Hi Brilliant! Just like what Aditya Kumar and Anastasiya Romanova conducted earlier, this year we would also like to conduct an integration contest for juniors.
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.
Eligibility:- People should fulfill either of the 2 following
17 years or below
Level 4 or below in Calculus
Eligible people 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.
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
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, digamma function,Harmonic numbers, trigonometric integral, Wallis' integral,
Please post your solution and your proposed problem in a single new thread.
Format your post is as follows:
1 2 3 4 5 6 7 |
|
The comments will be easiest to follow if you sort by "Newest":
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.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestProblem 18:
Prove that: ∫01(logx1+1−x1−21)1−xdx=−21+21ln(2π)−21γ.
The Problem has been solved by Ishan Singh
Log in to reply
Lemma 1 : r=1∑nHr=(n+1)Hn−n
Proof : r=1∑nHr=∫01r=1∑nx−1xr−1dx
=∫01(x−1)2xn+1−x−nx+ndx
=∫01(x−1)2xn+1−xdx−∫01x−1ndx
=A+B
Using Integration By Parts on the first integral A, we have,
A=−[x−1xn+1−x]01+∫01x−1(n+1)xn−1dx
=−n+∫01x−1(n+1)xn−1dx
⟹A+B=−n+(n+1)⋅∫01x−1xn−1dx
=(n+1)Hn−n
Lemma 2 : ∫01lnxxn−1dx=ln(n+1)
Proof : ∫01lnxxn−1 dx=∫01∫0nxy dy dx
=∫0n∫01xy dx dy
=∫0ny+11 dy
=ln(n+1)
Lemma 3 : ∫01(ln(x)1+1−x1)dx=γ
Proof : ∫01(1−x1+ln(x)1)dx=n→∞lim∫01(1−t1−tn−lnttn−1−1dt)
=n→∞lim(Hn−lnn)
=γ
Now,
I=∫01(lnx1+1−x1−21)x−1dx
=n→∞limr=1∑n∫01(lnxxr−1+1−xxr−1−2xr−1)
Using Lemma 2 and Lemma 3 , we have,
I=n→∞limr=1∑n(γ+lnr−Hr−1−2r1)
=n→∞lim(nγ+ln(n!)−r=1∑nHr−1−2Hn)
Using Lemma 1,
I=n→∞lim(nγ+ln(n!)−nHn−1+n−1−2Hn)
Using Stirling's Approximation, we have,
I=n→∞lim(nγ+21ln(2π)−n+21ln(n)+nln(n)−nHn−1+n−1−2Hn)
Simplifying using n→∞lim(Hn−lnn)=γ, we have,
I=21ln(2π)−21γ−21
Log in to reply
wow,that was probably why i gave up halfway through lol
i'm sorry i didn't attend the integration contest until now,there was a problem with my wifi and it got fixed
Log in to reply
Next problem please.
Log in to reply
@Aditya Kumar May post another question if he wants.
Log in to reply
Nice. I used Binet's formula.
Log in to reply
Let us start with an easy Problem.
Problem 1
Find the value of the following ∫−∞∞e−x2dx∫−∞∞(x+1)2e−(x+1)2dx
This problem has been solved by Aditya Kumar.
Log in to reply
Solution to Problem 1: I1=∫−∞∞(x+1)2e−(x+1)2dx
Substitute: (x+1)2=t
Therefore, we get: I1=21∫−∞∞(t)21e−tdt
This is symmetric about x=0. Therefore, we can write it as: I1=∫0∞(t)21e−tdt
I1=Γ(23)=21Γ(21)=2π
Similarly, I2=π
Hence, I2I1=21
Log in to reply
Problem 2:
Prove that: ∫01x2+2(x2+1)arctan(x2+2) dx=965π2
This problem has been solved by Harsh Shrivastava.
