zeta(3)\]

The Problem has been solved by Vighnesh Shenoy. Aareyan Manzoor · 6 months, 1 week ago

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@Aareyan Manzoor Lemma :

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

Consider the integral,
\( \displaystyle \int_{0}^{1} x^{m}dx = \dfrac{1}{m+1} \)

Differentiate n times with respect to m,
\( \displaystyle \int_{0}^{1} x^{m}\ln^{n}x dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}} \)

\( I = \displaystyle \int_{0}^{1} \dfrac{ln^{2}(1-x)}{x}dx = \int_{0}^{1} \dfrac{\ln^{2}(x)}{1-x} dx \)

Using the Taylor series for \( \dfrac{1}{1-x} \)

\( I = \displaystyle \sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln^{2}(x)dx \).
Using the Lemma,
\( I = \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^{2} 2!}{(k+1)^{3}} = \sum_{k=1}^{\infty} \dfrac{2}{k^{3}} = 2\zeta(3) \) Vighnesh Shenoy · 6 months, 1 week ago

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@Vighnesh Shenoy Post the next problem. Harsh Shrivastava · 6 months ago

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Problem 16:

Find the closed form of the following indefinite integral,

\[ \displaystyle \int \dfrac{\arctan (x)}{x^{11}} dx \]

This problem has been solved by Rajdeep Dhingra. Vighnesh Shenoy · 6 months, 1 week ago

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@Vighnesh Shenoy Solution to Problem 16

It is a long tedious one. I will just mention some main steps.

Apply integration by parts with 1st function as \(\arctan{(x)}\) and 2nd one as \(\frac{dx}{x^{11}}\).

Now you will get an integral \(\displaystyle \int{\frac1{10x^{10}(x^2 + 1)} \ dx} \).

Apply partial fractions you will get \[\frac1{x^{10}} - \frac1{x^8} + \frac1{x^6} - \frac1{x^4} + \frac1{x^2} - \frac1{x^2 +1}\] Rest all is trivial Integration. Rajdeep Dhingra · 6 months, 1 week ago

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@Rajdeep Dhingra Instead of partial fracs, you can substiture \( \dfrac{1}{x} = t \) Vighnesh Shenoy · 6 months, 1 week ago

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@Vighnesh Shenoy Is the answer \(\displaystyle \large \sum_{k = 1}^{5}{\dfrac{{(-1)}^k}{10(2k-1){x}^{2k-1}}} - \frac{\arctan{(x)}}{10} - \frac{\arctan{(x)}}{10x^{10}}\) ? Rajdeep Dhingra · 6 months, 1 week ago

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Problem 13:

Evaluate \[ \displaystyle \int_{0}^{\pi /2} \ln (\cos x) dx \]

The Problem has been solved by Rajdeep Dhingra and Vighnesh Shenoy. Vighnesh solved it late by 1second. Still credit is given to Vighnesh for the next Question. Harsh Shrivastava · 6 months, 1 week ago

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@Harsh Shrivastava Solution to Problem 13

Let us start with \[I = \int_{0}^{\pi/2} { \log{(\cos{(x)})} \ dx} \] Using properties we know that \[I = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx} \] Adding both we get \[2I = \int_{0}^{\pi/2} { \log{(\sin{(x)}\cos{(x)})} \ dx}\] Using little manipulation we get that \[2I = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx} - \int_{0}^{\pi/2} { \log{(2)} \ dx} \] Using lemma we get \[2I = I - \dfrac{\pi}2 \log{(2)} \\ \Rightarrow \boxed{I = - \dfrac{\pi}2 \log{(2)} }\]


Proof of Lemma Used

\[I_1 = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx} \] Substitute 2x = t. You will get \[I_1 = \frac12 \int_{0}^{\pi} { \log{(\sin{(t)})} \ dt} \\I_1 = \frac12 \left( \int_{0}^{\pi/2} { \log{(\sin{(t)})} \ dt} + \int_{\pi/2}^{\pi} { \log{(\sin{(x)})} \ dx} \right) \] Now let us substitute \(x - \frac{\pi}2 = t\) in the 2nd integral. Then using properties and manipulating. we get \[I_1 = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx}\] Rajdeep Dhingra · 6 months, 1 week ago

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@Harsh Shrivastava \( I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\cos x) dx \)
\( \therefore I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\sin x) dx \)
Adding those expressions,

\( 2I = I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\cos x \sin x) dx \)

\( 2I =I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\sin 2x) dx - \int_{0}^{\dfrac{\pi}{2}} \ln(2)dx\)
In the first expression , substitute 2x = t .

