Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 2(Part 1)
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.
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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.
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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, Dirichlet eta function, dilogarithm, digamma function, trigonometric integral, Wallis' integral, Bessel function, contour integration and Ramanujan's Master Theorem (including Mellin Transform).
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UPDATE:
Tanishq Varshney has been banned from this contest indefinitely.
Contour integration is allowed in the contest.
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Top NewestProblem 44:
Evaluate ∫0∞(ln2(x)+π2ln(1+x))x2dx.
Due to time constraint, the author decided to post their own solution.
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Lemma : I=∫0∞y2+π2e−kydy=π(−1)k∫kπ∞xsin(x)dx , for k is a whole number.
Proof : We being with noting that :
w1∫0∞e−stsinwtdt=s2+w21
So we can rewrite the integral on the left side as :
I=π1∫0∞e−kydy∫0∞e−tysinπtdt=π1∫0∞sinπt∫0∞e−(k+t)ydydt
⇒I=π1∫0∞t+ksinπtdt
⇒I=π1∫k∞tsin(πt−πk)dt
⇒I=(−1)k∫k∞tsin(πt)dt
⇒I=(−1)k∫kπ∞tsin(t)dt
Hence proved
Solution :
We start with x=e−y to get out integral as :
I=∫−∞∞y2+π2eyln(1+e−y)dy
Break in into two parts :
I=∫0∞y2+π2eyln(1+e−y)dy+∫−∞0y2+π2eyln(1+e−y)dy
In the second part put y=−x to get :
I=∫0∞y2+π2eyln(1+e−y)dy+∫0∞x2+π2e−xln(1+ex)dy
Manipulating it we have :
I=∫0∞y2+π2eyln(1+e−y)dy+∫0∞y2+π2e−yln(1+e−y)dy+∫0∞y2+π2ye−ydy
⇒I=∫0∞y2+π2(ey+e−y)ln(1+e−y)dy+∫0∞y2+π2ye−ydy
We write I=J+K
J=∫0∞x2+π2(ex+e−x)ln(1+e−x)dx
Wrting ln(1+e−x) in it's taylor series we have :
J=∫0∞x2+π2(ex+e−x)r=1∑∞r(−1)r−1e−rxdx
We interchange the integral and sum :
J=r=1∑∞r(−1)r−1(∫0∞x2+π2e−(r−1)xdx+∫0∞x2+π2e−(r+1)xdx)
Using the lemma we have :
J=π1r=1∑∞r1(∫(r−1)π∞xsin(x)dx+∫(r+1)π∞xsin(x)dx)
Now it is worthy noting that :
∫rπ∞xsin(x)dx=∫1∞xsin(rπx)dx , for all positive r, what if it was applied for r=0, then the left integral is 2π, while the write one is 0, so for r=0, we can write it as :
∫rπ∞xsin(x)dx=∫1∞xsin(rπx)dx+2π, for r=0, having said that we proceed further :
J=π1r=1∑∞r1(∫21∞xsin(2(r−1)πx)+sin(2(r+1)πx)dx)+21
⇒J=π2r=1∑∞r1(∫21∞xcos(2πx)sin(2rπx)dx)+21
Again changing sum and integral we have :
⇒J=π2∫21∞cos(2πx)r=1∑∞rxsin(2rπx)dx+21
It is a well known result that :
r=1∑∞rπsin(2rπx)=⌊x⌋+21−x
Using this we have :
J=2∫21∞xcos(2πx)(−x+⌊x⌋+21)dx+21
Leaving this let's evaluate K first :
Using the same techniques as used in proving the lemma, we can show that :
K=∫0∞x2+π2xe−xdx=−∫1/2∞xcos(2πx)dx
Now I=J+K=2∫21∞xcos(2πx)(−x+⌊x⌋+21)dx+21−∫1/2∞xcos(2πx)dx=21+2∫1/2∞xcos(2πx){x}dx
I=21+2∫1∞xcos(2πx){x}dx+2∫211cos(2πx)dx=21+2∫1∞xcos(2πx){x}dx
Now we will be evaluating :
M=∫1∞xcos(2πx){x}dx
It can be written as :
M=n=1∑∞∫nn+1xcos(2πx)(x−n)dx
⇒M=−n=1∑∞n∫nn+1xcos(2πx)dx
M can also be written as :
M=−n=1∑∞∫n∞xcos(2πx)dx
This manipulation can easily be justified with the help of properties of double summations :
Using integration by parts we have :
M=n=1∑∞∫n∞2πx2sin(2πx)dx−n=1∑∞2πnsin(2πn)
⇒M=n=1∑∞∫n∞2πx2sin(2πx)dx
⇒M=n=1∑∞∫1∞2πnx2sin(2πnx)dx
Again changing sum and integral we have :
⇒M=∫1∞n=1∑∞2πnx2sin(2πnx)dx
Using the result we have used above we have :
M=21∫1∞2x21−x2{x}dx
⇒M=21∫1∞2x21−x2{x}dx
For solving H=∫1∞x2{x}dx
write it as a sum :
H=n=1∑∞∫nn+1x1−x2ndx
⇒H=n=1∑∞ln(n+1)−ln(n)−n+11
H=n→∞lim1+ln(n+1)−Hn+1=1−γ
Hence M=2γ−41
Finally we got :
I=γ
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Problem 26:
Show that ∫0πe2cosxsin2(sinx)dx=2π(J0(2i)−1).
