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.
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@Ronak Agarwal
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Nicely done, +1. Now you must post a new problem, but before that. I think it would be nice if you post this as a new post, no need to reply @Sudeep Salgia. See the rules & the format post. Thank you. :)
@Ronak Agarwal
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Where is your proposed problem? We are waiting it. Please don't let us wait too long like this. You should post your solution and your proposed problem at once.
Nicely done, +1! But you forget to post your problem anyway. You have a privilege for that so that this contest keeps going on. Post PROBLEM 2 in your thread solution. Thanks :)
Well, there must be a rule for people like Ronak! Solving a problem doesn't give you the right to withhold the contest. In such a case, someone should be allowed to interfere, right?
@Satyam Bhardwaj
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Sorry If that troubled all because actually I have to go to coaching centre, and my internet connection got down and I went to my coaching centre, I have returned just now and has posted the question.
@Satyam Bhardwaj
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OK, if no-one posts his/her own problem after solving a problem within a day. The one who has a right to post a problem is the last solver before him/her. Is this fair?
Keep going fella, +1! But please post your solution & proposed problem in a new thread like I did. No need to reply other threads to post a solution & a problem. Anyway, I've posted the solution of your problem & also proposed a new problem.
Ingenious! But I have a doubt - In your fourth step, how is it guaranteed that the function 2kcos(2kx) is positive on (0,∞)? PS : you forgot to write the dx in the third step.
@Pratik Shastri
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I don't know what did you mean by positive/negative on (0,∞), but if you meant to swap between integral & summation sign, it is valid because 1+x2cos(2kx) is continuous, finite & integrable, therefore swapping those two signs can be justified by Fubini's theorem.
Note that
∫01xadx=a+11for α>−1.
Differentiating equation above k times w.r.t. a we have
∫01∂ak∂k(xa)dx=∫01xalnkxdx=(a+1)k+1(−1)kk!Q.E.D.
Now, we will evaluate the general case of
∫0∞ebx−1xa−1dx=∫0∞1−e−bxxa−1e−bxdxfor a,b>0
Set y=e−bx, then
∫0∞ebx−1xa−1dx=ba(−1)a−1∫011−ylna−1ydy
Use a geometric series for 1−y1 then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have
Since @Pranav Arora is unable to propose a problem (I hope it's only temporary) and to make this contest sustains, then according to rule 4, I, as the last solver, have a right to propose a new one. Here is the problem:
Solution -Problem - 9
We substitute,
t=1−x1+x⟹x=1+t1−t
Doing the substitution and , simplifying , gives,
I=−∫01(t)(1−t)ln(t)dt
Now we substitute, t=sin2(x)I=−4∫02πcos(x)ln(sin(x))dx
Using, Problem -4I=−4×8−π2=2π2
I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I did @Shivang Jindal
Wait a sec!? Could you rectify your solution, because the original problem is written as arctan(2x). And one thing, what did you mean by "Now we know that cos(arctanx)=1+4x21"? Could you elaborate?
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Top NewestHere is the solution: I=0∫aax−x2ln xdx
I=0∫aax−x2ln (a−x)dx
⇒2I=0∫aax−x2ln (ax−x2)dx
Put x=2a(1−sint)⇒ax−x2=4a2cos2t
Therefore, the integral becomes, 2I=2π∫2−π4a2cos2tln (4a2cos2t)(2acost)(−dt)
Rearranging, we obtain,
2I=40∫2πln (2acost)dt
The value of 0∫2πln (cosθ)dθ is 2−π ln 2 (which I have calculated separately and I can post if it is required).
Thus, we obtain, I=π ln (2a)−22π ln 2=π ln (4a)
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Find a closed form expression for the integral : I=∫sin(2015x)sin2013x dx
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Solutionofproblem2
Split sin(2015x) as sin(2014x+x) and our integral becomes :
I=∫(sin2014x.cos(2014x)+sin(2014x)cos(x)sin2013(x))dx
Multiply and divide it by 2014 to get :
20141∫(sin2014x.(2014cos(2014x))+sin(2014x)(2014sin2013(x)cos(x)))dx
⇒I=20141∫sin2014(x)dsin(2014x)+sin(2014x)(dsin2014x)
⇒I=20141∫d(sin2014(x)sin(2014x))
I=2014sin2014(x)sin(2014x)+C
Problem3
Evaluate I=∫0∞(xsin(x))2dx
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@Sudeep Salgia. See the rules & the format post. Thank you. :)
Nicely done, +1. Now you must post a new problem, but before that. I think it would be nice if you post this as a new post, no need to replyLog in to reply
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@Anastasiya Romanova I have posted my problem sorry for holding the contest.
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Nicely done, +1! But you forget to post your problem anyway. You have a privilege for that so that this contest keeps going on. Post PROBLEM 2 in your thread solution. Thanks :)
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Well, there must be a rule for people like Ronak! Solving a problem doesn't give you the right to withhold the contest. In such a case, someone should be allowed to interfere, right?
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Solution of Problem 3 :
Use integration by parts by taking u=sin2x and dv=x2dx, we get
∫0∞x2sin2xdx=−xsin2x∣∣∣∣0∞+∫0∞x2sinxcosxdx=0+∫0∞xsin2xdx=∫0∞tsintdt⇒t=2x
Now consider
I(a)=∫0∞te−atsintdta≥0
so that I(∞)=0 and our considered integral is I(0). Differentiating w.r.t. a and then integrating back, we get
I′(a)I(a)=−∫0∞e−atsintdt=−1+a21⇒integration by parts twice=−∫1+a21da=−arctan(a)+C
For I(∞)=0, implying C=2π. Hence
I(a)=∫0∞te−atsintdt=2π−arctan(a)
and
I(0)=∫0∞tsintdt=2π
Thus
Problem 4 :
Prove
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SolutionofProblem4
Firstly we will prove a general result
Result
∫01xaln(x)dx=(1+a)2−1
Proof
I=∫01xaln(x)dx
Integrating by parts u=ln(x),dv=xadx
I=ln(x)a+1xa+10∣1−a+11∫01xadx
I=(1+a)2−1
Hence proved
Now we have :
I=∫02πsin(x)ln(cos(x))dx
Take cos(x)=y to get our integral as :
∫011−y2ln(y)dy
Now 1−y21=n=0∑∞y2n
Our integral becomes :
I=n=0∑∞∫01y2nln(y)dy
Using our proved result we get :
I=−n=0∑∞(2n+1)21
Also the summation can we written as :
I=−(n=1∑∞n21−n=1∑∞(2n)21)
I=−(ζ(2)−41ζ(2))=4−3ζ(2)=8−π2
Where ζ(x) is the zeta function
Problem5
Find closed form of I=∫0∞1+xndx
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Keep going fella, +1! But please post your solution & proposed problem in a new thread like I did. No need to reply other threads to post a solution & a problem. Anyway, I've posted the solution of your problem & also proposed a new problem.
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Solution of Problem 7:
Proposition :
Proof :
It can be proven by using Weierstrass substitution: t=tan(2x), then
∫0πp+cosxdxdx=∫0∞p+1+(p−1)t22dt⇒t=p−1p+1tany=p2−12arctant∣∣∣∣∣0∞=p2−1πQ.E.D.
Differentiating the proposition w.r.t. p twice and setting p=10, we have
∂p2∂2∫0πp+cosx1dx∫0π(p+cosx)32dx∫0π(10+cosx)31dx=∂p2∂2[p2−1π]=(p2−1)5π(2p2+1)=1627π
Problem 8 :
Prove
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Solution of Problem 8:
Substitute tanx↦x, then the integral is:
∫0∞(1+x2)(1+8sin2x)dx=∫0∞(1+x2)(5−4cos(2x))dx
=∫0∞1+x21(31(1+2k=1∑∞2kcos(2kx)))=6π+32k=1∑∞2k1∫0∞1+x2cos(2kx)dx
=6π+3πk=1∑∞(2e21)k=6π+3π2e2−11=6π(2e2−12e2+1)
I have used the following result:
∫0∞x2+a2cos(mx)dx=2aπe−am
I cannot think of a challenging problem at the moment, I request somebody else to post one. Thanks!
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Ingenious! But I have a doubt - In your fourth step, how is it guaranteed that the function 2kcos(2kx) is positive on (0,∞)? PS : you forgot to write the dx in the third step.
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(0,∞), but if you meant to swap between integral & summation sign, it is valid because 1+x2cos(2kx) is continuous, finite & integrable, therefore swapping those two signs can be justified by Fubini's theorem.
I don't know what did you mean by positive/negative onLog in to reply
@Anastasiya Romanova .
If pranav refuses to put up a question then who put will put it upLog in to reply
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Sorry for very late response. Using elementary techniques,-
Solution - Problem 7
∫0πp−cosxdx
2I=∫0πp2−cos2x2p dx
I=∫0πp2(1+tan2x)−1p dx
I=p∫0πp2tan2x+(p2−1)2sec2x dx
tanx = t
I=2p∫0∞p2t2+(p2−1)2dt
I=p2−12p×p1[0∞tan−1p2−1pt
I=p2−1π
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Solution to problem 10
First integrate by parts -
I=∫0∞x2sin3xdx=∫0∞x3sin2xcosxdx
Then, use the property of the laplace transform that L{tf(t)}(s)=∫s∞F(p)dp (in our case, s→0).
I=3∫0∞L{sin2tcost}(p) dp
I=3∫0∞(s2+1)(s2+9)2sds=43log3 The last integral can be found by using partial fractions.
Note : L{f(t)}(s)=∫0∞e−stf(t)dt
Problem 11
Find
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Solution of Problem 13 :
First we prove the following proposition:
Proposition :
Proof :
Note that ∫01xa dx=a+11for α>−1. Differentiating equation above k times w.r.t. a we have ∫01∂ak∂k(xa) dx=∫01xalnkx dx=(a+1)k+1(−1)kk!Q.E.D.
Now, we will evaluate the general case of
∫0∞ebx−1xa−1dx=∫0∞1−e−bxxa−1e−bxdxfor a,b>0
Set y=e−bx, then
∫0∞ebx−1xa−1dx=ba(−1)a−1∫011−ylna−1ydy
Use a geometric series for 1−y1 then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have
∫0∞ebx−1xa−1dx=ba(−1)a−1∫01k=0∑∞yklna−1ydy=ba(−1)a−1k=0∑∞∫01yklna−1ydy=ba(−1)a−1k=0∑∞(k+1)a(−1)a−1(a−1)!=ba(a−1)!k=1∑∞ka1=baΓ(a)ζ(a)
where Γ(a) is the gamma function and ζ(a) is the Riemann zeta function.
Hence, by setting a=n+1 and b=1, we have
∫0∞ex−1xndx=Γ(n+1)ζ(n+1)
Problem 14 :
Prove that
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Since @Pranav Arora is unable to propose a problem (I hope it's only temporary) and to make this contest sustains, then according to rule 4, I, as the last solver, have a right to propose a new one. Here is the problem:
PROBLEM 9
Prove that
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Solution -Problem - 9 We substitute, t=1−x1+x⟹x=1+t1−t Doing the substitution and , simplifying , gives, I=−∫01(t)(1−t)ln(t)dt Now we substitute, t=sin2(x) I=−4∫02πcos(x)ln(sin(x))dx Using, Problem -4 I=−4×8−π2=2π2
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I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I did @Shivang Jindal
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@Shivang Jindal
You can try my posted question.Log in to reply
I am able to write the integral as a series - I=πn=0∑∞22n(n!)2(2n+1)(2n)!
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Solutionofproblem9
I=∫01ln(1−x1+x)x1−x2dx
Put x=cos(θ) to get our integral as :
∫02πln(cot22θ)cosθdθ
⇒I=(−2)∫02πln(tan2θ)cosθdθ
Put tan(θ/2)=x to get our integral as :
(−4)∫011−t2ln(t)dt
I proved in my solution to problem 4 that :
∫011−t2ln(t)dt=8−π2
Using this I get :
I=2π2
Problem10
Find ∫0∞x2sin3xdx
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Solution of Problem 11 :
Using substitution u=xα2⇒x=uα2⇒dx=−u2α2 du, then
∫0∞x2+α2lnx dxI(α)I(α)=∫0∞(uα2)2+α2ln(uα2)⋅u2α2 du=∫0∞α2+u22lnα−lnu du=2lnα∫0∞α2+u21 du−∫0∞u2+α2lnu du=2lnα∫0∞α2+u21 du−I(α)=lnα∫0∞α2+u21 du.
The last integral can easily be evaluated by using substitution u=αtanθ, then
∫0∞x2+α2lnx dx=αlnα∫02π dθ=2απlnα
Problem 12 :
Prove
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The last substitution should be u=αtanθ :)
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Fixed it. Sorry, I'm too hasty. Thanks... :)
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This is a simple kind of NCERT problem. It can be easily done by substituting t=tan−1αx and then applying by parts
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SolutionofProblem12
It's a very good disguised integral.
First note that the function is an even function hence our integral can be written as :
I=2∫0∞(1+x2)1+4x2cos(arctan(2x))dx
Now we know that cos(arctan(2x))=1+4x21. Using this our integral becomes :
I=2∫0∞(1+x2)(1+4x2)1dx
Splliting it by partial fractions we get :
I=32(∫0∞(41+x2)dx−∫0∞1+x2dx)
I=32(2tan−1(2x)0∣∞−tan−1(x)0∣∞)
I=3π
Problem13
Find closed form of I=∫0∞ex−1xndx
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Wait a sec!? Could you rectify your solution, because the original problem is written as arctan(2x). And one thing, what did you mean by "Now we know that cos(arctanx)=1+4x21"? Could you elaborate?
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That was my typing mistake, also Letarctan(2x)=θ, hence 2x=tan(θ), hence cos(θ)=1+4x21
⇒cos(arctan(2x))=1+4x21
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Solutionofproblem14
Lemma
∫2−π2πln(a2cos2x+b2sin2x)dx=2πln(2a+b)
Proof :
I(a)=∫2−π2πln(a2cos2x+b2sin2x)dx
Differentiating with respect to a we get :
I′(a)=2a∫2−π2πa2cos2x+b2sin2xcos2xdx
Put tan(x)=t to get our integral as :
I′(a)=2a∫2−π2π(a2+b2t2)(1+t2)dt
=2a∫2−π2π(a2+b2t2)(1+t2)dt=b2−a22a(∫2−π2πa2+b2t2b2dt−∫2−π2π1+t2dt)
Solving it we get :
I′(a)=a+b2π
⇒I(a)=2πln(a+b)+C
Put a=b=1 to get C=−2πln(2)
Hence I(a)=2πln(2a+b)
In our integral in the given question put x=21−sinθ to get the integral as :
I=∫2−π2πln(7−sin(θ))dx−∫2−π2πln(5+sin(θ))dx
Using the property ∫abf(x)dx=∫abf(a+b−x)dx we get :
I=∫2−π2πln(7+sin(θ))dx−∫2−π2πln(5−sin(θ))dx
Adding these two we get :
I=21(∫2−π2πln(49−sin2θ)dx−∫2−π2πln(25−sin2(θ))dx)
I=21(∫2−π2πln(49cos2θ+48sin2θ)dx−∫2−π2πln(25cos2θ+24sin2θ)dx)
Using our given lemma we get :
I=21(2πln(27+43)−2πln(25+26))=πln(5+267+43)
Problem15
Evaluate I=∫01ln(x)1+1−x1dx
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According to timeline, you're faster than @jatin yadav to post your solution, then the right to propose a new problem is yours.
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_ Solution of problem 14 _
I=∫01ln(3−x3+x)x(1−x2)1dx
= ∫01ln(1+x/3)x(1−x)1dx
= ∫0π/22ln(1+31sin2θ)dθ−∫0π/22ln(1−31sin2θ)dθ
Lemma ∫0π/2ln(1+asin2θ)=πln(21+a+1)
Proof
Let I(a)=∫0π/2