Brilliant Integration Contest - Season 3

Hi Brilliant! Here is the most awaited Integration contest! See Season 1 and Season 2.

The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows:

  • I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  • You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  • Only make substantial comment that will contribute to the discussion.

  • Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  • If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  • The scope of questions is only computation of integrals either definite or indefinite integrals.

  • You are NOT allowed to post a multiple integrals problem.

  • It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

  • Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest

  • Direct problems on contour integration are not allowed. However, solutions which use contour integration are allowed.

  • Also don't forget to upvote good problems and solutions!

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":

Continue here for part 2 of the contest.

Note by Aditya Kumar
2 years, 9 months ago

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Problem 6:

Prove that 0lnx(x+a)2+b2dx  =  1btan1balna2+b2 \int_0^\infty \frac{\ln x}{(x+a)^2 + b^2}\,dx \; = \; \tfrac{1}{b}\,\tan^{-1}\tfrac{b}{a}\,\ln\sqrt{a^2+b^2} for all a,b>0a,b > 0.

This problem has been solved by Aditya Sharma.

Mark Hennings - 2 years, 9 months ago

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Solution to problem 6 :

Let us denote the integral by II .

I=0lnx(x+a)2+b2dx=0lnx(x+a+ib)(x+aib)\displaystyle I = \int_0^\infty \frac{\ln x}{(x+a)^2+b^2}dx = \int_0^\infty \frac{\ln x}{(x+a+ib)(x+a-ib)}

With the substitution xy=a2+b2xy=a^2+b^2 we have,

I=ln(a2+b2)0dyy2+2ay+a2+b20lny(y+a)2+b2dyI\displaystyle I=\ln(a^2+b^2)\int_0^\infty \frac{dy}{y^2+2ay+a^2+b^2} - \underbrace{\int_0^\infty \frac{\ln y}{(y+a)^2+b^2}dy}_I

I=ln(a2+b2)0dx(x+a)2+b2=1btan1balna2+b2\displaystyle I=\ln(\sqrt{a^2+b^2})\int_0^\infty \frac{dx}{(x+a)^2+b^2} = \frac{1}{b}\tan^{-1}\frac{b}{a}\ln\sqrt{a^2+b^2}.

Aditya Narayan Sharma - 2 years, 9 months ago

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As an alternative to Aditya's proof...

Consider the cut plane C\[0,)\mathbb{C} \backslash [0,\infty), so that 0<Argz<2π0 < \mathrm{Arg}\,z < 2\pi for all zz, and consider the integral of f(z)  =  (logz)2(z+a)2+b2 f(z) \; = \; \frac{(\log z)^2}{(z+a)^2 + b^2} around the keyhole contour γ1+γ2γ3γ4\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4, where

  • γ1\gamma_1 is the line segment z=xe0iz \,=\, x e^{0i}, for ε<x<R\varepsilon < x < R, just above the cut,
  • γ2\gamma_2 is the circular contour z=Reiθz \,=\, Re^{i\theta} for 0<θ<2π0 < \theta < 2\pi,
  • γ3\gamma_3 the the line segment z=xe2πiz \,=\, x e^{2\pi i}, for ε<x<R\varepsilon < x < R, just below the cut,
  • γ4\gamma_4 is the circular contour z=εeiθz \,=\, \varepsilon e^{i\theta} for 0<θ<2π0 < \theta < 2\pi.