Where Jn(z) denote the Bessel Function of First Kind.
This problem has been solved by Tanishq Varshney.
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Solution to Problem 26
Firstly some common things
2sin2x=1−cos(2x)
πIn(z)=∫0πezcosθcos(nθ)dθ
where In(z) is modified Bessel function of first kind
Also In(z)=e2−inπJn(iz) where i2=−1 and Jn(z) is Bessel function of first kind.
In this question we have n=0
Now in the integral
∫0πe2cosx(21−cos(2sinx))dx
On separating
=2πI0(2)−21∫0πe2cosxcos(2sinx)dx
Let L=∫0πe2cosxcos(2sinx)dx
L=ℜ(∫0πe2cosxe2isinxdx)
L=ℜ(∫0πe2(cosx+isinx)dx)
L=ℜ(∫0πe2eixdx)
L=ℜ⎝⎛∫0πk=0∑∞k!(2eix)kdx⎠⎞
L=∫0πk=0∑∞k!2kcos(kx)dx
Notice that ∫0πcos(kx)dx=0∀k≥1k is an integer
Thus only k=0 case is considered and hence L=π
On combining we have
2π(I0(2)−1)
2π(J0(2i)−1)
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Problem 36:
Evaluate ∫0∞ex−1sinxdx.
This problem has been solved by Deep Seth.
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Problem 27:
Evaluate
∫02(2x2−x3)1/3dx.
This problem has been solved by Rajorshi Chaudhuri.
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Substitute x=2t So the equation becomes ∫01 t−2/3(1−t)−1/3dt This comes out as Γ(31)Γ(32) which is equal to 32π
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Problem 28:
Evaluate ∫0∞(1+x2)3x5sin(x)dx.
This problem has been solved by Sudeep Salgia.
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Solution to Problem 28:
The following result is directly used in the solution:∫0∞(a2+x2)2cos(mx)dx=4a3πe−am(1+am).
This can be obtained easily by differentiating the integral w.r.t. a evaluated in solution of problem 15 of this contest. Consider,
I(b)=∫0∞(1+x2)3xsin(bx)dx. The required integral is ∂b4∂4I(b). Integrating by parts, we get,
I(b)=−4(1+x2)2sin(bx)∣∣∣∣0∞+4b∫0∞(1+x2)2cos(bx)dx=16πe−b(b+b2)
Differentiate four times w.r.t b and put b=1 to get the value of the integral as 8eπ.
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Problem 29:
Show that ∫01x(x+1)ln(cos(2πx))dx=21(ln2)2−(lnπ)(ln2).
This problem has been solved by Tanishq Varshney.
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Solution to Problem 29
∫01xln(cos(2πx))dx−∫01x+1ln(cos(2πx))dx
=∫01xln(2sin(2πx)sin(πx))dx−∫01x+1ln(cos(2πx))dx
=∫01xln(2sin(πx))dx−∫01xln(sin(2πx))dx−∫01x+1ln(cos(2πx))dx
=∫01xln(2sin(πx))dx−∫011−xln(sin(2π(1−x)))dx−∫01x+1ln(cos(2πx))dx
In the second integral x→−x
=∫01xln(2sin(πx))dx−∫−101+xln(cos(2πx))dx−∫01x+1ln(cos(2πx))dx
=∫01xln(2sin(πx))dx−∫−11x+1ln(cos(2πx))dx
Now in the second integral x+1→x
=∫01xln(2sin(πx))dx−∫02xln(sin(2πx))dx
=a→0lim⎝⎛∫a1xln(2sin(πx))dx−∫a2xln(sin(2πx))dx⎠⎞
Again in the second integral 2x→x
=a→0lim⎝⎛−ln2∫a1x1dx+∫a1xln(sin(πx))dx−∫2a1xln(sin(πx))dx⎠⎞
=a→0lim⎝⎛−ln2∫a1x1dx−∫2aaxln(sin(πx))dx⎠⎞
We know sinθ≈θ when θ approaches zero.
=a→0lim⎝⎛−ln2∫a1x1dx−∫2aaxln(πx)dx⎠⎞
=a→0lim(ln2lna−lnπln2−21((lna)2−(ln2a)2))
using a2−b2=(a−b)(a+b) and on simplifying the terms containing a get cancelled and we finally have
21(ln2)2−(lnπ)(ln2)
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CHEATER!!!
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Problem 30:
Prove that ∫0π/8ln(tan2x)dx=−2G.
Where G denotes the Catalan's constant, G=n=0∑∞(2n+1)2(−1)n≈0.91596559.
This problem has been solved by Surya Prakesh (first) and Julian Poon (second) almost at the same time.
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Lemma : ∫01taln(t)dt=−(a+1)21
This lemma is already proved in many problems in this contest.