Log in to reply
Solution to Problem 2
I=∫01x2+2(x2+1)arctan(x2+2) dx=965π2
Lets split I as A and B by using the property that arctan(x)=π/2−arctan(1/x)
Thus our integral becomes I=A−B,
where A=π/2∫01(1+x2)(2+x2)dx
and B=∫01(1+x2)(2+x2)arctan(x2+21)
Evaluating A,
Let's evaluate A, without the limits,
π/2∫(1+x2)(2+x2)dx=π/2arctanx2+2x
Which on substituting limits gives A = π2/12
Evaluating B,
Using 1/aarctan(1/a)=∫0ay2+a2dy,
Let a=x2+2.
We get B=∫01∫01(1+x2)(2+x2+y2)dxdy
B=∫01∫01(1+x2)(1+y2)dxdy−∫01∫01(1+x2)(2+x2+y2)dxdy
2B=∫01∫01(1+x2)(1+y2)dxdy
2B=(∫011+x2dx)2
Thus B=π2/32
Hence I=A−B=965π2
Log in to reply
Generalization
Log in to reply
Problem 14:
If ∫0∞1+eαxx3dx=641
Find α given that , α is a positive constant and ζ(4)=90π4
Problem has been solved by Rajdeep Dhingra and Aareyan Manzoor at the same instant. Aareyan has been given credits due to his solution.
Log in to reply
I=∫0∞x3n=1∑∞(−e−αx)ndx I=n=1∑∞(−1)n∫0∞x3e−αnxdx=n=1∑∞(−1)n(αn)46 I=α46(1−2−3)ζ(4)=641 the result follows.
Log in to reply
Is the answer α=π⎝⎜⎛1556⎠⎟⎞41 ?
Log in to reply
Problem 6:
Find a closed form of the indefinite integral ∫Hxdx
Where Hx denotes harmonic number
This problem has been solved by Aditya Kumar.
Log in to reply
Solution to Problem 6:
I'll use the definition: Hx=ψ(x+1)+γ
∫Hxdx=∫(ψ(x+1)+γ)dx
We know that ψ(x)=dxd(x!). Hence, using this, we get: ∫Hxdx=ln(x!)+γx+c
Log in to reply
Problem 10
Let S=sin2(x)+32sin4(x)+52sin6(x)+⋯ Upto Infinity
Prove that ∫0π/2S dx=4π2−2π
The Problem has been solved by Ishan Singh.
Log in to reply
I have come up with several ways to prove it. Here is one of the more interesting methods.
Lemma: f(a)=1+a1+r=1∑∞[(2r+a+1)1k=1∏r(2k2k−1)]=2a+1πΓ2(2a+2)Γ(a+1);a>−1
Proof: I have proved it here
Now,
I=∫0π/2S dx=r=1∑∞∫01(2r−1)2sin2rdx
=21r=1∑∞(2r−1)2B(r+21,21)
=21r=1∑∞(2r−1)2Γ(r+1)Γ(r+21)Γ(21)
=2πr=1∑∞(2r−1)2Γ(21)Γ(r+1)Γ(r+21)
Changing the index of summation and using Γ(x+1)=xΓ(x), we have,
I=4πr=0∑∞(r+1)(2r+1)Γ(21)Γ(r+1)Γ(r+21)
=2π[r=0∑∞(2r+1)Γ(21)Γ(r+1)Γ(r+21)−r=0∑∞(2r+2)Γ(21)Γ(r+1)Γ(r+21)]
Using my solution here, we have,
I=2π[1+r=1∑∞(2r+1)1k=1∏r(2k2k−1)−21−r=1∑∞(2r+2)1k=1∏r(2k2k−1)]
=2π[f(0)−f(1)]
=4π2−2π
Log in to reply
What was your intended solution?
Log in to reply
More or less the same.
Log in to reply
@Ishan Singh @Rajdeep Dhingra Do any of you know another method to solve this question?
Log in to reply
Ishan wrote that he has come up with several ways so I guess has many. @Ishan Singh
Log in to reply
Log in to reply
Note to the Participants:
Please refrain from posting solution through images. It makes this note slow to load.
