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!
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Top NewestProblem 6:
Prove that ∫0∞(x+a)2+b2lnxdx=b1tan−1ablna2+b2 for all a,b>0.
This problem has been solved by Aditya Sharma.
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Solution to problem 6 :
Let us denote the integral by I .
I=∫0∞(x+a)2+b2lnxdx=∫0∞(x+a+ib)(x+a−ib)lnx
With the substitution xy=a2+b2 we have,
I=ln(a2+b2)∫0∞y2+2ay+a2+b2dy−I∫0∞(y+a)2+b2lnydy
I=ln(a2+b2)∫0∞(x+a)2+b2dx=b1tan−1ablna2+b2.
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As an alternative to Aditya's proof...
Consider the cut plane C\[0,∞), so that 0<Argz<2π for all z, and consider the integral of f(z)=(z+a)2+b2(logz)2 around the keyhole contour γ1+γ2−γ3−γ4, where
where we assume that 0<ε<a2+b2<R. Now ∫γ1f(z)dz∫γ3f(z)dz(∫γ1−∫γ3)f(z)dz=∫εR(x+a)2+b2(lnx)2dx=∫εR(x+a)2+b2(lnx+2πi)2dx=∫εR(x+a)2+b24π2−4πilnxdx We also note that ∫γ2f(z)dz=O(Rln2R)∫γ4f(z)dz=O(εln2ε) as R→∞ and ε→0. Letting R→∞ and ε→0, we deduce that ∫0∞(x+a)2+b24π2−4πilnxdx=2πi[Resz=−a+ibf(z)+Resz=−a−ibf(z)] Now f(z) has simple poles at each of −a±ib, and Resz=−a±ibf(z)Resz=−a+ibf(z)+Resz=−a−ibf(z)=±2ib(lna2+b2+iπ∓itan−1ab)2=2ib−4i(lna2+b2+iπ)tan−1ab=−b2tan−1ab(lna2+b2+iπ) and hence ∫0∞(x+a)2+b24π2−4πilnxdx=b4πtan−1ab(π−ilna2+b2) Taking imaginary parts of this equation gives the result.
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Problem 16:
Evaluate the integral ∫01(x+p)a+bxa−1(1−x)b−1dx for all a,b,p>0.
This problem has been solved by Aditya Sharma.
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If we use the substitution, x=u+11 then
I=∫0∞(up+p+1)a+bub−1du
Again using the substitution, u=y(pp+1) the integral transforms to,
I=pb(1+p)a1∫0∞(1+y)a+byb−1dy=(pb(1+p)aβ(a,b))
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The substitution z=p+x(p+1)x Is even more direct. Not a hard problem, but a very elegant result.
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Problem 14 : Prove That
∫0∞lnx[ln(2x+1)−x+11−ψ(2x+1)]dx=2ln22+ln2⋅lnπ−1
Notation : ψ(x) denotes the Digamma Function.
This problem has been solved by Mark Hennings.
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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)=lnx−2x1−2∫0∞(t2+x2)(e2πt−1)tdt we deduce that ln(2x+1)+x+11−ψ(2x+1)=2∫0∞(t2+41(x+1)2)(e2πt−1)tdt=8∫0∞(4t2+(x+1)2)(e2πt−1)tdt=2∫0∞(t2+(x+1)2)(eπt−1)tdt and hence ∫0∞lnx(ln(2x+1)+x+11−ψ(2x+1))dx=2∫0∞(∫0∞(x+1)2+t2lnxdx)eπt−1tdt=2∫0∞(t1tan−1tlnt2+1)eπt−1tdt=∫0∞eπt−1tan−1tln(t2+1)dt=21ln22+ln2lnπ−1 using the results of Problems 5 and 6.
Ishan, please set another one!
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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 ∫0∞1+x2log(1−e−2aπx)dx=π[21log(2aπ)+a(loga−1)−log(Γ(a+1))] wrt a. I have proved it here.
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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 5 and 6) here. If we can evaluate it using other methods, it'll also give us an alternate solution for Problem 5.
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Problem 1:
Show that: ∫01xcosh(alog(x))log(1+x)dx=2a1(πcsc(πa)−a1), a<1
This problem has been solved by Aditya Sharma.
