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, 11 months ago

No vote yet
1 vote

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hey @Rajdeep Dhingra i'm 16 yrs in age but level 5 in calculus , can i participate ?

A Former Brilliant Member - 3 years, 3 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, 10 months ago

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Problem 26:

Prove that:

0cosxex2x dx=γ2\int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx = -\frac{\gamma}{2}

Aditya Kumar - 3 years, 10 months ago

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

Proof : 0cosxexxdx=00(eyxcosxex(y+1))dx dy\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

=0(yy2+11y+1)dy\displaystyle = \int_{0}^{\infty} \left( \dfrac{y}{y^2 + 1} - \dfrac{1}{y+1} \right) \mathrm{d}y

=[ln(y2+1y+1)]y=0y=0\displaystyle = \left[\ln \left( \dfrac{\sqrt{y^2+1}}{y+1} \right)\right]_{y=0}^{y \to \infty} = 0

Proposition : 011exxdx1exxdx=γ\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \gamma

Proof : 011exxdx1exxdx\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x

=001(1ex)eyxdx dy01ex(y+1)dx dy\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

=0(1eyy+e(y+1)1y+1)dy0e(y+1)y+1dy\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

=0(ey1y+1)dyy\displaystyle = -\int_{0}^{\infty} \left(e^{-y} - \dfrac{1}{y+1} \right)\dfrac{\mathrm{d}y}{y}

=0(ex1x+1)dxx\displaystyle = -\int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x}

Using the Lemma,

=0(cosx1x+1)dxx\displaystyle = - \int_{0}^{\infty} \left(\cos x - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x}

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

Now,

I=0cosxex2xdx\displaystyle \text{I} = \int_{0}^{\infty} \dfrac{\cos x - e^{-x^2}}{x} \mathrm{d}x

=0exex2xdx(Using Lemma)\displaystyle = \int_{0}^{\infty} \dfrac{e^{-x} - e^{-x^2}}{x} \mathrm{d}x \quad (\text{Using Lemma})

=01exxdx+01ex2xdx\displaystyle = - \int_{0}^{\infty} \dfrac{1-e^{-x}}{x} \mathrm{d}x + \int_{0}^{\infty} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x

=011exxdx11exxdx+011ex2xdx+11ex2xdx\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

=γ+011ex2xdx1ex2xdx(Using Proposition)\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 x2xx^2 \mapsto x

    I=γ+12(011exxdx1exxdx)\displaystyle \implies \text{I} = -\gamma + \dfrac{1}{2} \left( \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x}}{x} \mathrm{d}x \right)

=γ+γ2(Using Proposition) \displaystyle = -\gamma + \dfrac{\gamma}{2} \quad (\text{Using Proposition})

=γ2 \displaystyle = \boxed{-\dfrac{\gamma}{2}}

Ishan Singh - 3 years, 10 months ago

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Post the next problem.

Aditya Kumar - 3 years, 10 months 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 - 3 years, 10 months ago

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Nice one! There's a shorter method that uses rmt!

Aditya Kumar - 3 years, 10 months ago

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@Aditya Kumar R.M.T. on which function?

Ishan Singh - 3 years, 10 months ago

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@Ishan Singh RMT on both. You'll get lims0(Γ(s)cos(πs2)12Γ(s2))=γ2\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 - 3 years, 10 months ago

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@Aditya Kumar Nice!

Ishan Singh - 3 years, 10 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, 11 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, 10 months ago

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Problem 24:

Prove that

ex2dx=π\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}

The Problem has been solved by Rajdeep Dhingra.

Samuel Jones - 3 years, 11 months ago

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Solution to Problem 24 :

Let I=ex2 dxI = \int_{-\infty}^{\infty} { e^{- x^2} \ dx} We can write it as I=I2I = \sqrt{I^2} I=ey2 dyex2 dx I = \sqrt{\int_{-\infty}^{\infty} { e^{- y^2} \ dy} \int_{-\infty}^{\infty} { e^{- x^2} \ dx}} I=ex2+y2 dy dxI = \sqrt{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} { e^{- x^2 + y^2} \ dy \ dx }} Switching to Polar Coordinates of one integral. I=02π0er2r dr dθI = \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=πI = \sqrt{ \pi}

Rajdeep Dhingra - 3 years, 11 months ago

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Nice! Please post the next problem.