where we assume that 0<ε<a2+b2<R0 < \varepsilon < \sqrt{a^2+b^2} < R. Now γ1f(z)dz=εR(lnx)2(x+a)2+b2dxγ3f(z)dz=εR(lnx+2πi)2(x+a)2+b2dx(γ1γ3)f(z)dz=εR4π24πilnx(x+a)2+b2dx\begin{aligned} \int_{\gamma_1} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x)^2}{(x+a)^2 + b^2}\,dx \\ \int_{\gamma_3} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x + 2\pi i)^2}{(x+a)^2 + b^2}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,f(z)\,dz & = \int_{\varepsilon}^R \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \end{aligned} We also note that γ2f(z)dz  =  O(ln2RR)γ4f(z)dz  =  O(εln2ε) \int_{\gamma_2} f(z)\,dz \; = \; O\big(\tfrac{\ln^2R}{R}\big) \hspace{2cm} \int_{\gamma_4} f(z)\,dz \; = \; O\big(\varepsilon \ln^2\varepsilon\big) as RR \to \infty and ε0\varepsilon \to 0. Letting RR \to \infty and ε0\varepsilon \to 0, we deduce that 04π24πilnx(x+a)2+b2dx  =  2πi[Resz=a+ibf(z)+Resz=aibf(z)] \int_0^\infty \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \; = \; 2\pi i\Big[ \mathrm{Res}_{z=-a+ib} f(z) \,+\, \mathrm{Res}_{z=-a-ib} f(z)\Big] Now f(z)f(z) has simple poles at each of a±ib-a \pm ib, and Resz=a±ibf(z)=(lna2+b2+iπitan1ba)2±2ibResz=a+ibf(z)+Resz=aibf(z)=4i(lna2+b2+iπ)tan1ba2ib=2btan1ba(lna2+b2+iπ)\begin{aligned} \mathrm{Res}_{z = -a\pm ib} f(z) & = \frac{\Big(\ln\sqrt{a^2+b^2} + i\pi \mp i\tan^{-1}\frac{b}{a}\Big)^2}{\pm 2ib} \\ \mathrm{Res}_{z = -a+ib} f(z) + \mathrm{Res}_{z = -a-ib} f(z) & = \frac{-4i(\ln\sqrt{a^2+b^2}+i\pi)\tan^{-1}\frac{b}{a}}{2ib} \\ & = - \frac{2}{b}\tan^{-1}\frac{b}{a}(\ln\sqrt{a^2+b^2} + i\pi) \end{aligned} and hence 04π24πilnx(x+a)2+b2dx  =  4πbtan1ba(πilna2+b2) \int_0^\infty \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \; = \; \frac{4\pi}{b}\tan^{-1}\frac{b}{a}\big(\pi - i\ln\sqrt{a^2+b^2}\big) Taking imaginary parts of this equation gives the result.

Mark Hennings - 2 years, 9 months ago

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Problem 16:

Evaluate the integral 01xa1(1x)b1(x+p)a+bdx \int_0^1 \frac{x^{a-1}(1-x)^{b -1}}{(x+p)^{a+b}}\,dx for all a,b,p>0a,b,p > 0.

This problem has been solved by Aditya Sharma.

Mark Hennings - 2 years, 9 months ago

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If we use the substitution, x=1u+1\displaystyle x=\frac{1}{u+1} then

I=0ub1(up+p+1)a+bdu\displaystyle I =\int_0^\infty \frac{u^{b-1}}{(up+p+1)^{a+b}}\,du

Again using the substitution, u=y(p+1p)\displaystyle u=y\left(\frac{p+1}{p}\right) the integral transforms to,

I=1pb(1+p)a0yb1(1+y)a+bdy=(β(a,b)pb(1+p)a)\displaystyle I =\frac{1}{p^b(1+p)^a}\int_0^\infty \frac{y^{b-1}}{(1+y)^{a+b}}\,dy = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right)

Aditya Narayan Sharma - 2 years, 9 months ago

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The substitution z=(p+1)xp+x z = \frac{(p+1)x}{p+x} Is even more direct. Not a hard problem, but a very elegant result.

Mark Hennings - 2 years, 9 months ago

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@Mark Hennings Yes, Although this substitution emerges by combining the chain of substitutions used, or it's tough to put it directly without knowing why it is the reqd. substitution.

Aditya Narayan Sharma - 2 years, 9 months ago

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Problem 14 : Prove That

0lnx[ln(x+12)1x+1ψ(x+12)]dx=ln222+ln2lnπ1 \int_{0}^{\infty} \ln x\left[\ln \left( \dfrac{x+1}{2} \right) - \dfrac{1}{x+1} - \psi \left( \dfrac{x+1}{2} \right) \right] \mathrm{d}x = \dfrac{\ln^2 2}{2}+\ln2\cdot\ln\pi-1

Notation : ψ(x) \psi(x) denotes the Digamma Function.

This problem has been solved by Mark Hennings.

Ishan Singh - 2 years, 9 months ago

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This is a really nice use of previous results of this round of the contest!