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If we denote the integral by I & using 2cosh(alnx)=xa+x−a & ln(1+x)=n≥1∑n(−1)n−1xn the integral can be transformed into,
I=n≥1∑n(−1)n−1∫01(xn+a−1+xn−a−1)=n≥1∑n(−1)n−1(n−a1+n+a1)=2a1n≥1∑(−1)n−1(n−a1−n+a1)
For removing the alternating factor (−1)n−1 this can be written as ,
2aI=n≥1∑(n−a1−n+a1)−2n≥1∑(2n−a1−2n+a1)
which simplifies to ,
2aI=n≥1∑(n−a1−n+a1−n−2a1+n+2a1)
Using Ha=n≥1∑(n1−n+a1) we have,
I=2a1(Ha−H−a−H2a+H−2a)
Again using reflection formula for harmonic numbers as a<1
i.e. H1−a−Ha=πcot(πa)−a1+1−a1 it simplifies to,
I=2a1(−πcot(πa)+πcot(π2a)−a1)=2a1(πcsc(πa)−a1)
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Problem 5:
Prove that:
∫0∞eπx−1ln(x2+1)arctanxdx=2ln22+ln2⋅lnπ−1.
This problem has been solved by Mark Hennings.
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Solution of Problem 5
Starting with the Hermite formula for the generalized Riemann zeta function: ζ(s,a)=21a−s+a1−s(s−1)−1+2∫0∞e2πy−1(a2+y2)−21ssin(stan−1ay)dy we have ζ(s,21)=2s−1(1+s−11)+2∫0∞e2πy−1(41+y2)−21ssin(stan−12y)dy=2s−1s−1s+2s+1∫0∞e2πy−1(1+4y2)−21ssin(tan−12y)dy=2s−1s−1s+2s∫0∞eπt−1(1+t2)−21ssin(stan−1t)dt so that F(s)=∫0∞eπt−1(1+t2)−21ssin(stan−1t)dt=2−sζ(s,21)−2(s−1)s Now since F′′(s)=∂s2∂2∫0∞eπt−1(1+t2)−21ssin(stan−1t)dt=∫0∞eπt−1(1+t2)−21s{41(ln(1+t2))2sin(stan−1t)−ln(1+t2)tan−1tcos(stan−1t)−(tan−1t)2sin(stan−1t)}dt we deduce that F′′(0)=−∫0∞eπt−1ln(1+t2)tan−1tdt and hence that the desired integral is ∫0∞eπt−1ln(1+t2)tan−1tdt=−F′′(0)=∂s2∂2[2(s−1)s−2−sζ(s,21)]∣∣∣s=0 which simplifies to give ∫0∞eπt−1ln(1+t2)tan−1tdt=−1−21ln22+ln2⋅ln2π=−1+21ln22+ln2lnπ
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Problem 8:
Show that ∫021πln2(asinθ)dθ=241π3+61π[ln2(a2)−2ln(a2)ln(2a)] for any a>0.
This problem has been solved by Jasper Braun.
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Solution to Problem 8: I=∫02πln2a+2lnalnsinx+ln2sinxdx=2πln2a−lnaπln2+∫02πln2sinxdx
∫02πln2sinxdx=∫011−x2ln2xdx(by u=sinx)
Now using B(m,n)=2∫01x2m−1(1−x2)n−1dx and thus ∫011−x2ln2xdx=dm2d22∗4B(m,21)∣∣21
Using Γ′(m)=Γ(m)ψ(m),Γ′′(m)=Γ(m)ψ(m)2+Γ(m)ψ(1)(m)
8πdm2d2Γ(m+21)Γ(m)∣∣∣21=2πln22+4πζ(2)
I=24π3+2πln22+2πln2a−πlnaln2
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Problem 13:
Evaluate ∫02πecosθcos(nθ−sinθ)dθ for any integer n.
This problem has been solved by Ishan Singh.
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It's simplest using contour integration...
Note that ∫02πecosθcos(sinθ−nθ)dθ=Re(∫02πecosθei(sinθ−nθ)dθ)=Re(∫02πeeiθ−inθdθ) With the substitution z=eiθ, this becomes the contour integral Re(∫∣z∣=1ezz−nizdz)=Re(i1∫∣z∣=1zn+1ezdz)=2πRe(Resz=0zn+1ez)={n!2π0n≥0n<0
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I=∫02πecostcos(nt−sint) dt=ℜ(∫02πeint⋅e(cost−isint) dt)
=ℜ(∫02πeint⋅ee−it dt)
=n!2π+ℜ⎝⎛r≥0 ; r=n∑(r!1∫02πeit(n−r) dt)⎠⎞
Since last integral is 0, we have,
I=n!2π
If n is negative, we can write n=−k ; k∈Z+, so that,
J=∫02πecostcos(kt+sint) dt
By the above method, we can similarly show that J=0.
