Hi Brilliant! Just like what Anastasiya Romanova conducted last year, this year I would also like to conduct an integration contest.
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.
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.
You are NOT allowed to post a multiple integrals problem.
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).
Please post your solution and your proposed problem in a single new thread.
Format your post is as follows:
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The comments will be easiest to follow if you sort by "Newest":
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Top NewestProblem 1:
Evaluate: ∫011−x2xln(x)dx
This problem has been solved by Surya Prakash.
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Another solution: Let (1−x2)=t
(1−x2)−xdx=dt
So integration becomes
−∫10ln(1−t2)dt
−21∫10ln(1−t2)dt
21∫01ln(1−t2)dt
21∫01(ln(1−t)+ln(1+t))dt
This can be easily evaluated and comes out to be ln2−1
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Solution to Problem 1:
∫011−x2xln(x)dx Take t=arcsin(x)⟺dt=1−x2dx and the limits changes to t=0 to t=2π. ∫02πsin(t)ln(sin(t))dt
Consider,
B(x,y)=∫01tx−1(1−t)y−1dt
Taking the transformation t=sin2(z)⟺dt=2sin(z)cos(z)dz, the integral transforms into
B(x,y)=2∫0π/2sin2x−1(z)cos2y−1(z)dz
But B(x,y)=Γ(x+y)Γ(x)Γ(y)
So, ∫0π/2sin2x−1(z)cos2y−1(z)dz=21Γ(x+y)Γ(x)Γ(y)
Differentiate it w.r.t x
2∫0π/2sin2x−1(z)cos2y−1(z)ln(sin(z))dz=21∂x∂Γ(x+y)Γ(x)Γ(y)∫0π/2sin2x−1(z)cos2y−1(z)ln(sin(z))dz=41Γ(x+y)Γ(x)Γ(y)(ψ(x)−ψ(x+y))
Take x=1 and y=21.
∫0π/2sin(z)ln(sin(z))dz=41Γ(3/2)Γ(1)Γ(1/2)(ψ(1)−ψ(3/2))=ln(2)−1
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Problem 2:
Evaluate:
∫0∞exp(x3)xcos(x3)dx
This problem has been solved by Tanishq Varshney.
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Solution to Problem 2
Firstly I would add a proof of my solution
∫0∞e−axxn−1dx=anΓ(n) where Γ(n) is Gamma function
Replace a→a+ib where i=−1
∫0∞e−axe−ibxxn−1dx=(a+ib)nΓ(n)
Put a=rcosy and b=rsiny.
So that r2=a2+b2 and y=arctan(ab)
Using de moviers theorem
∫0∞e−ax(cosbx−isinbx)xn−1dx=rnΓ(n)(cosny+sinny)−1
Comparing the real and imaginary parts we finally get
∫0∞xn−1e−ax(cosbx)dx=rnΓ(n)(cosny)
now in the given integral put x3=t
The integral now becomes
31∫0∞t−31e−tcos(t)dt
here n=32;a=1 and b=1
Thus we get 31231Γ(32)cos(6π)
3.2343Γ(32)
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Problem 23:
Evaluate:∫01log(1+x)log(1−x3)dx
Due to time constraint, Aditya Kumar decided to post a solution himself.
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Solution to Problem 23:
Substitute: x⟶1+x1−x
I=∫01ln(1+x)ln(1−x3)dx=2∫01ln((1+x)32x(3+x2))ln(1+x2)(1+x)2dx
On separating the integrands,
ln((1+x)32x(3+x2))ln(1+x2)=ln22−ln2lnx+ln2ln(3+x2)−4ln2ln(1+x)−lnxln(1+x)+3ln2(1+x)−ln(1+x)ln(3+x2)
1st integral:
\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12
2nd integral:
∫01(1+x)2lnxdx=n≥1∑n(−1)n=−ln2
3rd integral:
\int_0^1 \frac{\ln(1+x)}{(1+x)^2}dx=-\frac{\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac{dx}{(1+x)^2}dx=\frac12-\frac{\ln2}{2}
4th integral:
\int_0^1 \frac{\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2x}{(1+x)(3+x^2)}dx \\=-\ln2+\ln3+2\operatorname{Re} \int_0^1 \frac{dx}{(1+x)(x+i\sqrt{3})}=-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}
5th integral:
\int_0^1 \frac{\ln x \ln(1+x)}{(1+x)^2}dx=-\frac{\ln x\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac1{1+x}\left(\frac{\ln(1+x)}{x}+\frac{\ln x}{1+x}\right)dx \\=-\ln2+\int_0^1 \frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{\pi^2}{12}-\frac12\ln^22-\ln2
6th integral:
\int_0^1 \frac{\ln^2(1+x)}{(1+x)^2}dx=-\frac{\ln^2(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2\ln(1+x)}{(1+x)^2}dx \\=1-\ln2-\frac12\ln^22
7th integral:
\int_0^1 \frac{\ln(1+x)\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(1+x)\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac1{1+x}\left(\frac{2x\ln(1+x)}{3+x^2}+\frac{\ln(3+x^2)}{1+x}\right)dx \\=-\ln^22-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}+2J
where,
J=∫01(1+x)(3+x2)xln(1+x)dx=Re∫01(1+x)(x+i3)ln(1+x)dx=−81ln22−Rei3−11∫01x+i3ln(1+x)dx
Now on using the formula:
∫01x+aln(1+x)dx=ln2lna−1a+1+Li2(1−a2)−Li2(1−a1)
Putting a=i3 then gives, after taking the real part, and assuming the principle value of the logarithm,
J=−81ln22+12π3ln2+41ReLi2(eiπ/3)−43ImLi2(eiπ/3)−41ReLi2(21eiπ/3)+43ImLi2(21eiπ/3)
Using:
ReLi2(eiπ/3)=36π2,
ImLi2(eiπ/3)=231ψ1(31)−33π2,
Li2(21eiπ/3)=−Li2(−3i)−21ln2(43−i43),
and
ReLi2(−3i)=n=1∑∞(2n)23n(−1)n=41Li2(−31)
we have ReLi2(21eiπ/3)=72π2−81ln2(43)−41Li2(−31)
and ImLi2(21eiπ/3)=12πln(43)−ImLi2(−3i).