Starting with the identity ψ(x)  =  lnx12x20t(t2+x2)(e2πt1)dt \psi(x) \; = \; \ln x - \frac{1}{2x} - 2\int_0^\infty \frac{t}{(t^2+x^2)(e^{2\pi t}-1)}\,dt we deduce that ln(x+12)+1x+1ψ(x+12)=20t(t2+14(x+1)2)(e2πt1)dt=80t(4t2+(x+1)2)(e2πt1)dt=20t(t2+(x+1)2)(eπt1)dt\begin{aligned} \ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big) & = 2\int_0^\infty \frac{t}{\big(t^2 + \frac14(x+1)^2\big)(e^{2\pi t} - 1)}\,dt \\ & = 8\int_0^\infty \frac{t}{(4t^2 + (x+1)^2)(e^{2\pi t} - 1)}\,dt \\ & = 2\int_0^\infty \frac{t}{(t^2 + (x+1)^2)(e^{\pi t}-1)}\,dt \end{aligned} and hence 0lnx(ln(x+12)+1x+1ψ(x+12))dx=20(0lnx(x+1)2+t2dx)teπt1dt=20(1ttan1tlnt2+1)teπt1dt  =  0tan1tln(t2+1)eπt1dt=12ln22+ln2lnπ1\begin{aligned} \int_0^\infty \ln x & \left(\ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big)\right)\,dx \\ & = 2\int_0^\infty \left(\int_0^\infty \frac{\ln x}{(x+1)^2 + t^2}\,dx\right)\,\frac{t}{e^{\pi t}-1}\,dt \\ & = 2\int_0^\infty \left(\frac{1}{t} \tan^{-1}t \ln\sqrt{t^2 + 1}\right)\,\frac{t}{e^{\pi t}-1}\,dt \; = \; \int_0^\infty \frac{\tan^{-1}t \ln(t^2+1)}{e^{\pi t} - 1}\,dt \\ & = \tfrac12\ln^22 + \ln 2 \ln\pi - 1 \end{aligned} using the results of Problems 55 and 66.

Ishan, please set another one!

Mark Hennings - 2 years, 9 months ago

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Yes indeed! This integral arose from my attempt to find an alternate solution of Problem 5 :) The first identity can be proved (using real analysis) by differentiating the identity 0log(1e2aπx)1+x2dx=π[12log(2aπ)+a(loga1)log(Γ(a+1))]\int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right] wrt aa. I have proved it here.

Ishan Singh - 2 years, 9 months ago

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I have posted a question on Math Stack Exchange asking for alternate methods to evaluate the above integral (independently of Problems 55 and 66) here. If we can evaluate it using other methods, it'll also give us an alternate solution for Problem 55.

Ishan Singh - 2 years, 9 months ago

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Problem 1:

Show that: 01cosh(alog(x))log(1+x)xdx=12a(πcsc(πa)1a),   a<1\int_{0}^{1}\frac{\cosh(a\log(x))\log(1+x)}{x}dx=\frac{1}{2a}(\pi \csc(\pi a)-\frac{1}{a}), \;\ a<1

This problem has been solved by Aditya Sharma.

Aditya Kumar - 2 years, 9 months ago

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If we denote the integral by II & using 2cosh(alnx)=xa+xa2cosh(a\ln x)=x^a+x^{-a} & ln(1+x)=n1(1)n1nxn\displaystyle \ln(1+x)=\sum_{n\ge 1}\frac{(-1)^{n-1}}{n}x^n the integral can be transformed into,

I=n1(1)n1n01(xn+a1+xna1)=n1(1)n1n(1na+1n+a)=12an1(1)n1(1na1n+a)\displaystyle I = \sum_{n\ge 1} \frac{(-1)^{n-1}}{n} \int_0^1 (x^{n+a-1}+x^{n-a-1})=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n}(\frac{1}{n-a}+\frac{1}{n+a})=\frac{1}{2a}\sum_{n\ge 1} (-1)^{n-1} (\frac{1}{n-a}-\frac{1}{n+a})

For removing the alternating factor (1)n1(-1)^{n-1} this can be written as ,

2aI=n1(1na1n+a)2n1(12na12n+a)\displaystyle 2a I = \sum_{n\ge 1} (\frac{1}{n-a}-\frac{1}{n+a})-2\sum_{n\ge 1} (\frac{1}{2n-a}-\frac{1}{2n+a})

which simplifies to ,

2aI=n1(1na1n+a1na2+1n+a2)\displaystyle 2a I = \sum_{n\ge 1} (\frac{1}{n-a}-\frac{1}{n+a}-\frac{1}{n-\frac{a}{2}}+\frac{1}{n+\frac{a}{2}})

Using Ha=n1(1n1n+a)\displaystyle H_a = \sum_{n\ge 1}(\frac{1}{n}-\frac{1}{n+a}) we have,

I=12a(HaHaHa2+Ha2)\displaystyle I=\frac{1}{2a}(H_a-H_{-a}-H_{\frac{a}{2}}+H_{-\frac{a}{2}})

