This is Brilliant Integration Contest - Season 1 (Part 2) as a continuation of the previous contest (Part 1). There is a major change in the rules of contest, so please read all of them carefully before take part in this contest.
I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. 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 (s)he 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.
- Please make a substantial comment.
- Make sure you know how to solve your own problem before posting it in case there is no one can answer it within a week, 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 as well as a complex integral problem.
- 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, Dirichlet eta function, dilogarithm, digamma function, and trigonometric integral.
Format your post is as follows:
SOLUTION OF PROBLEM xxx (number of problem) :
[Post your solution here]
PROBLEM xxx (number of problem) :
[Post your problem here]
Remember, put them separately.
Please share this note so that lots of users here know this contest and take part in it. (>‿◠)✌
Okay, let the contest part 2 begin!
P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 3).
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Top NewestThanks for doing this. There is a lot to learn from these integration questions that you have shared.
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Thank you for your help. You're too kind to me. I really appreciate it ⌣¨
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Problem 20
∫01xsinh−1(x)−log[(2−1)x+1]dx=log(2)log(1+2)−24π2
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Solving this problem is really tedious job. Honestly, I'm unwilling to answer it too (even if I know how to solve it). I don't know what is the OP's motivation by posting two well-defined integrals in a single problem. If the OP has an elementary and a clever method than mine, please do share to us. Okay, here is an attempt using a cannon.
Let split the integral into two parts
I−J=∫01xarcsinhxdx−∫01xln(1+(2−1)x)dx
Perform integration by parts for I by taking u=arcsinhx and dv=xdx.
∫01xarcsinhxdxI=arcsinhxlnx∣∣∣∣01−∫011+x2lnxdx⇒x=tant=∫0π/4costln(cost)−ln(sint)dt=∫0π/4costln(cost)dt−∫0π/4costln(sint)dt=21∫0π/4costln(1−sin2t)dt−∫0π/4costln(sint)dt
Putting y=sint and a=21, we get
∫01xarcsinhxdx=∫0a1−y2ln(1−y2)dy−∫0a1−y2lnydy=∫0a(1−y)(1+y)ln(1−y)+ln(1+y)dy−∫0a(1−y)(1+y)lnydy=I1+I2
Performing partial fractions decomposition we get
I1=21∫0a1+yln(1+y)dy+21∫0a1+yln(1−y)dy+21∫0a1−yln(1+y)dy+21∫0a1−yln(1−y)dy=4ln2(1+a)+21∫0a1+yln(1−y)dy+21∫0a1−yln(1+y)dy−4ln2(1−a)=41ln2(1−a1+a)+21∫0a1+yln(1−y)dy+21∫0a1−yln(1+y)dy⇒2z=1+y=ln2(1+2)+21∫cbzln(2−2z)dz+21∫cb1−zlnzdz=ln2(1+2)+2ln2∫cbzdz+21∫cbzln(1−z)dz+21∫1−c1−bzln(1−z)dz=ln2(1+2)+2ln2ln(21+2)+21∫cbzln(1−z)dz+21∫1−c1−bzln(1−z)dz
and
I2=21∫0a1−ylnydy+21∫0a1+ylnydy=21∫11−ayln(1−y)dy+2lnyln(1+y)∣∣∣∣0a−∫0ayln(1+y)dy=21∫11−ayln(1−y)dy+2lnaln(1+a)−∫0ayln(1+y)dy=21∫11−ayln(1−y)dy−∫0ayln(1+y)dy−4ln2ln(21+2)
where b=221+2 and c=21.
Now, let us evaluate J. Set x=t2, we get
J=∫01xln(1+(2−1)x)dx=2∫01tln(1+(2−1)t)dt
Here is the cannon, recall a special function dilogarithm.