Finally, putting everything together gives: I=ln22−81ln23+2ln2ln3−23ln3−6ln2+6−43π(2+ln3)−7237π2+21ψ1(31)−41Li2(−31)+3ℑLi2(−3i)
...
This was not original. I found it very awesome, so I thought to share it here.
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I don't know what's wrong with the rendering of the Latex, since it sometimes render and sometimes it doesn't. So I'm going to post in pictures the one's that didn't render. I think it's better to move this discussion to another note.
1st Integral:
3rd Integral:
4th Integral:
5th Integral:
6th Integral:
7th Integral:
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Whoops, I accidentally deleted Tanishq's comment:
Problem 5:
∫0π/2tan−1(1729cos2x)dx
This problem has been solved by Sudeep Salgia.
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I'll post the outline of the solution as it is a bit cumbersome. Let
I(a)=∫02πarctan(acos2x)dx⇒I′(a)=1+a2cos4xcos2xdx=sec4x+a2sec2xdx Substitute tanx=t, to get I′(a)=∫0∞t4+2t2+(a2+1)1dt . Put a2+1=k2 and with some simple rearrangement, we get,
2kI′(a)=∫0∞(t−tk)2+2(k+1)1+t2k−∫0∞(t+tk)2+2(−k+1)1−t2k Substitute t−tk=z and t+tk=y in the respective integrals and evaluate them easily (They come out to be arctan of something ). After all this we can write,
I(a)=22π∫a2+11a2+1+11da
This can be rearranged to be written as
I(a)=22π∫1+2a2+1−11×22a2+1−11×21×a2+1ada
This evaluates to 2πarctan⎝⎛2a2+1−1⎠⎞
The constant of integration is 0 from I(0)=0. Put a=1729 to get the answer.
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Problem 24:
Prove that
∫0π/2sin8x+cos8x1dx=210−2π.
This problem has been solved by Tanishq Varshney.
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Solution to Problem 24
∫0∞1+xxm−1dx=sinmππ0<m<1
Refer to problem 18 for this step.
Now in the integral tanx→t
∫0∞1+t8(1+t2)3dt
t8→z
81(∫0∞1+zz87−1+1+z3z85−1+1+z3z83−1+1+zz81−1dz)
→4π(sin(8π)1+cos(8π)3)
=210−2π
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Problem 3
Evaluate:
∫02πtan53xdx
This problem has been solved by Aditya Kumar.
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Solution to Problem 3: ∫02π(tanx)53dx=∫02π(sinx)53(cosx)5−3dx=2B(54,51)...........{B(x,y)isbetafunction}=2Γ(1)Γ(54)Γ(51)=2sin(5π)π...............{byeuler′sreflectionformula}=10−252π
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Problem 4:
Let P(x) be a polynomial such that P(x)P(x+1)=x2−x+1x2+x+1 and P(2)=3, evaluate∫01(tan−1(P(x)).tan−1(1−xx))dx.
This problem has been solved by Tanishq Varshney.
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Solution to Problem 4
Clearly P(x)=x2−x+1 and P(1−x)=P(x)
Also tan−1(x)=cot−1(x1)
tan−1(x)+cot−1(x)=2π
I=∫01(tan−1(P(x))cot−1(1−xx))dx
2I=2π∫01tan−1(x2−x+1)dx
2I=2π(2π−∫01tan−1(1+x(x−1)1))
2I=2π(2π−∫01tan−1(1+x(x−1)x−(x−1))dx)
2I=2π(2π−(∫01tan−1(x)dx−∫01tan−1(x−1)dx))
2I=2π(2π−(2π−2ln(2)))
I=2πln(2)
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Problem 9:
Evaluate the indefinite integral: ∫(xsinx+5cosx)2x2+20dx.