Again using reflection formula for harmonic numbers as a<1a<1

i.e. H1aHa=πcot(πa)1a+11a\displaystyle H_{1-a}-H_a = \pi \cot(\pi a)-\frac{1}{a}+\frac{1}{1-a} it simplifies to,

I=12a(πcot(πa)+πcot(πa2)1a)=12a(πcsc(πa)1a)\displaystyle I = \frac{1}{2a} (-\pi\cot(\pi a)+\pi\cot(\pi\frac{a}{2})-\frac{1}{a}) = \frac{1}{2a}(\pi\csc(\pi a)-\frac{1}{a})

Aditya Narayan Sharma - 2 years, 9 months ago

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Problem 2:

Evaluate 01(x2+1)tan13xxdx\large \displaystyle \int_{0}^{1} \frac{(x^{2}+1)\tan^{-1}3x}{x} \, dx.

This problem has been solved by Aditya Sharma.

Rohith M.Athreya - 2 years, 9 months ago

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The integral can be written as : I=01xtan1(3x)dxI1+03tan1xxdxI2\displaystyle I=\underbrace{\int_0^1 x\tan^{-1} (3x) dx}_{I_1}+\underbrace{\int_0^3 \frac{\tan^{-1}x}{x}dx}_{I_2}

A simple integration by parts would yield I1=59tan1316\displaystyle I_1 = \frac{5}{9}\tan^{-1}3 - \frac{1}{6}

I2I_2 is the Inverse Tangent Integral and we represent it by dilogarithms.

Ti2(3)=12i(Li2(3i)Li2(3i))\displaystyle Ti_2(3)=\frac{1}{2i}(Li_2(3i)-Li_2(-3i))

So, I=59tan1316+12i(Li2(3i)Li2(3i))\displaystyle I = \frac{5}{9}\tan^{-1}3 - \frac{1}{6} +\frac{1}{2i}(Li_2(3i)-Li_2(-3i))

Aditya Narayan Sharma - 2 years, 9 months ago

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Problem 5:

Prove that:

0ln(x2+1)arctanxeπx1dx=ln222+ln2lnπ1.{\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\arctan x}{e^{\pi x}-1}dx=\frac{\ln^22}2+\ln2\cdot\ln\pi-1.

This problem has been solved by Mark Hennings.

Aditya Kumar - 2 years, 9 months ago

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Solution of Problem 5

Starting with the Hermite formula for the generalized Riemann zeta function: ζ(s,a)  =  12as+a1s(s1)1+20(a2+y2)12ssin(stan1ya)e2πy1dy \zeta(s,a) \; = \; \tfrac12a^{-s} + a^{1-s}(s-1)^{-1} + 2\int_0^\infty \frac{(a^2+y^2)^{-\frac12s} \, \sin\big(s \tan^{-1}\frac{y}{a}\big)}{e^{2\pi y} - 1}\,dy we have ζ(s,12)=2s1(1+1s1)+20(14+y2)12ssin(stan12y)e2πy1dy=2s1ss1+2s+10(1+4y2)12ssin(tan12y)e2πy1dy=2s1ss1+2s0(1+t2)12ssin(stan1t)eπt1dt \begin{aligned} \zeta(s,\tfrac12) & = 2^{s-1}\left(1 + \tfrac{1}{s-1}\right) + 2\int_0^\infty \frac{(\frac14 + y^2)^{-\frac12s}\,\sin\big(s \tan^{-1}2y\big)}{e^{2\pi y}-1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^{s+1}\int_0^\infty \frac{(1 + 4y^2)^{-\frac12s} \, \sin\big(\tan^{-1}2y\big)}{e^{2\pi y} - 1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^s \int_0^\infty \frac{(1+t^2)^{-\frac12s}\,\sin\big(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \end{aligned} so that F(s)  =  0(1+t2)12ssin(stan1t)eπt1dt  =  2sζ(s,12)s2(s1) F(s) \; = \; \int_0^\infty \frac{(1+t^2)^{-\frac12s}\,\sin(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \; = \; 2^{-s}\zeta(s,\tfrac12) - \frac{s}{2(s-1)} Now since F(s)=2s20(1+t2)12ssin(stan1t)eπt1dt=0(1+t2)12seπt1{14(ln(1+t2))2sin(stan1t)ln(1+t2)tan1tcos(stan1t)(tan1t)2sin(stan1t)}dt \begin{aligned} F''(s) & = \frac{\partial^2}{\partial s^2}\int_0^\infty \frac{(1 + t^2)^{-\frac12s} \, \sin(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \\ & = \int_0^\infty \frac{(1+t^2)^{-\frac12s}}{e^{\pi t} - 1}\left\{ \begin{array}{l} \tfrac14\big(\ln(1+t^2)\big)^2 \sin(s\tan^{-1}t) - \ln(1 + t^2) \tan^{-1}t \cos\big(s \tan^{-1}t\big) \\ - \big(\tan^{-1}t\big)^2 \sin\big(s \tan^{-1}t\big)\end{array}\right\}\,dt \end{aligned} we deduce that F(0)  =  0ln(1+t2)tan1teπt1dt F''(0) \; = \; -\int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt and hence that the desired integral is 0ln(1+t2)tan1teπt1dt  =  F(0)  =  2s2[s2(s1)2sζ(s,12)]s=0 \int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt \; = \; -F''(0) \; = \; \frac{\partial^2}{\partial s^2}\Big[\tfrac{s}{2(s-1)} - 2^{-s}\zeta(s,\tfrac12)\Big] \Big|_{s=0} which simplifies to give 0ln(1+t2)tan1teπt1dt  =  112ln22+ln2ln2π  =  1+12ln22+ln2lnπ \int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt \; = \; -1 - \tfrac12\ln^22 + \ln2 \cdot \ln2\pi \; = \; -1 + \tfrac12\ln^22 + \ln2\ln\pi