Li2(z)=−∫0ztln(1−t)dt
Hence, the rest integrals can be easily evaluated by using dilogarithm and an elementary substitution, i.e. t=kx, where k is a constant. We may also utilize these identities Li2(z)+Li2(−z)Li2(1−z)+Li2(1−z1)Li2(z)+Li2(1−z)=21Li2(z2)=−2ln2z=6π2−lnz⋅ln(1−z) and special values Li2(−1)Li2(0)Li2(21)Li2(1)=−12π2=0=12π2−2ln22=6π2 Performing a cumbersome and a tedious calculation, we will get the announced result.
I hope you understand my feelings while trying to solve and to write it down. So, please do not ever post a problem like this again. LOL
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The reason for taking the difference of the two integrals is that the result is much simpler than the two integrals separately (which involve dilogarithms), which I found beautiful. It only works if both terms are exactly as they are (including the weird factor (2−1) ).
Your solution can be significantly simplified. Both terms in the integral have a relatively simple antiderivative in terms of dilogarithms. After plugging in the limits, it then boils down to showing
Li2(2−1)−Li2(1−2)=8π2−21log2(2−1)
using the dilogarithm identities you posted.
To find the antiderivative of the second term, just substitute (2−1)x=u. For the first term, substitute u=(x+1+x2)2. This will reduce the integral to
∫u2−1lnudu,
which I am sure you can calculate in a few lines.
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Edit : Aha! I get it. Use this relation: arcsinhx=ln(x+x2+1). Very clever!
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The term with arcsinhx/x can be evaluated by substituting u=(x+1+x2)2. The result is something with a dilogarithm. The other term also gives a dilogarithm, but the dilogarithm terms cancel precisely, leading to an elementary result. I hope you can appreciate the beauty of the problem :)
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PROBLEM 16 :
Prove
where B(x,y) is the beta function.
PS : POST YOUR SOLUTION BELOW EACH PROBLEM THREAD AND POST YOUR PROPOSED PROBLEM AS A NEW THREAD. PUT THEM IN SEPARATED THREAD. SO THAT THE POSTS LOOK MORE ORGANIZED. THANKS.
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Isn't this problem too difficult to high school students Anna? I decide to answer Problem 16 because I'm afraid if this continues till a week, this contest will lose its interest. IMHO, you should propose an easy problem so that this contest will be fun as the stated aims of it. So, here is a solution:
SOLUTION OF PROBLEM 16 :
Rewrite the integral as follows ∫02πcosv−1xcosax dx=21∫−2π2π(2eix+e−ix)v−1cosax dx=2v1∫−2π2π(1+e2ix)v−1e−i(v−1)xcosax dx=2v1∫−2π2πn=0∑v−1(nv−1)e2inx⋅e−i(v−1)xcosax dx=2v1n=0∑v−1(nv−1)∫−2π2πei(2n−v+1)xcosax dx.(1) Consider f(x)=⎩⎨⎧eiωx0for −2π<x<2πotherwise The Fourier transform of f(x) is F[f(x)]∫−2π2πeiωxcosαx dx−i∫−2π2πeiωxsinαx dxℜ(F[f(x)])∫−2π2πcosαx dx=∫−∞∞f(x) e−iαx dx=∫−2π2πeiωx e−iαx dx=∫−2π2πei(ω−α)x dx=[i(ω−α)ei(ω−α)x]x=−2π2π=[ω−αsin(ω−α)x]x=−2π2π=ω−α2sin(ω−α)2π.(2) Using (2), then (1) turns out to be ∫02πcosv−1xcosax dx=2v−11n=0∑v−1(nv−1)2n−v+1−asin(2n−v+1−a)2π=2v1n=0∑v−1(nv−1)n−2v−1+asin(n−2v−1+a)π.(3) Now, let us express (zy) in term of beta function that can be related to (3). (zy)=z!(y−z)!y!=Γ(1+z)Γ(1+y−z)y!=zΓ(z)Γ(1−z)(y−z)⋯(1−z)y!=πzsin(πz)⋅(y−z)⋯(1−z)y!=πzsin(πz)n=0∑y(ny)(−1)nz−nn=n=0∑y(ny)π(z−n)sinπ(z−n).(4) Using (4), then (3) turns out to be ∫02πcosv−1xcosax dx=2vπ(2v−1+av−1)=2vπΓ(2v+a+1)Γ(2v−a+1)Γ(v)=2v v B(2v+a+1,2v−a+1)π■
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OK, fine. I'll post high school integral problems from now. -_-"
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Your solution is valid only if v is an integer, whereas the identity holds in general also.