Up-vote good solutions and problems. I have see that people are not up-voting at all. This is a wrong sign in the contest. Up-voting encourages the users to improve.
Log in to reply
Very True.I will change the note after it crosses 50 questions.
Log in to reply
Maybe you should do it now, my browser has hanged twice loading this note!
Log in to reply
Change it after 20. 50 is too much.
Log in to reply
Problem 17
Prove that if c<1, ∫0π/2arcsin(ccos(x)) dx=12c+32c3+52c5+⋯ (Upto Infinity)
The Problem has been solved by Aditya Kumar
Log in to reply
Solution to Problem 17:
Maclaurin series of arcsin(x) is: arcsin(x)=n=0∑∞π(2n+1)n!Γ(n+21)x2n+1
Here x=ccos(x)
∫02πarcsin(ccos(x))dx=∫02πn=0∑∞π(2n+1)n!Γ(n+21)(ccos(x))2n+1dx
∫02πarcsin(ccos(x))dx=n=0∑∞π(2n+1)n!Γ(n+21)c2n+1∫02π(cos(x))2n+1dx
Therefore, by beta function we get:
∫02πarcsin(ccos(x))dx=n=0∑∞(π(2n+1)n!Γ(n+21)c2n+121Γ(n+23)Γ(21)Γ(n+1))
Hence, on simplifying, we get: ∫0π/2arcsin(ccos(x)) dx=12c+32c3+52c5+⋯ (Upto Infinity)
Log in to reply
Problem 9
Prove ∫0π/2xln(tan(x))dx=87ζ(3)
The Problem has been solved by Rajdeep Dhingra.
Log in to reply
Solution to Problem 9
Log in to reply
The second page is not at all clear. Please LaTeX your solution.
Log in to reply
Problem 23
Prove that for an integer n≥2
∫01{nx1}dx=n−1n−ζ(n)
hint:let x=un1
Log in to reply
Let I=∫01{nx1}dx
Substitute x=tn1
⟹I=n∫1∞tn+1{t}dt
=nr=1∑∞∫rr+1tn+1t−rdt
=nr=1∑∞[1−nt1−n+rnt−n]rr+1
=−nr=1∑∞(n(n+1)(r+1)n−11+n(r+1)n1)
=n−1n−ζ(n)
Log in to reply
Does the Curly brackets mean fractional part ?
Log in to reply
It is pointless to continue the contest. It hasn't garnered attention like the previous original integration contests.
Log in to reply
@Hummus a will post.
No Problem. Still it is open. Please post the next question if you want or if don't want to post, thenLog in to reply
Problem 3:
Prove that ∫01x(1−x2)(ln(1+x)−ln(1−x))dx=2π2
This problem has been solved by Rajdeep Dhingra.
Log in to reply
Solution to Problem 3
Let us start with the integral I(a)=∫01x1−x2ln(1+ax) dx. Differentiate with respect to a. I′(a)=∫01(1+ax)1−x21 dx This can easily be solved by substituting 1+ax=t1.We now get I′(a)=1−a2arccos(a) Now we find I(a). I(a)=∫1−a2arccos(a) da=−21arccos2(x)+C Now we put a = 1. We get I(1)=C Now we put a = -1. We get I(−1)=−2π2+C We get I(1)−I(−1)=2π2
Q.E.D
Log in to reply
Problem 8
Prove that ∫02πecos(θ)cos(sin(θ)) dθ=2π
Not Original
The Problem has been solved by Aareyan Manzoor.
Log in to reply
solution to problem 8
Considering the function f(t)=∫02πetcos(θ)cos(tsin(θ))dθ, we see that tf′(t)===t∫02πetcosθ[cosθcos(tsinθ)−sinθsin(tsinθ)]dθ∫02π∂θ∂[etcosθsin(tsinθ)]dθ[etcosθsin(tsinθ)]02π=0f′(t)=0→f(t)=C Since f(0)=2π, the desired integral is f(1)=2π.