Mark Hennings - 2 years, 9 months ago

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Problem 8:

Show that 012πln2(asinθ)dθ  =  124π3+16π[ln2(2a)2ln(2a)ln(a2)] \int_0^{\frac12\pi} \ln^2\big(a\sin\theta\big)\,d\theta \; = \; \tfrac{1}{24}\pi^3 + \tfrac{1}{6}\pi\Big[\ln^2\big(\tfrac{2}{a}\big) - 2\ln\big(\tfrac{2}{a}\big)\ln\big(\tfrac{a}{2}\big)\Big] for any a>0a > 0.

This problem has been solved by Jasper Braun.

Mark Hennings - 2 years, 9 months ago

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Solution to Problem 8: I=0π2ln2a+2lnalnsinx+ln2sinxdx=πln2a2lnaπln2+0π2ln2sinxdx\displaystyle I=\int_0^\frac{\pi}{2}\ln^2{a}+2\ln{a}\ln{\sin{x}}+\ln^2{\sin{x}}dx=\frac{\pi\ln^2{a}}{2}-\ln{a}\pi\ln{2}+\int_0^\frac{\pi}{2}\ln^2{\sin{x}}dx

0π2ln2sinxdx=01ln2x1x2dx(by u=sinx)\displaystyle\int_0^\frac{\pi}{2}\ln^2{\sin{x}}dx = \int_0^1\frac{\ln^2{x}}{\sqrt{1-x^2}}dx\quad\text{(by }u= \sin{x})

Now using B(m,n)=201x2m1(1x2)n1dx and thus 01ln2x1x2dx=d2dm2B(m,12)2412\displaystyle\text{Now using }B(m,n) = 2\int_0^1x^{2m-1}(1-x^2)^{n-1}dx \text{ and thus }\int_0^1\frac{\ln^2{x}}{\sqrt{1-x^2}}dx = \frac{d^2}{dm^2}\frac{B(m,\frac{1}{2})}{2*4}\big|_\frac{1}{2}

Using Γ(m)=Γ(m)ψ(m),Γ(m)=Γ(m)ψ(m)2+Γ(m)ψ(1)(m)\displaystyle\text{Using }\Gamma'(m) = \Gamma(m)\psi(m),\qquad\Gamma''(m) = \Gamma(m)\psi(m)^2+\Gamma(m)\psi^{(1)}(m)

π8d2dm2Γ(m)Γ(m+12)12=πln222+πζ(2)4\displaystyle\frac{\sqrt{\pi}}{8}\frac{d^2}{dm^2}\frac{\Gamma(m)}{\Gamma(m+\frac{1}{2})}\Big|_\frac{1}{2} = \frac{\pi\ln^2{2}}{2}+\frac{\pi\zeta(2)}{4}

I=π324+πln222+πln2a2πlnaln2\displaystyle I=\boxed{\frac{\pi^3}{24}+\frac{\pi\ln^2{2}}{2}+\frac{\pi\ln^2{a}}{2}-\pi\ln{a}\ln{2}}

Jasper Braun - 2 years, 9 months ago

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Problem 13:

Evaluate 02πecosθcos(nθsinθ)dθ \int_0^{2\pi} e^{\cos\theta} \cos\big(n\theta - \sin\theta\big)\,d\theta for any integer nn.