Samuel Jones - 3 years, 11 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, 11 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, 11 months ago

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Does the Curly brackets mean fractional part ?

Rajdeep Dhingra - 3 years, 11 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, 11 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, 11 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, 11 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, 11 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, 11 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, 11 months ago

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Problem 21:

Prove That:

011xxdx=n=11nn \int_{0}^1\dfrac {1}{x^x} dx = \sum_{n=1}^{\infty} \dfrac {1}{n^n}

Samuel Jones - 3 years, 11 months ago

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Problem 20 prove 01ln(x2+x+1)xdx=2ζ(2)3\int_0^1 \dfrac{\ln(x^2+x+1)}{x} dx=\dfrac{2\zeta(2)}{3}

Aareyan Manzoor - 3 years, 11 months ago

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01ln(x2+x+1)xdx=01ln(1x3)xdx01ln(1x)xdx\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

=n=101x3n1ndx+n=101xn1ndx\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

=23ζ(2)\displaystyle = \dfrac {2}{3} \zeta (2)

Samuel Jones - 3 years, 11 months ago

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well done, you are correct.

Aareyan Manzoor - 3 years, 11 months ago

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Awesome!

Post the next problem.

Harsh Shrivastava - 3 years, 11 months ago

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Problem 19

Prove that 01ln2(1x)xdx=2ζ(3)\int_0^1 \dfrac{\ln^2(1-x)}{x}dx=2\zeta(3)

The Problem has been solved by Vighnesh Shenoy.

Aareyan Manzoor - 3 years, 11 months ago

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Lemma :

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

Consider the integral,
01xmdx=1m+1 \displaystyle \int_{0}^{1} x^{m}dx = \dfrac{1}{m+1}

Differentiate n times with respect to m,
01xmlnnxdx=(1)nn!(m+1)n+1 \displaystyle \int_{0}^{1} x^{m}\ln^{n}x dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}}

I=01ln2(1x)xdx=01ln2(x)1xdx 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 11x \dfrac{1}{1-x}

I=k=001xkln2(x)dx I = \displaystyle \sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln^{2}(x)dx .
Using the Lemma,
I=k=0(1)22!(k+1)3=k=12k3=2ζ(3) I = \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^{2} 2!}{(k+1)^{3}} = \sum_{k=1}^{\infty} \dfrac{2}{k^{3}} = 2\zeta(3)

A Former Brilliant Member - 3 years, 11 months ago

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Post the next problem.

Harsh Shrivastava - 3 years, 11 months ago

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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, 11 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, 11 months ago

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Nice. I used Binet's formula.

Aditya Kumar - 3 years, 11 months ago

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Next problem please.

Harsh Shrivastava - 3 years, 11 months ago

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

Ishan Singh - 3 years, 11 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, 11 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, 11 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, 11 months ago

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Problem 16:

Find the closed form of the following indefinite integral,

arctan(x)x11dx \displaystyle \int \dfrac{\arctan (x)}{x^{11}} dx

This problem has been solved by Rajdeep Dhingra.

A Former Brilliant Member - 3 years, 11 months ago

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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)\arctan{(x)} and 2nd one as dxx11\frac{dx}{x^{11}}.

Now you will get an integral 110x10(x2+1) dx\displaystyle \int{\frac1{10x^{10}(x^2 + 1)} \ dx} .

Apply partial fractions you will get 1x101x8+1x61x4+1x21x2+1\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 - 3 years, 11 months ago

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Is the answer k=15(1)k10(2k1)x2k1arctan(x)10arctan(x)10x10\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 - 3 years, 11 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, 11 months ago

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

Rajdeep Dhingra - 3 years, 11 months ago

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Change it after 20. 50 is too much.