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PROBLEM 18
Show that
∫0π/4tan1/3xdx=61(π3−3log2)
My bad, it should be 3log2 indeed. Kinshuk's result is correct. Sorry for the confusion.
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There should be 3log(2) instead of 2log(2)
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Solutionofproblem18:I=∫04πtan31x.dxputtan31x=tdx=1+t63t2.dtI=∫1+t63t3.dtputt2=u2t.dt=duI=23∫1+u3u.duusingpartialfraction,ourintegrationturnsouttobe:I=2−1∫1+udu+21∫u2−u+1(1+u)duaftersolvingandapplyinglimits:I=∫04πtan31x.dx=61(π3−3log2)
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PROBLEM 21 :
Show that ∫0∞x4+2cos(2θ)x2+1dx=4cosθπ
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We have
∫0∞x4+ax2+b2dx=2b2b+aπ,
which I proved on MSE. Plugging in a=2cos2θ and using that 2cos2(θ)=1+cos2θ immediately gives the answer.
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Oh I know you're now. You're user111187. I thought you're an old man. Haha
Nice to meet you here Ruben. It seems you'll be a tough opponent because you're a Math SE and I&S user. ⌣¨
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Expecting a question from you @Ruben Doornenbal
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Problem23
Find ∫04πln(tan(x))dx
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Solution 23
A well-known problem. Sub tanx=u to get
∫01du1+u2dulnu=k≥0∑(−1)k∫01dulnuu2k=k≥0∑(−1)k(2k+1)21=G.
The penultimate equality follows from integration by parts.
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@Ruben Doornenbal Can we have problem 24?
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Sir can you elaborate I did'nt understood this one @Ruben Doornenbal
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1+u21 in a geometric series and interchange summation and integration. The last equality is just the definition of Catalan's constant.
The idea is to expand the factorLog in to reply
@Ronak Agarwal @Ruben Doornenbal The answer given is wrong!!! It should be -G!!! You must have forgotten the negative sign.......
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Problem30
Find ∫1∞xx−⌊x⌋−0.5dx
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I=n→∞limr=1∑r=n−1∫rr+1xx−r−21dx I=n→∞limr=1∑r=n−11−(r+21)ln(rr+1) 2I=n→∞limr=1∑r=n−12−(2r+1)ln(rr+1) 2I=n→∞limr=1∑r=n−12−(2r+2)ln(rr+1)+ln(rr+1) 2I=n→∞limr=1∑r=n−12−(2r+2)ln(r+1)+(2r+2)ln(r)+ln(rr+1) 2I=n→∞limr=1∑r=n−12−2((r+1)ln(r+1)−rln(r))+ln(r+1)+ln(r) 2I=n→∞limr=1∑r=n−12−2((r+1)ln(r+1)−rln(r))+ln(r+1)+ln(r) 2I=n→∞lim2(n−1)−2nln(n)+2ln(n!)−ln(n) 2I=n→∞lim2(n−1)−2nln(n)+(2n+1)ln(n)−2n+ln(2π)−ln(n) 2I=ln(2π)−2 I=2ln(2π)−1
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This one is easy too :) 2ln2π−1
@Shivang Jindal : Sorry, I was kidding & I am busy right now so I have no time to write down my answer. Could you elaborate yours then you're good to go (propose your problem). Sorry for the inconvenience...
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Expecting a question from you @Tunk-Fey Ariawan , also please give a proof of your answer.
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Trick , is to break the integral from (1,2),(2,3)...(n−1,n). Then, we compute the sum in terms of n . and then use Stirling approximation :) .
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Exactly, you got it perfectly right.
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PROBLEM 22
This one is particularly beautiful, in my opinion.
∫0acosxcos(a−x)xdx=sinaalnseca.