This problem has been solved by Ishan Singh.

Mark Hennings - 2 years, 9 months ago

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It's simplest using contour integration...

Note that 02πecosθcos(sinθnθ)dθ  =  Re(02πecosθei(sinθnθ)dθ)  =  Re(02πeeiθinθdθ)\int_0^{2\pi} e^{\cos\theta}\,\cos(\sin\theta - n\theta)\,d\theta \; = \; \mathfrak{Re}\left(\int_0^{2\pi} e^{\cos\theta} e^{i(\sin\theta - n\theta)}\,d\theta\right) \; = \; \mathfrak{Re}\left(\int_0^{2\pi} e^{e^{i\theta} - in\theta}\,d\theta\right) With the substitution z=eiθz = e^{i\theta}, this becomes the contour integral Re(z=1ezzndziz)  =  Re(1iz=1ezzn+1dz)  =  2πRe(Resz=0ezzn+1)  =  {2πn!n00n<0 \mathfrak{Re}\left(\int_{|z|=1}e^z z^{-n} \frac{dz}{iz}\right) \; = \; \mathfrak{Re}\left(\frac{1}{i}\int_{|z|=1}\frac{e^z}{z^{n+1}}\,dz\right) \; = \; 2\pi\mathfrak{Re}\left(\mathrm{Res}_{z=0} \frac{e^z}{z^{n+1}}\right) \; = \; \left\{ \begin{array}{ll} \frac{2\pi}{n!} & n \ge 0 \\ 0 & n < 0 \end{array} \right.

Mark Hennings - 2 years, 9 months ago

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I=02πecostcos(ntsint) dt=(02πeinte(costisint) dt) \text{I} = \int_{0}^{2 \pi} e^{\cos t} \cos (nt - \sin t) \ \mathrm{d}t = \Re{ \left(\int_{0}^{2 \pi} e^{i n t} \cdot e^{(\cos t - i \sin t)} \ \mathrm{d}t \right)}

=(02πeinteeit dt) = \Re{ \left(\int_{0}^{2 \pi} e^{i n t} \cdot e^{\Large e^{ -i t}} \ \mathrm{d}t \right)}

=2πn!+(r0 ; rn(1r!02πeit(nr) dt)) = \dfrac{2 \pi}{n!} + \Re{ \left(\sum_{r \geq 0 \ ; \ r \neq n} \left( \dfrac{1}{r!} \int_{0}^{2 \pi} e^{i t (n-r)} \ \mathrm{d}t \right) \right)}

Since last integral is 00, we have,

I=2πn! \text{I} = \dfrac{2 \pi}{n!}

If n n is negative, we can write n=k ; kZ+n = -k \ ; \ k \in \mathbb{Z}^+, so that,

J=02πecostcos(kt+sint) dt \text{J} = \int_{0}^{2 \pi} e^{\cos t} \cos (kt + \sin t) \ \mathrm{d}t

By the above method, we can similarly show that J=0 \text{J} = 0 .

Therefore,

02πecostcos(ntsint) dt={2πn!n00n<0 \int_{0}^{2 \pi} e^{\cos t} \cos (nt - \sin t) \ \mathrm{d}t = \left\{ \begin{array}{ll} \frac{2\pi}{n!} & n \ge 0 \\ 0 & n < 0 \end{array} \right.

Ishan Singh - 2 years, 9 months ago

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Problem 03:

Show that 01lnxln2(1+x)xdx=π4244Li4(12)7ζ(3)ln22+π2ln226ln426\displaystyle \int_0^1 \frac{\ln x \ln^2(1+x)}{x}\, dx = \dfrac{\pi^4}{24}-4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}.

This problem has been solved by Jasper Braun.