Log in to reply
Problem 11
Prove that ∫0π/2(sin(x)x)2 dx=πlog(2)
The Problem has been solved Aareyan Manzoor.
Log in to reply
solution to problem 11 ∫uv′=uv−∫u′v I=∫0π/2x2csc2(x)dx u=x2→u′=2x,v′=csc2x→v=−cot(x) I=−x2cot(x)∣0π/2+2∫0π/2xcot(x)dx=2∫0π/2xcot(x)dx u=x→u′=1,v′=cot(x)→v=−ln(sin(x)) I=2xln(sin(x))∣0π/2+2∫0π/2ln(sin(x))dx=∫0π/2ln(sin2(x))dx B(a,1/2)=∫0π/22sin2a−1(x)dx B′(a,1/2)=∫0π/22ln(sin2(x))sin2a−1(x)dx B′(a,1/2)=B(a,1/2)(ψ(a)−ψ(a+1/2)) a=1/2 ∫0π/22ln(sin2(x))dx=B(1/2,1/2)(ψ(1/2)−ψ(1))=π(−γ−(−γ−2ln(2)))=2πln(2) I=∫0π/2ln(sin2(x))dx=πln(2)
Log in to reply
Problem 22 :
Prove That
∫0∞xsin(2xx2+1)cos(2xx2−1)dx=2π
This problem has been solved by Aditya Kumar.
Log in to reply
Solution to Problem 22:
Use the formula 2sin(A)cos(B)=sin(A+B)−sin(A−B)
Therefore, we get: I=21{∫0∞xsin(x)+∫0∞xsin(x1)dx}
It is easy to see that both the integrals are same. Hence, I=∫0∞xsin(x)=2π
Log in to reply
@Rajdeep Dhingra post the next problem. I can't come up with one of this level.
Log in to reply
@Samuel Jones@Harsh Shrivastava@Aareyan Manzoor
I don't have anymore problems. Any one can post the Next Question.Log in to reply
Problem 25
Let k be a positive real number
then prove
∫01∫01{yxk}dxdy=(k+1)22k+1−k+1γ
Hint:let t=yxk
Log in to reply
Solution to Problem 25:
I=∫01∫01{yxk}dxdy
I=∫01(∫01{yxk}dy)dx
By IBP, we get: I=(k+1xk+1∫xk∞t2{t}dt)01+k+1k∫01xkdx
Hence on simplifying, we get: ∫01∫01{yxk}dxdy=(k+1)22k+1−k+1γ
Log in to reply
Problem 27
Evaluate
∫01∫01{x+y1}ndxdy
with n∈N
Hint:let x+y=t
Log in to reply
Problem 4
Prove ∫0∞(1+x)(2+x)(3+x)x dx=2π(−22−1+3)
This problem has been solved by Harsh Shrivastava.
Log in to reply
Log in to reply
Problem 5:
Evaluate
∫01xln(1+x)
This problem was first solved by Aaryen Manzoor using a disallowed method and then solved by Aditya Kumar using a legal method. Credit is still given to Aaryen.
Log in to reply
solution to problem 5
replace x with -x to have ∫0−1xln(1−x)dx n≥1∑nxn=ln(1−x) dividing by x and integrating from 0 to a ∫0axln(1−x)dx=n≥1∑n2an We have ∫0−1xln(1−x)dx=n≥1∑n2(−1)n=n≥1∑n2(−1)n=n≥1∑n21−2n≥1∑(2n)21=21n≥1∑n21=12π2
Log in to reply
Yeah there is a hell easy method.
:)
Log in to reply
Alternate solution to Problem 5:Using MacLaurin expansion of ln(1+x):ln(1+x)=r=1∑∞(−1)r+1rxr⟹xln(1+x)=r=1∑r=∞(−1)r+1rxr−1⟹I=∫01(r=1∑r=∞(−1)r+1rxr−1)dx⟹I=r=1∑∞(−1)r+1r21=12π2.
Log in to reply
Solution to Problem 5: