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?
@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.
We have
∫01(1−x1+lnx1)dx=∫01(1−x)lnxlnx+1−xdx
Proposition :
I(s)=∫01(1−x)lnxslnx+1−xsdx=lnΓ(s+1)+γs
Proof :
Let
I(s)=∫01(1−x)lnxslnx+1−xsdx
then
I′(s)I′′(s)I′(s)I′(s)I′(s)=∫011−x1−xsdx=−∫011−xxslnxdx=−∫01n=0∑∞xn+slnxdx=−n=0∑∞∂s∫01xn+sdx=−n=0∑∞∂s[n+s+11]=n=0∑∞(n+s+1)21=ψ1(s+1)=∫ψ1(s+1)ds=∫∂s∂[ψ(s+1)]ds=ψ(s+1)+C
For s=0, we have I′(0)=0. Implying C=−ψ(1)=γ, then
I′(s)I(s)=ψ(s+1)+γ=∫ψ(s+1)ds+γs+C=∫∂s∂[lnΓ(s+1)]ds+γs+C=lnΓ(s+1)+γs+C
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.
Can you please post a all together new note of the integration contest it's getting messy and messier, Please move this contest to a new note @Anastasiya Romanova
Nice answer, +1! But I'm really sorry, the one who deserves to propose the next problem is Ronak since according to the timeline his answer comes first than the other answers. Maybe, you can try to answer his proposed problem. ⌣¨
@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.
I(a)=∫0π(a−cos(x))1,a>1
Use Weierstrass substitution, its not hard to show that , this integral =a2−1π
Now , we differentiate, both sides with respect to a.
I′(a)=∫0π−(a−cos(x))21=dada2−1π
Again differentiating both sides,
I′′(a)=∫0π(a−cos(x))32=d2ada2−1π
Taking, 2 to RHS gives,
I′′(a)=∫0π(a−cos(x))31=d2ad2a2−1π
Note that, we need I′′(10) .
Plugging, a=10 , and evaluating the expression in RHS gives, 1627π
@Shivang Jindal
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-_- . Solution was already there :||||. Please, make this thread little structured. I didn't saw any solution after problem problem -7th, so wrote solution here. After posting here , i realized that, there is exactly same solution posted before.
@Shivang Jindal
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I can do nothing since the order of the threads on Brilliant.org is based on number of upvotes and downvotes. The only thing I can do is numbering the problem & the solution. If you're interesting in participating on this contest, you can try to answer the problem (PROBLEM 9) that I've just posted below.
It looks like there's someone here who is interesting in knowing the closed-form of your previous problem 7 and he posted it on Math S.E.. I am wondering, does that really have a closed-form? I tried to solve it for hours but no success & you really gave me lots of trouble cracking that integral.
@Anastasiya Romanova
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Well ,actually I had to change the problem as I didn't know its solution. Actually, few days back, at the same website I tried answering this problem and got struck at it.
@Jatin Yadav
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OMG! Why did you propose then here on this contest? Anyway, just ignore that user. He always ask two or three hard integrals in a single question on Math S.E. I can answer one of his question but only gets 3 upvotes. It's not worth enough. Luckily, he accepts my answer. I also answer one of his question on Integrals and Series, but of course not for free. LOL
@Anastasiya Romanova
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My bad! Didn't read the rules carefully. By the way , seeing a 14 year old student having so much knowledge of calculus is really impressive! I knew only basic integration properly when I was 14.
@Jatin Yadav
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No worries. It's so-so in my country. I learned from my late grandpa, my big brother (I hate him!), and lots of people on internet since I was 7. You might be interested in seeing this. You're very good at all math subjects & physics too. That's truly impressive!!
Anyway, I've just remembered that the same user (Samurai, the one who posts your previous problem 7 on Math S.E.) also posted one of my problem on Brilliant at Math S.E. too. What a cheater!
@Anastasiya Romanova
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If I'm not mistaken, you only solve / post problems in the calculus section. Why are you specifically interested in Calculus? How about other fields of Math?