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SolutionofProblem22
I=∫0acos(x)cos(a−x)xdx=∫0acos(x)cos(a−x)(a−x)dx
Adding these two forms we get :
I=2a∫0acos(x)cos(a−x)dx
Multiplying and dividing by sin(a) we get :
I=2sin(a)a∫0acos(x)cos(a−x)sin(x+(a−x))dx
I=2sin(a)a∫0a(tan(x)+tan(a−x))dx
Also since ∫0atan(x)dx=∫0atan(a−x)dx
We get I=sin(a)a∫0atan(x)dx
I=sin(a)aln(sec(a))
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PROBLEM 24
∫01arcsechxarcsinxdx=8π2−ln2.
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Solutionofproblem24
First note that :
arcsech(x)=ln(x1+1−x2)
In our integral put x=sin(θ)
I=∫02πθcosθln(sinθ1+cosθ)dθ
Applying integration by parts we get , u=ln(sinθ1+cosθ),dv=θcosθdθ
I=(θsinθ+cosθ)ln(sinθ1+cosθ)0∣2π+∫02π(θsinθ+cosθ)cosecθdθ
I=(θsinθ+cosθ)ln(sinθ1+cosθ)+2θ2+ln(sinθ)0∣2π
Which on evaluating we get :
I=8π2−ln(2)
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Can you elaborate the first line please , it will be great to learn from you
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PROBLEM 26 :
Prove
∫0∞coshxlnxdx=2πln(πΓ4(43))
P.S. You may use any well-known expressions.
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SOLUTION 26
Consider
I(a)=∫0∞dxcoshxxa−1=2k≥0∑(−1)k∫0∞dxxa−1e−(2k+1)x=2Γ(a)β(a).
Our integral is
I′(1)=2Γ′(1)β(1)+2Γ(1)β′(1)=2(−γ)(π/4)+24π[γ+2ln2+3lnπ−4lnΓ41].
Here we used a result from Mathworld. Using the Euler reflection formula,
Γ(1/4)=π2(Γ(3/4))−1.
Collecting all the terms gives I′(1)=−2πlnπ+2πlnΓ(3/4),
which equals the stated result.
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PROBLEM 27
My last two integrals were clearly too easy. By finding an antiderivative or otherwise, show that
∫0∞dxln2tanhx=47ζ(3).
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Solution of Problem 27
Set t=tanhx, we have
∫0∞ln2(tanhx)dx=∫011−t2ln2tdt=∫01n=0∑∞t2nln2tdt=n=0∑∞∫01t2nln2tdt=2n=0∑∞(2n+1)31⇒see solution of Problem 13=2[n=1∑∞n31−n=1∑∞(2n)31]=47n=1∑∞n31=47ζ(3)
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Problem 28
Prove
∫−∞∞sinhπxsinh2xcos2x dx=cos2+cosh2sin2
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Solution 28
The integral equals
I=ℜ∫−∞∞dxe2πx−1(e2x−e−2x)eπxe2ix.
Substitute eπx=u. We get
I=J+−J−,
where
J±=ℜπ1∫0∞dxu2−1u2(i±1)/π=−ℜ21cot[2π(2(i±1)/π+1)]=ℜ21tan(i±1).
Here we made use of the well-known integral
PV∫0∞1−xbxa−1=bπcotbπa.
Now using the identity
tan2A+B=cosA+cosBsinA+sinB
gives
J±=±21cosh2+cos2sin2,
which gives the desired result.
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Wait!? For PV∫0∞1−xbxa−1dx=bπcot(bπa) could you prove it without using contour integration or residue method? See the rules.
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b=1 in the proof. The general result follows from a substitution. Separate the integrals over (0,1) and over (1,∞). Put u=1/x in the second integral. The result is
Of course, my dear. It is clear that we can take∫01dx1−xxa−1−x−a=ψ(1−a)−ψ(a)=πcotπa,
as was to be proven. Here we used a result derived by real methods here.
You have sharp eye for integrals that I normally derive with residues :p
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