\( 2I = \displaystyle \dfrac{1}{2} \cdot \int_{0}^{\pi}{\ln(\sin t) \ dt} - \dfrac{\pi \ln(2)}{2} \)
\( 2I =\displaystyle \dfrac{1}{2} \cdot 2 \cdot \int_{0}^{\dfrac{\pi}{2}} \ln(\sin t) dt - \dfrac{\pi \ln(2)}{2} \)
\( 2I = I - \dfrac{\pi \ln(2)}{2} \)
\( I = \dfrac{-\pi \ln(2)}{2} \) Vighnesh Shenoy · 6 months, 1 week ago

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@Harsh Shrivastava Delete this question! This question is on brilliant. See this. Aditya Kumar · 6 months, 1 week ago

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Problem 12

\[\int_0^1 \ln(x)\ln(1-x)x(1-x) dx\]

Problem has been solved by Harsh Shrivastava. Aareyan Manzoor · 6 months, 1 week ago

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@Aareyan Manzoor Simple derivative of beta function.

\(B(a,b) = \displaystyle \int_{0}^{1} x^{m-1} (1-x)^{n-1}dx \)

Required integral is \(\dfrac{\partial B(m,n)}{\partial m \partial n}\) at \(m=n=2\)

This evaluates to\(\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (((\psi (m)-\psi (m+n))(\psi (n)-\psi (m+n))-\psi '(m+n))\).

On putting the values of m and n, the final answer evaluates to \(0.0684\). Harsh Shrivastava · 6 months, 1 week ago

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@Harsh Shrivastava This is not a complete answer. Please show how you got 0.0684. Aditya Kumar · 6 months, 1 week ago

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Problem 7:

Prove that \[\int_0^1 \log \Gamma(x)\cos (16\pi x)\, dx=\frac{1}{32}\]

This problem has been solved by Rajdeep Dhingra. Aditya Kumar · 6 months, 1 week ago

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@Aditya Kumar Let us start with \[I = \int_{0}^{1}{ \log{(\Gamma{(x)})}cos(16\pi x) \ dx}\] We can also write \[I = \int_{0}^{1}{ \log{(\Gamma{(1-x)})}cos(16\pi - 16\pi x) \ dx} = \int_{0}^{1}{ \log{(\Gamma{(1-x)})}cos(16\pi x) \ dx}\] Adding both we get \[2I = \int_{0}^{1}{ \log{(\Gamma{(1-x)}\Gamma{(x)})}cos(16\pi x) \ dx}\] Using Euler Reflection Formula we get \[2I = \int_{0}^{1}{ \log{(\pi)}cos(16\pi x) \ dx - \int_{0}^{1}{ \log{(\sin{(\pi x)})}cos(16\pi x) \ dx}}\] Both are standard computations. \[2I = 0 - \left(-\dfrac1{16}\right)\] \[\boxed{I = \frac{1}{32}}\] Q.E.D Rajdeep Dhingra · 6 months, 1 week ago

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Problem 15:

Time for an indefinite integral:

\[\int \dfrac{\sin^9 x}{\cos^8 x} dx\]

This problem has been solved by Vighnesh Shenoy and Aditya Kumar at almost the same time. Credits has been given to Vighnesh. Aareyan Manzoor · 6 months, 1 week ago

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@Aareyan Manzoor Solution to Problem 15:

\[I=\int { \frac { \sin ^{ 9 }{ x } }{ \cos ^{ 8 }{ x } } dx } \]

Substitute: \(\cos x=t\)

\[I=-\int { \left( { t }^{ -8 }-{ 4t }^{ -6 }+{ 6t }^{ -4 }-{ 4t }^{ -2 }+1 \right) } dt\]

\[\therefore I=-\cos { \left( x \right) } +\frac { \sec ^{ 7 }{ \left( x \right) } }{ 7 } -\frac { 4\sec ^{ 5 }{ \left( x \right) } }{ 5 } +2\sec ^{ 3 }{ \left( x \right) } -4\sec { \left( x \right) } \] Aditya Kumar · 6 months, 1 week ago

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@Aareyan Manzoor \( \cos x = t \)
\( \therefore -\sin x dx = dt \)
\( I = \displaystyle \int \dfrac{(1-t^{2})^{4}}{t^{8}}\cdot -dt \)
\( I =- \displaystyle \int \dfrac{1-4t^{2} + 6t^{4} -4t^{6} + t^{8}}{t^{8}} dt \)
\( I = - \left( \dfrac{-1}{7t^{7}} + \dfrac{4}{5t^{5}} - \dfrac{6}{3t^{3}} + \dfrac{4}{t} + t \right) \) + c \( I = - \left( \dfrac{-1}{7\cos^{7} x} + \dfrac{4}{5\cos^{5} x} - \dfrac{6}{3\cos^{3}x}+ \dfrac{4}{\cos x} + \cos x \right) + c \) Vighnesh Shenoy · 6 months, 1 week ago

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@Vighnesh Shenoy Lol I was typing. So I didn't see. Post the next problem. Aditya Kumar · 6 months, 1 week ago

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@Vighnesh Shenoy Oh no! I was about to start typing... Nihar Mahajan · 6 months, 1 week ago

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