Aditya Kumar - 3 years, 11 months ago

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

Akshay Yadav - 3 years, 11 months ago

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Problem 15:

Time for an indefinite integral:

sin9xcos8xdx\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 - 3 years, 11 months ago

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Solution to Problem 15:

I=sin9xcos8xdxI=\int { \frac { \sin ^{ 9 }{ x } }{ \cos ^{ 8 }{ x } } dx }

Substitute: cosx=t\cos x=t

I=(t84t6+6t44t2+1)dtI=-\int { \left( { t }^{ -8 }-{ 4t }^{ -6 }+{ 6t }^{ -4 }-{ 4t }^{ -2 }+1 \right) } dt

I=cos(x)+sec7(x)74sec5(x)5+2sec3(x)4sec(x)\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 - 3 years, 11 months ago

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cosx=t \cos x = t
sinxdx=dt \therefore -\sin x dx = dt
I=(1t2)4t8dt I = \displaystyle \int \dfrac{(1-t^{2})^{4}}{t^{8}}\cdot -dt
I=14t2+6t44t6+t8t8dt I =- \displaystyle \int \dfrac{1-4t^{2} + 6t^{4} -4t^{6} + t^{8}}{t^{8}} dt
I=(17t7+45t563t3+4t+t) I = - \left( \dfrac{-1}{7t^{7}} + \dfrac{4}{5t^{5}} - \dfrac{6}{3t^{3}} + \dfrac{4}{t} + t \right) + c I=(17cos7x+45cos5x63cos3x+4cosx+cosx)+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

A Former Brilliant Member - 3 years, 11 months ago

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Lol I was typing. So I didn't see. Post the next problem.

Aditya Kumar - 3 years, 11 months ago

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Oh no! I was about to start typing...

Nihar Mahajan - 3 years, 11 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, 11 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, 11 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, 11 months ago

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Problem 13:

Evaluate 0π/2ln(cosx)dx \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 - 3 years, 11 months ago

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Delete this question! This question is on brilliant. See this.

Aditya Kumar - 3 years, 11 months ago

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Solution to Problem 13

Let us start with I=0π/2log(cos(x)) dxI = \int_{0}^{\pi/2} { \log{(\cos{(x)})} \ dx} Using properties we know that I=0π/2log(sin(x)) dxI = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx} Adding both we get 2I=0π/2log(sin(x)cos(x)) dx2I = \int_{0}^{\pi/2} { \log{(\sin{(x)}\cos{(x)})} \ dx} Using little manipulation we get that 2I=0π/2log(sin(2x)) dx0π/2log(2) dx2I = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx} - \int_{0}^{\pi/2} { \log{(2)} \ dx} Using lemma we get 2I=Iπ2log(2)I=π2log(2)2I = I - \dfrac{\pi}2 \log{(2)} \\ \Rightarrow \boxed{I = - \dfrac{\pi}2 \log{(2)} }


Proof of Lemma Used

I1=0π/2log(sin(2x)) dxI_1 = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx} Substitute 2x = t. You will get I1=120πlog(sin(t)) dtI1=12(0π/2log(sin(t)) dt+π/2πlog(sin(x)) dx)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 xπ2=tx - \frac{\pi}2 = t in the 2nd integral. Then using properties and manipulating. we get I1=0π/2log(sin(x)) dxI_1 = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx}

Rajdeep Dhingra - 3 years, 11 months ago

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Problem 12

01ln(x)ln(1x)x(1x)dx\int_0^1 \ln(x)\ln(1-x)x(1-x) dx

Problem has been solved by Harsh Shrivastava.

Aareyan Manzoor - 3 years, 11 months ago

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Simple derivative of beta function.

B(a,b)=01xm1(1x)n1dxB(a,b) = \displaystyle \int_{0}^{1} x^{m-1} (1-x)^{n-1}dx

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

This evaluates toΓ(m)Γ(n)Γ(m+n)(((ψ(m)ψ(m+n))(ψ(n)ψ(m+n))ψ(m+n))\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 0.06840.0684.

Harsh Shrivastava - 3 years, 11 months ago

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This is not a complete answer. Please show how you got 0.0684.

Aditya Kumar - 3 years, 11 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, 11 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)