Aditya Narayan Sharma - 2 years, 9 months ago

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Solution to Problem 3:

Hn(k)=j=1Hn(k)xiinH(x)=ln(1x)1x\displaystyle H_n^{(k)} = \sum_{j=1}^{\infty}\frac{H_n^{(k)}x^i}{i^n}\quad H(x) = \frac{-\ln{(1-x)}}{1-x}

I=0ln2(1+x)lnxxdx=01ln(1+x)ln2x1+xdx=by parts\displaystyle I = \int_{0}^{\infty}\frac{\ln^2{(1+x)}\ln{x}}{x}dx = -\int_{0}^{1}\frac{\ln{(1+x)}\ln^2{x}}{1+x}dx=\quad\text{by parts}

12ln2(x1)lnxxdx=(by u=1+x)\displaystyle -\int_{1}^{2}\frac{\ln^2{(x-1)}\ln{x}}{x}dx=\quad\text{(by } u = 1+x)

121ln2(1x1)ln1xxdx=121lnxln2(1x)x2ln2xln(1t)x+ln3xxdx(by u=1/x)\displaystyle -\int_{\frac{1}{2}}^{1}\frac{\ln^2{(\frac{1}{x}-1)}\ln{\frac{1}{x}}}{x}dx=\int_{\frac{1}{2}}^{1}\frac{\ln{x}\ln^2{(1-x)}}{x}-\frac{2\ln^2{x}\ln{(1-t)}}{x}+\frac{\ln^3{x}}{x}dx\quad\text{(by }u = 1/x)

lnxln2(1x)x=ln2x2ln2(1x)+ln2xln(1x)1xdx\displaystyle\int\frac{\ln{x}\ln^2{(1-x)}}{x} = \frac{\ln^2{x}}{2}\ln^2{(1-x)}+\int\frac{\ln^2{x}\ln{(1-x)}}{1-x}dx

ln2xln(1x)1xdx=n=1Hnd2dn2xndx=n=1Hn(xn+1ln2xn+12xn+1lnx(1+n)2+2xn+1(1+n)3)=\displaystyle \int\frac{\ln^2{x}\ln{(1-x)}}{1-x}dx = -\sum_{n=1}^{\infty}H_n\frac{d^2}{dn^2}\int x^ndx = -\sum_{n=1}^{\infty}H_n\big(\frac{x^{n+1}\ln^2{x}}{n+1}-\frac{2x^{n+1}\ln{x}}{(1+n)^2}+\frac{2x^{n+1}}{(1+n)^3}\big)=

Now use Hn=Hn+11n+1H_n = H_{n+1} - \frac{1}{n+1} to obtain:

(H1(x)ln2xLi2(x)ln2x2H2(x)lnx+2Li3(x)lnx+2H3(x)2Li4(x))\displaystyle -\Big(H_1(x)\ln^2{x}-Li_2(x)\ln^2{x}-2H_2(x)\ln{x}+2Li_3(x)\ln{x}+2H_3(x)-2Li_4(x)\Big)

Next 2ln2xln(1x)xdx=2n=11nd2dn2xn1=\displaystyle\textbf{Next }-2\int\frac{\ln^2{x}\ln{(1-x)}}{x}dx = 2\sum_{n=1}^{\infty}\frac{1}{n}\frac{d^2}{dn^2}\int x^{n-1} =

2n=1xnln2xn22xnlnxn3+2xnn4=2(Li2(x)ln2x2Li3(x)lnx+2Li4(x))\displaystyle 2\sum_{n=1}^{\infty}\frac{x^n\ln^2{x}}{n^2}-\frac{2x^n\ln{x}}{n^3}+\frac{2x^n}{n^4} = 2\big(Li_2(x)\ln^2{x}-2Li_3(x)\ln{x}+2Li_4(x)\big)

I=ln4x4121+ln2x2ln2(1x)1212H3(1)+2Li4(1)+\displaystyle I = \frac{\ln^4{x}}{4}\Big|^1_{\frac{1}{2}} + \frac{\ln^2{x}}{2}\ln^2{(1-x)}\Big|^1_{\frac{1}{2}}-2H_3(1)+2Li_4(1)+

+H1(12)ln22Li2(12)ln22+2H2(12)ln22Li3(12)ln2+2H3(12)2Li4(12)+\displaystyle +H_1(\frac{1}{2})\ln^2{2}-Li_2(\frac{1}{2})\ln^2{2}+2H_2(\frac{1}{2})\ln{2}-2Li_3(\frac{1}{2})\ln{2}+2H_3(\frac{1}{2})-2Li_4(\frac{1}{2})+

4Li4(1)2Li2(12)ln224Li3(12)ln24Li4(12)=\displaystyle4Li_4(1)-2Li_2(\frac{1}{2})\ln^2{2}-4Li_3(\frac{1}{2})\ln{2}-4Li_4(\frac{1}{2}) =

