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
3 years, 7 months ago

No vote yet
1 vote

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Problem 18:

Prove that: 01(1logx+11x12)dx1x=12+12ln(2π)12γ.\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

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

Proof : r=1nHr=01r=1nxr1x1dx\displaystyle \sum_{r=1}^{n} H_{r} = \int_{0}^{1} \sum_{r=1}^{n} \dfrac{x^r-1}{x-1} \mathrm{d}x

=01xn+1xnx+n(x1)2dx\displaystyle = \int_{0}^{1} \dfrac{x^{n+1} - x - nx + n}{(x-1)^2} \mathrm{d}x

=01xn+1x(x1)2dx01nx1dx\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

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

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

A=[xn+1xx1]01+01(n+1)xn1x1dx\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

=n+01(n+1)xn1x1dx\displaystyle = -n + \int_{0}^{1}\dfrac{(n+1)x^n - 1}{x-1} \mathrm{d}x

    A+B=n+(n+1)01xn1x1dx\displaystyle \implies \text{A} +\text{B} = -n + (n+1) \cdot \int_{0}^{1}\dfrac{x^n - 1}{x-1} \mathrm{d}x

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

Lemma 2 : 01xn1lnxdx=ln(n+1)\displaystyle \int_{0}^{1} \dfrac{x^n - 1}{\ln x} \mathrm{d}x = \ln (n +1)

Proof : 01xn1lnx dx=010nxy dy dx\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

=0n01xy dx dy\displaystyle = \int_{0}^{n} \int_{0}^{1} x^y \ \mathrm{d}x \ \mathrm{d}y

=0n1y+1 dy\displaystyle = \int_{0}^{n} \dfrac{1}{y+1} \ \mathrm{d}y

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

Lemma 3 : 01(1ln(x)+11x)dx=γ\displaystyle \int_{0}^{1} \left(\dfrac{1}{\ln (x)} + \dfrac{1}{1-x}\right) \mathrm{d}x = \gamma

Proof : 01(11x+1ln(x))dx=limn01(1tn1ttn11lntdt)\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)

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

=γ\displaystyle = \gamma

Now,

I=01(1lnx+11x12)dxx1\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}

=limnr=1n01(xr1lnx+xr11xxr12)\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,

I=limnr=1n(γ+lnrHr112r)\displaystyle \text{I} = \lim_{n \to \infty} \sum_{r=1}^{n} \left(\gamma + \ln r - H_{r-1} -\dfrac{1}{2r} \right)

=limn(nγ+ln(n!)r=1nHr1Hn2)\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,

I=limn(nγ+ln(n!)nHn1+n1Hn2)\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,

I=limn(nγ+12ln(2π)n+12ln(n)+nln(n)nHn1+n1Hn2)\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 limn(Hnlnn)=γ\displaystyle \lim_{n \to \infty} (H_{n} - \ln n) = \gamma, we have,

I=12ln(2π)12γ12\displaystyle \text{I} = \boxed{\dfrac{1}{2} \ln (2\pi) -\dfrac{1}{2} \gamma - \dfrac{1}{2}}

Ishan Singh - 3 years, 7 months ago

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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

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Next problem please.

Harsh Shrivastava - 3 years, 7 months ago

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@Harsh Shrivastava @Aditya Kumar May post another question if he wants.

Ishan Singh - 3 years, 7 months ago

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Nice. I used Binet's formula.

Aditya Kumar - 3 years, 7 months ago

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Let us start with an easy Problem.

Problem 1

Find the value of the following (x+1)2e(x+1)2dxex2dx\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

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Solution to Problem 1: I1=(x+1)2e(x+1)2dx{ I }_{ 1 }=\int _{ -\infty }^{ \infty }{ { \left( x+1 \right) }^{ 2 }{ e }^{ { -\left( x+1 \right) }^{ 2 } } } dx

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

Therefore, we get: I1=12(t)12etdt{ I }_{ 1 }=\frac { 1 }{ 2 }\int _{ -\infty }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt

This is symmetric about x=0x=0. Therefore, we can write it as: I1=0(t)12etdt{ I }_{ 1 }=\int _{ 0 }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt

I1=Γ(32)=12Γ(12)=π2{ I }_{ 1 }=\Gamma \left( \frac { 3 }{ 2 } \right) =\frac { 1 }{ 2 } \Gamma \left( \frac { 1 }{ 2 } \right) =\frac { \sqrt { \pi } }{ 2 }

Similarly, I2=π{ I }_{ 2 }=\sqrt { \pi }

Hence, I1I2=12\boxed{\frac{{ I }_{ 1 }}{{ I }_{ 2 }}=\frac{1}{2}}

Aditya Kumar - 3 years, 7 months ago

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Problem 2:

Prove that: 01arctan(x2+2)x2+2(x2+1) dx=5π296\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 - 3 years, 7 months ago

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Solution to Problem 2

I=01arctan(x2+2)x2+2(x2+1) dx=5π296I = \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 II as A and BA \text{ and } B by using the property that arctan(x)=π/2arctan(1/x)\arctan(x) = \pi /2 -\arctan(1/x)

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

where A=π/201dx(1+x2)(2+x2)A = \displaystyle \pi /2\int_{0}^{1}\dfrac{dx}{(1+x^{2})(\sqrt{2+x^{2}})}

and B=01arctan(1x2+2)(1+x2)(2+x2)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,

π/2dx(1+x2)(2+x2)=π/2arctanxx2+2\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 = π2/12\pi ^{2}/12

Evaluating B,

Using 1/aarctan(1/a)=0adyy2+a21/a \arctan(1/a) = \int_{0}^{a} \frac{dy}{y^{2}+a^{2}},

Let a=x2+2a = \sqrt{x^2+2}.

We get B=0101dxdy(1+x2)(2+x2+y2)B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(2+x^{2} + y^{2})}

B=0101dxdy(1+x2)(1+y2)0101dxdy(1+x2)(2+x2+y2)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=0101dxdy(1+x2)(1+y2)2B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(1+y^{2})}

2B=(01dx1+x2)22B = \displaystyle (\int_{0}^{1}\dfrac{dx}{1+x^{2}})^{2}

Thus B=π2/32B = \pi^{2} / 32

Hence I=AB=5π296I = A-B=\dfrac{5\pi^{2}}{96}

Harsh Shrivastava - 3 years, 7 months ago

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Generalization

Ishan Singh - 3 years, 7 months ago

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Problem 14:

If 0x31+eαxdx=164 \displaystyle \int_{0}^{\infty} \dfrac{x^{3}}{1+e^{\alpha x}}dx = \dfrac{1}{64}
Find α \alpha given that , α \alpha is a positive constant and ζ(4)=π490 \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

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I=0x3n=1(eαx)ndxI=\int_0^\infty x^3 \sum_{n=1}^\infty (-e^{-\alpha x})^n dx I=n=1(1)n0x3eαnxdx=n=1(1)n6(αn)4I= \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=6α4(123)ζ(4)=164I=\dfrac{6}{\alpha^4} (1-2^{-3}) \zeta(4)=\dfrac{1}{64} the result follows.

Aareyan Manzoor - 3 years, 7 months ago

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Is the answer α=π(5615)14\huge \displaystyle \alpha = \pi {\left(\frac{56}{15}\right)}^{\frac14} ?

Rajdeep Dhingra - 3 years, 7 months ago

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Problem 6:

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

Where HxH_x denotes harmonic number

This problem has been solved by Aditya Kumar.

Aareyan Manzoor - 3 years, 7 months ago

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Solution to Problem 6:

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

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

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

Aditya Kumar - 3 years, 7 months ago

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Problem 10

Let S=sin2(x)+sin4(x)32+sin6(x)52+ Upto Infinity S = \sin^2{(x)} + \dfrac{sin^4{(x)}}{3^2} + \dfrac{sin^6{(x)}}{5^2} + \cdots \text{ Upto Infinity }

Prove that 0π/2S dx=π24π2\int_{0}^{\pi /2} {S \ dx} = \dfrac{{\pi}^2}{4} - \dfrac{\pi}2

The Problem has been solved by Ishan Singh.