Therefore,
∫02πecostcos(nt−sint) dt={n!2π0n≥0n<0
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Problem 2:
Evaluate ∫01x(x2+1)tan−13xdx.
This problem has been solved by Aditya Sharma.
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The integral can be written as : I=I1∫01xtan−1(3x)dx+I2∫03xtan−1xdx
A simple integration by parts would yield I1=95tan−13−61
I2 is the Inverse Tangent Integral and we represent it by dilogarithms.
Ti2(3)=2i1(Li2(3i)−Li2(−3i))
So, I=95tan−13−61+2i1(Li2(3i)−Li2(−3i))
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Problem 03:
Show that ∫01xlnxln2(1+x)dx=24π4−4Li4(21)−27ζ(3)ln2+6π2ln22−6ln42.
This problem has been solved by Jasper Braun.
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Solution to Problem 3:
Hn(k)=j=1∑∞inHn(k)xiH(x)=1−x−ln(1−x)
I=∫0∞xln2(1+x)lnxdx=−∫011+xln(1+x)ln2xdx=by parts
−∫12xln2(x−1)lnxdx=(by u=1+x)
−∫211xln2(x1−1)lnx1dx=∫211xlnxln2(1−x)−x2ln2xln(1−t)+xln3xdx(by u=1/x)
∫xlnxln2(1−x)=2ln2xln2(1−x)+∫1−xln2xln(1−x)dx
∫1−xln2xln(1−x)dx=−n=1∑∞Hndn2d2∫xndx=−n=1∑∞Hn(n+1xn+1ln2x−(1+n)22xn+1lnx+(1+n)32xn+1)=
Now use Hn=Hn+1−n+11 to obtain:
−(H1(x)ln2x−Li2(x)ln2x−2H2(x)lnx+2Li3(x)lnx+2H3(x)−2Li4(x))
Next −2∫xln2xln(1−x)dx=2n=1∑∞n1dn2d2∫xn−1=
2n=1∑∞n2xnln2x−n32xnlnx+n42xn=2(Li2(x)ln2x−2Li3(x)lnx+2Li4(x))
I=4ln4x∣∣∣211+2ln2xln2(1−x)∣∣∣211−2H3(1)+2Li4(1)+
+H1(21)ln22−Li2(21)ln22+2H2(21)ln2−2Li3(21)ln2+2H3(21)−2Li4(21)+
4Li4(1)−2Li2(21)ln22−4Li3(21)ln2−4Li4(21)=
4−3ln42−2H3(1)+H1(21)ln22+2H2(21)ln2+2H3(21)−6Li4(21)+6Li4(1)−3Li2(21)ln22−6Li3(21)ln2
Now using this for the evaluation of Hn(x)
−2H3(1)=36−π4,ln22H1(21)=2ln42+ln22Li221,2ln2H221=2ln2ζ(3)−ζ(2)ln22,2H3(21)=360π4+12ln42−4ζ(3)ln2+2Li4(21)
Adding all this together and using formulas for Li3 and Li2(I have checked on a calculator) gets the desired answer. Anyone may post the next problem.
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Problem 4:
Prove that ∫01x2Li32(−x)dx=−169ζ2(3)−43ζ(2)ζ(3)−41ζ2(2)+47ζ(4)−3ζ(3)ln2+6ζ(3)−6ζ(2)ln2+6ζ(2)−12ln22
This problem has been solved by Aditya Kumar.
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Solution to Problem 4:
I=∫01x2Li32(−x)dx
I=−xLi32(−x)∣∣∣∣∣01+∫01x22Li2(−x)Li3(−x)dx=−169ζ2(3)−x2Li2(−x)Li3(−x)∣∣∣∣∣01+∫01x22Li22(−x)dx−∫01x22Li3(−x)ln(1+x)dx
I=−169ζ2(3)−43ζ(2)ζ(3)−x2Li22(−x)∣∣∣∣∣01−∫01x24Li2(−x)ln(1+x)dx+x2Li3(−x)ln(1+x)∣∣∣∣∣01−∫01x(1+x)2Li3(−x)dx−∫01x22Li2(−x)ln(1+x)dx
I=−169ζ2(3)−43ζ(2)ζ(3)−21ζ2(2)+47ζ(4)−23