3ln4242H3(1)+H1(12)ln22+2H2(12)ln2+2H3(12)6Li4(12)+6Li4(1)3Li2(12)ln226Li3(12)ln2\displaystyle\frac{-3ln^4{2}}{4}-2H_3(1)+H_1(\frac{1}{2})\ln^2{2}+2H_2(\frac{1}{2})\ln{2}+2H_3(\frac{1}{2})-6Li_4(\frac{1}{2})+6Li_4(1)-3Li_2(\frac{1}{2})\ln^2{2}-6Li_3(\frac{1}{2})\ln{2}

Now using this for the evaluation of Hn(x)H_n(x)

2H3(1)=π436,ln22H1(12)=ln422+ln22Li212,2ln2H212=2ln2ζ(3)ζ(2)ln22,2H3(12)=π4360+ln4212ζ(3)ln24+2Li4(12)\displaystyle -2H_3(1) = \frac{-\pi^4}{36},\ln^2{2}H_1(\frac{1}{2})=\frac{\ln^4{2}}{2}+\ln^2{2}Li_2{\frac{1}{2}},2\ln{2}H_2{\frac{1}{2}} = 2\ln{2}\zeta(3)-\zeta(2)\ln^2{2},2H_3(\frac{1}{2})= \frac{\pi^4}{360}+\frac{\ln^4{2}}{12}-\frac{\zeta(3)\ln{2}}{4}+2Li_4(\frac{1}{2})

Adding all this together and using formulas for Li3Li_3 and Li2Li_2(I have checked on a calculator) gets the desired answer. Anyone may post the next problem.

Jasper Braun - 2 years, 9 months ago

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Problem 4:

Prove that 01Li32(x)x2dx=916ζ2(3)34ζ(2)ζ(3)14ζ2(2)+74ζ(4)3ζ(3)ln2+6ζ(3)6ζ(2)ln2+6ζ(2)12ln22\displaystyle \int_0^1 \frac{{\rm Li}_3^2(-x)}{x^2}\, dx = -\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{4}\zeta^2(2)+\frac{7}{4}\zeta(4)-3\zeta(3)\ln 2+6\zeta(3)-6\zeta(2)\ln 2+6\zeta(2)-12\ln^2 2

This problem has been solved by Aditya Kumar.

Aditya Narayan Sharma - 2 years, 9 months ago

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Solution to Problem 4:

I=01Li32(x)x2dx\displaystyle I=\int^1_0\frac{Li_3^2(-x)}{x^2}{\rm d}x

I=Li32(x)x01+012Li2(x)Li3(x)x2dx=916ζ2(3)2Li2(x)Li3(x)x01+012Li22(x)x2dx012Li3(x)ln(1+x)x2dx I=-\frac{Li_3^2(-x)}{x}\Bigg |^1_0+\int^1_0\frac{2Li_2(-x)Li_3(-x)}{x^2}{\rm d}x=-\frac{9}{16}\zeta^2(3)-\frac{2Li_2(-x)Li_3(-x)}{x}\Bigg |^1_0+\int^1_0\frac{2Li_2^2(-x)}{x^2}{\rm d}x-\int^1_0\frac{2Li_3(-x)\ln(1+x)}{x^2}{\rm d}x

I=916ζ2(3)34ζ(2)ζ(3)2Li22(x)x01014Li2(x)ln(1+x)x2dx+2Li3(x)ln(1+x)x01012Li3(x)x(1+x)dx012Li2(x)ln(1+x)x2dx I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{2Li_2^2(-x)}{x}\Bigg |^1_0-\int^1_0\frac{4Li_2(-x)\ln(1+x)}{x^2}{\rm d}x+\frac{2Li_3(-x)\ln(1+x)}{x}\Bigg |^1_0-\int^1_0\frac{2Li_3(-x)}{x(1+x)}{\rm d}x-\int^1_0\frac{2Li_2(-x)\ln(1+x)}{x^2}{\rm d}x

I=916ζ2(3)34ζ(2)ζ(3)12ζ2(2)+74ζ(4)32ζ(3)ln2+012Li3(x)1+xdx+6Li2(x) ln(1+x)x01016Li2(x)x(1+x)dx+016ln2(1+x)x2dx I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x+\frac{6Li_2(-x) \ \ln(1+x)}{x}\Bigg |^1_0-\int^1_0\frac{6Li_2(-x)}{x(1+x)}{\rm d}x+\int^1_0\frac{6\ln^2(1+x)}{x^2}{\rm d}x