Rajdeep Dhingra - 3 years, 7 months ago

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

Lemma: f(a)=11+a+r=1[1(2r+a+1)k=1r(2k12k)]=π2a+1Γ(a+1)Γ2(a+22);a>1 \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,

I=0π/2S dx=r=101sin2r(2r1)2dx\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

=12r=1B(r+12,12)(2r1)2\displaystyle = \dfrac{1}{2} \sum_{r=1}^{\infty} \dfrac{\operatorname{B}\left(r+\frac{1}{2} , \frac{1}{2} \right)}{(2r-1)^2}

=12r=1Γ(r+12)Γ(12)(2r1)2Γ(r+1)\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)}

=π2r=1Γ(r+12)(2r1)2Γ(12)Γ(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 Γ(x+1)=xΓ(x)\Gamma(x+1) = x\Gamma(x), we have,

I=π4r=0Γ(r+12)(r+1)(2r+1)Γ(12)Γ(r+1)\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)}

=π2[r=0Γ(r+12)(2r+1)Γ(12)Γ(r+1)r=0Γ(r+12)(2r+2)Γ(12)Γ(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,

I=π2[1+r=11(2r+1)k=1r(2k12k)12r=11(2r+2)k=1r(2k12k)]\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]

=π2[f(0)f(1)]\displaystyle = \dfrac{\pi}{2} \left[f(0) - f(1) \right]

=π24π2\displaystyle = \boxed{\dfrac{\pi ^2}{4} - \dfrac{\pi}{2}}

Ishan Singh - 3 years, 7 months ago

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What was your intended solution?

Samuel Jones - 3 years, 7 months ago

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More or less the same.

Rajdeep Dhingra - 3 years, 7 months ago

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

Samuel Jones - 3 years, 7 months ago

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

Rajdeep Dhingra - 3 years, 7 months ago

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

Ishan Singh - 3 years, 7 months 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 - 3 years, 7 months ago

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

Rajdeep Dhingra - 3 years, 7 months ago

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

Akshay Yadav - 3 years, 7 months ago

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Change it after 20. 50 is too much.

Aditya Kumar - 3 years, 7 months ago

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Problem 17

Prove that if c<1c < 1, 0π/2arcsin(ccos(x)) dx=c12+c332+c552+ (Upto Infinity)\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 - 3 years, 7 months ago

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Solution to Problem 17:

Maclaurin series of arcsin(x)\arcsin { \left( x \right) } is: arcsin(x)=n=0Γ(n+12)π(2n+1)n!x2n+1\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=ccos(x)x=c \cos(x)

0π2arcsin(ccos(x))dx=0π2n=0Γ(n+12)π(2n+1)n!(ccos(x))2n+1dx\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 }

0π2arcsin(ccos(x))dx=n=0Γ(n+12)c2n+1π(2n+1)n!0π2(cos(x))2n+1dx\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:

0π2arcsin(ccos(x))dx=n=0(Γ(n+12)c2n+1π(2n+1)n!12Γ(12)Γ(n+1)Γ(n+32))\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: 0π/2arcsin(ccos(x)) dx=c12+c332+c552+ (Upto Infinity)\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 - 3 years, 7 months ago

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Problem 9

Prove 0π/2xln(tan(x))dx=7ζ(3)8\int_0^{\pi/2} x\ln(\tan(x)) dx= \dfrac{7\zeta(3)}{8}

The Problem has been solved by Rajdeep Dhingra.

Aareyan Manzoor - 3 years, 7 months ago

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Solution to Problem 9

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Rajdeep Dhingra - 3 years, 7 months ago

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

Aditya Kumar - 3 years, 7 months ago

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Problem 23

Prove that for an integer n2n\ge2

01{1xn}dx=nn1ζ(n)\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx=\frac { n }{ n-1 } -\zeta (n)

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

Hamza A - 3 years, 7 months ago

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Let I=01{1xn}dx\displaystyle \text{I} = \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx

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

    I=n1{t}tn+1dt\displaystyle \implies \text{I} = n \int_{1}^{\infty} \dfrac{ \left \{t \right \} }{t^{n+1}} dt