This problem has been solved by Surya Prakesh.
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x2+20=(xsin(x)+5cos(x))2+(5sin(x)−xcos(x))2−5=(xsin(x)+5cos(x))2+(5sin(x)−xcos(x))2−5cos2(x)−5sin2(x)−xcos(x)sin(x)+xcos(x)sin(x)=(xsin(x)+5cos(x))(5cos(x)+xsin(x)−cos(x))+(5sin(x)−xcos(x))(5sin(x)−xcos(x)−sin(x))=(xsin(x)+5cos(x))(5cos(x)+xsin(x)−cos(x))−(5sin(x)−xcos(x))(xcos(x)+sin(x)−5sin(x))=(xsin(x)+5cos(x))dxd(5sin(x)−xcos(x))−(5sin(x)−xcos(x))dxd(xsin(x)+5cos(x))
Let u=xsin(x)+5cos(x) and v=5sin(x)−xcos(x). So, x2+20=udxdv−vdxdu=u2dxd(uv).
So, ∫(xsin(x)+5cos(x))2x2+20dx=∫u2u2dxd(uv)dx=uv=xsin(x)+5cos(x)5sin(x)−xcos(x)
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Problem 13: (Originally posted by Tanishq)
Evaluate ∫01xln(1−x1+x)dx.
This problem has been solved by Surya Prakesh.
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By using integration by parts,
∫xln(1−x1+x)dx=2x2ln(1−x1+x)+x−21ln(1+x)+21ln(1−x)
So,
∫01xln(1−x1+x)dx=a→0lim∫01−axln(1−x1+x)dx=a→0lim2(1−a)2ln(a2−a)+1−a−21ln(2−a)+21lna=1
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Problem 6:
Show that for positive reals, a,b where a<b, ∫abarccos((a+b)x−abx)dx=4πa+b(b−a)2
Statutory warning: The substitutions need not be very intuitive.
This problem has been solved Aditya Kumar.
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Solution to Problem 6:
I=∫abarccos((a+b)x−abx)dxI=∫abarccos((ab(a+b))x−1abx)dx(ab(a+b))x−1=tI=∫baab(a+b2tarccos(t(t2+1)ab))dtI=∫baab(a+b2tarccos((t+t1)ab))dt
Now using Integration By Parts,
I=∫baab((ab−ba)2−(t−t1)2t2+1)dtt=p+1+p2I=∫21(−ab+ba)21(ab−ba)⎝⎛4ba+ab+2−p22p2⎠⎞dp
Now,
p2=(4ba+ab+2)sinθ
This can be evaluated easily.
∴I=4π(a+b)(b−a)2
Moral of the question:
To whatever extent we learn anything, we must never forget our basics.
Note that this question is solved using very basic concepts.
Any elegant modification to the solution will be accepted.
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Problem 8:
Evaluate
f(ω)=2π1∫−∞∞e−a2t2eiωtdt
This problem has been solved by both Tanishq Varshney (first) and Surya Prakash (second) almost at the same time.
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Solution to problem 8
t→−t
adding this with the original equation
2f(w)=2π2∫−∞∞e−a2t2coswtdt
as eiwt+e−iwt=2coswt=2ℜ(eiwt)
f(w)=2π1ℜ⎝⎛∫−∞∞e−a2t2eiwtdt⎠⎞
Let at=z
f(w)=a2π1ℜ⎝⎛∫−∞∞e−z2eiawzdz⎠⎞
f(w)=a2π1ℜ⎝⎛∫−∞∞e−(z−2aiw)2e−4a2w2dz⎠⎞
Using Gaussian integral
f(w)=a21e−4a2w2
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f(ω)=2π1∫−∞∞e−a2t2eiωtdt
It implies that f(ω)=[F(F)](ω), where F(t)=e−a2t2.
Now, differentiate F(t), we get F′(t)=−2a2tF(t).
Applying fourier transformation on both sides gives iωf(ω)=−2ia2f′(ω).
⟹ωf(ω)=−2a2dωdf(ω)⟹f(ω)=ce−4a2ω2
Since c=f(0)=2π1∫−∞∞e−a2t2dt=22π1∫0∞e−a2t2dt=a2π1Γ(1/2)=a21.
Therefore, f(ω)=a21e−4a2ω2.
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Problem 10:
Evaluate
∫0π/2(9tan2(x)+16)31dx
This problem has been solved by Pi Han Goh.
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Let I denote the value of the integral. And let 3tan(x)=4tan(y), differentiate with respect to x: 3sec2(x)=4sec2(y)(dxdy)⇒dx=3sec2(x)4sec2(y)dy=9sec2(x)12sec2(y)dy=9tan2(x)+912sec2(y)dy=16tan2(y)+912sec2(y)dy
And of course, 9tan2(x)+16=(3tan(x))2+16=16(tan2(y)+1)=16sec2(y).
When x=0, y=0; when x→2π−, y→2π− as well. The integral becomes
I=====∫0π/2(16sec2(y))31⋅16tan2(y)+912sec2<