∫0π/8ln(tan(2x))dx=1/2∫011+t2ln(t)dt
I used the substitution t=tan(2x).
∫011+t2ln(t)dt=∫01i=0∑∞(−1)it2iln(t)dt=i=0∑∞(−1)i∫01t2iln(t)dt=−i=0∑∞(−1)i(2i+1)21=−β(2)=−G
Therefore,
∫0π/8ln(tan(2x))dx=2−G
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Not sure if this is considered but...
y=ln(tan(2x))
x=2tan−1(ey)
∫08πln(tan(2x))dx=∫−∞02tan−1(ey)dy=−21∫0∞tan−1(e−y)dy
A known identity of the catalan constant is G=∫0∞tan−1(e−y)dy
Hence the integral is −2G
The identity above can be deduced through the Taylor Series of tan−1(x)
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Problem 34:
Prove That ∫0∞(x2sin(x2))⋅(x2+x−1) dx=4π−8π.
This problem has been solved by Surya Prakesh.
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Lemma:
Consider the Gamma Function Γ(s)=∫0∞e−tts−1dt
Take the substitution, t=pun.
Γ(s)=nps∫0∞e−pununs−1du
Taking p=a+ib and comparing imaginary parts we get,
∫0∞e−aununs−1sin(bun)du=n∣p∣sΓ(s)sin(αs)
where p=a+ib and α is the argument of p.
1st Integral:
∫0∞sin(x2)dx
Comparing this with our lemma, we get a=0, b=1, n=2, s=1/2. Which gives us that ∣p∣=1 and α=π/2.
On substitution we get the integral evaluated to be I1=22π.
2nd Integral:
∫0∞xsin(x2)dx
Take the substitution, x=t1/2. We get,
21∫0∞tsin(t)dt
This is well known Integral (Dirichlet integral): ∫0∞xsin(x)dx=2π.
So, this integral evaluates to be I2=4π.
3rd Integral:
This can evaluated in similar way as we obtained for first integral. And this is evaluated to be I3=2π.
So, the overall integral I=I1+I2−I3=4π−22π
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Problem 40:
Prove that ∫0π/2x(sinxln(sinx)−2)cotx dx=−8π[Γ(41)]2
This problem has been solved by Surya Prakesh.
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∫0π/2x(sinxln(sinx)−2)cot(x)dx=x(∫(sinxln(sinx)−2)cot(x)dx)∣∣∣∣∣0π/2−∫0π/2(∫(sinxln(sinx)−2)cot(x)dx)dx
∫(sinxln(sinx)−2)cotxdx=∫(ttln(t)−2)dt=2∫(t)2lntd(t)−td(ln(t))=−2tln(t)=−2sin(x)ln(sin(x))
So,
x(∫(sinxln(sin(x))−2)cotxdx)∣∣∣∣∣0π/2=sin(x)−2xln(sin(x))∣∣∣∣∣0∞=0−a→0limsin(a)−2aln(sin(a))=a→0limsin(a)2aln(sin(a))=0
Above limit is evaluated using L'Hospital rule.
So,
∫0π/2x(sinxln(sinx)−2)cotxdx=−∫0π/2(∫(sinxln(sinx)−2)cotxdx)dx=2∫0π/2sin(x)ln(sin(x))dx
Let I(a)=∫0π/2sin2a−1(x)dx. So, our required integral is I′(1/4). By taking t=sin2(x), the integral transforms to I(a)=21∫01ta−1(1−t)−1/2dt=21B(a,1/2)=21Γ(a+1/2)Γ(a)Γ(1/2)
Differentiate it w.r.t a,
I′(a)=21Γ(a+1/2)Γ(a)Γ(1/2)(ψ(a)−ψ(a+1/2))
I′(1/4)=21Γ(3/4)Γ(1/4)Γ(1/2)(ψ(1/4)−ψ(3/4))
Using Euler's reflection formula. We get
Γ(1/4)Γ(3/4)=sin(π/4)π=2π⟹Γ(3/4)=Γ(1/4)2π
Also, ψ(1/4)−ψ(3/4)=πcot(3π/4)=−π.
Therefore the required integral evaluates to
−8π(Γ(41))2
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Problem 31:
Prove that n→∞lim∫011−xxn−x2ndx=ln(2).
This problem has been solved by Julian Poon.
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Using basic concepts in integration:
∫011−xxn−x2ndx=∫01xn1−x1−xndx=∫01xnk=0∑n−1xk dx=k=0∑n−1∫01xn+k dx=k=0∑n−1n+k+11=n1k=1∑n1+nk1
Then:
n→∞lim∫011−xxn−x2ndx=n→∞limn1k=1∑n1+nk1=∫011+x1dx=[ln(1+x)]01=ln2
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Substituting k=0∑∞xk=1−x1
∫01k=0∑∞xk+n−k=0∑∞xk+2ndx=k=1∑∞k+n1−k=1∑∞k+2n1
=H2n−Hn=ln(2)
The last step is from the approximation Hn=γ+ln(n) as n→∞