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

=nr=1[t1n1n+rtnn]rr+1\displaystyle = n \sum_{r=1}^{\infty} \left[ \dfrac{t^{1-n}}{1-n} +r \dfrac{t^{-n}}{n} \right]_{r}^{r+1}

=nr=1(1n(n+1)(r+1)n1+1n(r+1)n)\displaystyle = -n \sum_{r=1}^{\infty} \left( \dfrac{1}{n(n+1) (r+1)^{n-1}} + \dfrac{1}{n(r+1)^n} \right)

=nn1ζ(n)\displaystyle = \dfrac{n}{n-1} - \zeta(n)

Samuel Jones - 3 years, 7 months ago

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Does the Curly brackets mean fractional part ?

Rajdeep Dhingra - 3 years, 7 months ago

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

Samuel Jones - 3 years, 7 months 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 - 3 years, 7 months ago

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Problem 3:

Prove that 01(ln(1+x)ln(1x))x(1x2)dx=π22\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 - 3 years, 7 months ago

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Solution to Problem 3

Let us start with the integral I(a)=01ln(1+ax)x1x2 dxI(a) = \int_{0}^{1}{ \dfrac{\ln{(1+ax)}}{x\sqrt{1-x^2}} \ dx}. Differentiate with respect to a. I(a)=011(1+ax)1x2 dxI'(a) = \int_{0}^{1}{ \dfrac1{(1+ax)\sqrt{1-x^2}} \ dx} This can easily be solved by substituting 1+ax=1t1 + ax = \dfrac1t.We now get I(a)=arccos(a)1a2I'(a) = \dfrac{\arccos{(a)}}{\sqrt{1-a^2}} Now we find I(a). I(a)=arccos(a)1a2 da=12arccos2(x)+CI(a) = \int {\dfrac{\arccos{(a)}}{\sqrt{1-a^2}} \ da} = -\dfrac12\arccos^2{(x)} + C Now we put a = 1. We get I(1)=CI(1) = C Now we put a = -1. We get I(1)=π22+CI(-1) = -\dfrac{{\pi}^2}2 + C We get I(1)I(1)=π22I(1) - I(-1) = \dfrac{{\pi}^2}2

Q.E.D

Rajdeep Dhingra - 3 years, 7 months ago

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Problem 8

Prove that 02πecos(θ)cos(sin(θ)) dθ=2π \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 - 3 years, 7 months ago

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solution to problem 8

Considering the function f(t)=02πetcos(θ)cos(tsin(θ))dθf(t) = \int_{0}^{2\pi} e^{t \cos(\theta)} \cos(t\sin(\theta)) \, d\theta, we see that tf(t)=t02πetcosθ[cosθcos(tsinθ)sinθsin(tsinθ)]dθ=02πθ[etcosθsin(tsinθ)]dθ=[etcosθsin(tsinθ)]02π  =  0f(t)=0f(t)=C \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)=2πf(0) = 2\pi, the desired integral is f(1)=2πf(1) = \boxed{2\pi}.

Aareyan Manzoor - 3 years, 7 months ago

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Problem 11

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

The Problem has been solved Aareyan Manzoor.

Rajdeep Dhingra - 3 years, 7 months ago

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solution to problem 11 uv=uvuv\int uv'=uv-\int u'v I=0π/2x2csc2(x)dxI=\int_0^{\pi/2} x^2 \csc^2(x) dx u=x2u=2x,v=csc2xv=cot(x)u=x^2\to u'=2x, v'=\csc^2 x\to v=-cot(x) I=x2cot(x)0π/2+20π/2xcot(x)dx=20π/2xcot(x)dxI=-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=xu=1,v=cot(x)v=ln(sin(x))u=x\to u'=1, v'=\cot(x)\to v=-\ln(sin(x)) I=2xln(sin(x))0π/2+20π/2ln(sin(x))dx=0π/2ln(sin2(x))dxI=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)=0π/22sin2a1(x)dxB(a,1/2)=\int_0^{\pi/2} 2\sin^{2a-1}(x) dx B(a,1/2)=0π/22ln(sin2(x))sin2a1(x)dxB'(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)(ψ(a)ψ(a+1/2))B'(a,1/2)=B(a,1/2)(\psi(a)-\psi(a+1/2)) a=1/2a=1/2 0π/22ln(sin2(x))dx=B(1/2,1/2)(ψ(1/2)ψ(1))=π(γ(γ2ln(2)))=2πln(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=0π/2ln(sin2(x))dx=πln(2)I=\int_0^{\pi/2} \ln(\sin^2(x)) dx=\boxed{\pi\ln(2)}