@Christopher Boo
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Because I wanna beat my big brother. Seriously, he always says to me that calculus is the highest level mathematics, so that's why I'm really interested in calculus specially integrals & series and almost every weekend we have a competition of solving integrals & series problems although he always beats me easily. Of course I love math, it means I also love the other fields of math but I'm just not too interested in them. Besides, my first crush on math because of calculus. I don't wanna tell the story because it's too personal. Special for you, I have answered two problems here that don't relate to Calculus to show my capability other than Calculus (It's also because I'm bored). Here they are Algebra problem: Advanced System Of Equations and Geometry problem: All Three Concurrency. Answering Algebra problem makes me level up, so I must down my level again. I just want the other know me because I'm sorta good at Calculus, not the other subjects. Knowing Calculus in such young age also gives me lots of advantages on the other forums where I join in. Okay, I think it's enough. I hope I am not making any immodest statements or showing narcissism on my part. ⌣¨
@Anastasiya Romanova
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Tip for beating your older brother: Calculus is far from the hardest discipline of mathematics. Learn abstract algebra, or complex analysis, or dynamical systems. Then you'll be able to appreciate fields of mathematics where there is still a lot of work left to be done, even today, and you'll probably end up knowing more math than your big brother. ;)
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Here 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
@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.
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
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
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
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
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.I am able to write the integral as a series - I=πn=0∑∞22n(n!)2(2n+1)(2n)!
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
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... :)
This is a simple kind of NCERT problem. It can be easily done by substituting t=tan−1αx and then applying by parts
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
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.
_ 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/2ln(1+asin2x)dx
I′(a)=∫0π/21+asin2xsin2xdx
I′(a)=∫0π/2a1(1−1+(a+1)tan2xsec2xdx)
= 2aπ−a1(∫0∞1+(a+1)z2dz)
=2aπ(1−a+11)
Hence, I(a)−I(0)=∫0a2π(x1−xx+11)dx
= πln(x+1+1)∣∣∣∣0a=πln(21+a+1) (Note that I(0)=0)
Put a=1/3 , and −1/3, then substitute back in original integral to get:
I=2πln(1+2/31+4/3)=πln(5+267+43)
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Nice solution Jatin, +1! But you're a bit late about 10 min than Ronak, so the right to propose a new problem goes to Ronak.
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@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.
Solution of Problem 15 :
We have ∫01(1−x1+lnx1)dx=∫01(1−x)lnxlnx+1−xdx
Proposition :
Proof :
Let
I(s)=∫01(1−x)lnxslnx+1−xsdx
then
I′(s)I′′(s)I′(s)I′(s)I′(s)=∫011−x1−xsdx=−∫011−xxslnxdx=−∫01n=0∑∞xn+slnxdx=−n=0∑∞∂s∫01xn+sdx=−n=0∑∞∂s[n+s+11]=n=0∑∞(n+s+1)21=ψ1(s+1)=∫ψ1(s+1)ds=∫∂s∂[ψ(s+1)]ds=ψ(s+1)+C For s=0, we have I′(0)=0. Implying C=−ψ(1)=γ, then I′(s)I(s)=ψ(s+1)+γ=∫ψ(s+1)ds+γs+C=∫∂s∂[lnΓ(s+1)]ds+γs+C=lnΓ(s+1)+γs+C
where ψ(z) is the digamma function. For s=0, we have I(0)=0. Implying C=0, then
I(s)=∫01(1−x)lnxslnx+1−xsdx=lnΓ(s+1)+γs□
For s=1, we have
I(1)=∫01(1−x1+lnx1)dx=∫01(1−x)lnxlnx+1−xdx=γ
where γ is The Euler–Mascheroni constant.
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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Can you please post a all together new note of the integration contest it's getting messy and messier, Please move this contest to a new note @Anastasiya Romanova
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OK
The Brilliant Integration Contest - Season 1 (Part 1) moves to The Brilliant Integration Contest - Season 1 (Part 2). Enjoy it!! (>‿◠)✌
Solution 9: First, consider the following Lemma.
LEMMA : ∫011−x2lnxdx=−8π2
Proof: Since, 1−x21=n=0∑∞x2n, we have,
∫011−x2lnx=n=0∑∞∫01x2nlnxdx
=−n=0∑∞(2n+1)21 (Using Integration By Parts)
=n=0∑∞(2n)21−n=0∑∞(n)21
=4−3ζ(2)=−8π2
Now,
I=∫01ln(1−x1+x)x1−x2dx
Substituting x=1+y21−y2, we have,
I=−4×∫011−y2lnydx
=2π2 (Using the Lemma)
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Nice answer, +1! But I'm really sorry, the one who deserves to propose the next problem is Ronak since according to the timeline his answer comes first than the other answers. Maybe, you can try to answer his proposed problem. ⌣¨
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@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.
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Solution of Problem 5 :
I will prove the general form of
∫0∞1+xnxm−1dx=nsinnmππ
It can easily be proven by taking substitution y=1+xn1 and the integral becomes Beta function
n1∫01y1−nm−1 (1−y)nm−1dy=nΓ(1−nm)Γ(nm)=nsinnmππ
where the last part comes from Euler's reflection formula for the gamma function. Hence, by setting m=1, we have
Problem 6 :
Show that
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I=∫0π/2sinx+cosxsin2xdx
I=∫0π/2sinx+cosxcos2xdx (As ∫0af(x)dx=∫0af(a−x)dx )
Hence, 2I=∫0π/2sinx+cosx1dx
Hence, I=21∫0π/22tan(x/2)+1−tan2(x/2)1+tan2(x/2)dx
Putting tan2x=t,
I=∫012−(1−t)2dt
=221ln(2−t+12+t−1)∣∣∣∣01
= 22ln(3+22)
Problem 7
Prove that ∫0π(10+cosx)31=1627π
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I(a)=∫0π(a−cos(x))1,a>1 Use Weierstrass substitution, its not hard to show that , this integral =a2−1π Now , we differentiate, both sides with respect to a. I′(a)=∫0π−(a−cos(x))21=dada2−1π Again differentiating both sides, I′′(a)=∫0π(a−cos(x))32=d2ada2−1π Taking, 2 to RHS gives, I′′(a)=∫0π(a−cos(x))31=d2ad2a2−1π Note that, we need I′′(10) . Plugging, a=10 , and evaluating the expression in RHS gives, 1627π
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It looks like there's someone here who is interesting in knowing the closed-form of your previous problem 7 and he posted it on Math S.E.. I am wondering, does that really have a closed-form? I tried to solve it for hours but no success & you really gave me lots of trouble cracking that integral.
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this problem and got struck at it.
Well ,actually I had to change the problem as I didn't know its solution. Actually, few days back, at the same website I tried answeringLog in to reply
his question but only gets 3 upvotes. It's not worth enough. Luckily, he accepts my answer. I also answer one of his question on Integrals and Series, but of course not for free. LOL
OMG! Why did you propose then here on this contest? Anyway, just ignore that user. He always ask two or three hard integrals in a single question on Math S.E. I can answer one ofLog in to reply
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this. You're very good at all math subjects & physics too. That's truly impressive!!
No worries. It's so-so in my country. I learned from my late grandpa, my big brother (I hate him!), and lots of people on internet since I was 7. You might be interested in seeingAnyway, I've just remembered that the same user (Samurai, the one who posts your previous problem 7 on Math S.E.) also posted one of my problem on Brilliant at Math S.E. too. What a cheater!
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Advanced System Of Equations and Geometry problem: All Three Concurrency. Answering Algebra problem makes me level up, so I must down my level again. I just want the other know me because I'm sorta good at Calculus, not the other subjects. Knowing Calculus in such young age also gives me lots of advantages on the other forums where I join in. Okay, I think it's enough. I hope I am not making any immodest statements or showing narcissism on my part. ⌣¨
Because I wanna beat my big brother. Seriously, he always says to me that calculus is the highest level mathematics, so that's why I'm really interested in calculus specially integrals & series and almost every weekend we have a competition of solving integrals & series problems although he always beats me easily. Of course I love math, it means I also love the other fields of math but I'm just not too interested in them. Besides, my first crush on math because of calculus. I don't wanna tell the story because it's too personal. Special for you, I have answered two problems here that don't relate to Calculus to show my capability other than Calculus (It's also because I'm bored). Here they are Algebra problem:Log in to reply
Ans:ln4a/2a^2