Aareyan Manzoor - 3 years, 7 months ago

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Problem 22 :

Prove That

0sin(x2+12x)cos(x212x)xdx=π2\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 - 3 years, 7 months ago

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Solution to Problem 22:

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

Therefore, we get: I=12{0sin(x)x+0sin(1x)xdx} 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=0sin(x)x=π2I=\int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } =\frac{\pi}{2}

Aditya Kumar - 3 years, 7 months ago

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@Rajdeep Dhingra post the next problem. I can't come up with one of this level.

Aditya Kumar - 3 years, 7 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 - 3 years, 7 months ago

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Problem 25

Let kk be a positive real number

then prove

0101{xky}dxdy=2k+1(k+1)2γk+1\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=xkyt=\frac{x^k}{y}

Hamza A - 3 years, 7 months ago

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Solution to Problem 25:

I=0101{xky}dxdyI=\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } }

I=01(01{xky}dy)dxI=\int _{ 0 }^{ 1 }{ \left( \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dy } \right) dx }

By IBP, we get: I=(xk+1k+1xk{t}t2dt)01+kk+101xkdxI={ \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: 0101{xky}dxdy=2k+1(k+1)2γk+1\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 - 3 years, 7 months ago

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Problem 27

Evaluate

0101{1x+y}ndxdy\displaystyle\int _{0}^{1}{\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x+y } \right\} ^{ n }dxdy }}

with nNn\in\mathbb{N}

Hint:let x+y=tx+y=t

Hamza A - 3 years, 7 months ago

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Problem 4

Prove 0x(1+x)(2+x)(3+x) dx=π2(221+3)\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 - 3 years, 7 months ago

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Harsh Shrivastava - 3 years, 7 months ago

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Problem 5:

Evaluate

01ln(1+x)x\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 - 3 years, 7 months ago

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solution to problem 5

replace x with -x to have 01ln(1x)xdx\int_0^{-1} \dfrac{\ln(1-x)}{x} dx n1xnn=ln(1x)\sum_{n≥1} \dfrac{x^n}{n}=\ln(1-x) dividing by x and integrating from 0 to a 0aln(1x)xdx=n1ann2\int_0^{a} \dfrac{\ln(1-x)}{x} dx=\sum_{n≥1} \dfrac{a^n}{n^2} We have 01ln(1x)xdx=n1(1)nn2=n1(1)nn2=n11n22n11(2n)2=12n11n2=π212\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 - 3 years, 7 months ago

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Yeah there is a hell easy method.

:)

Harsh Shrivastava - 3 years, 7 months ago

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@Harsh Shrivastava Alternate solution to Problem 5:\text{Alternate solution to Problem 5:}Using MacLaurin expansion of ln(1+x)\ln(1+x):ln(1+x)=r=1(1)r+1xrr\ln(1+x)=\sum_{r=1}^{\infty} (-1)^{r+1}\dfrac{x^r}{r}ln(1+x)x=r=1r=(1)r+1xr1r\Longrightarrow \dfrac{\ln(1+x)}{x}=\sum_{r=1}^{r=\infty}(-1)^{r+1}\dfrac{x^{r-1}}{r}I=01(r=1r=(1)r+1xr1r)dx\Longrightarrow I=\int_{0}^{1}(\sum_{r=1}^{r=\infty}(-1)^{r+1}\dfrac{x^{r-1}}{r})dxI=r=1(1)r+11r2=π212\Longrightarrow I=\sum_{r=1}^{\infty}(-1)^{r+1}\dfrac{\color{#D61F06}{1}}{r^2}=\dfrac{{\pi}^2}{12}.

Adarsh Kumar - 3 years, 7 months ago

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