Brilliant Integration Contest - Season 2 (Part 2)

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.

  • 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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**SOLUTION OF PROBLEM xxx (number of problem) :**

**[Post your solution here]**

**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":


UPDATE:

  • Tanishq Varshney has been banned from this contest indefinitely.

  • Contour integration is allowed in the contest.

Note by Aditya Kumar
3 years, 10 months ago

No vote yet
1 vote

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Problem 44:

Evaluate 0(ln(1+x)ln2(x)+π2)dxx2.\displaystyle \int _{ 0 }^{ \infty }{ \left( \dfrac{ \ln { (1+x) } }{ { \ln }^{ 2 }(x)+{ \pi }^{ 2 } }\right) \frac { dx }{ {x}^{2} } } .

Due to time constraint, the author decided to post their own solution.

Ronak Agarwal - 3 years, 10 months ago

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Lemma : I=0ekyy2+π2dy=(1)kπkπsin(x)xdx \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ -ky } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } =\frac{{ (-1) }^{ k }}{\pi} \int _{ k\pi }^{ \infty }{ \frac { sin(x) }{ x } dx } , for kk is a whole number.

Proof : We being with noting that :

1w0estsinwtdt=1s2+w2 \displaystyle \frac { 1 }{ w } \int _{ 0 }^{ \infty }{ { e }^{ -st }sinwtdt } =\frac { 1 }{ { s }^{ 2 }+{ w }^{ 2 } }

So we can rewrite the integral on the left side as :

I=1π0ekydy0etysinπtdt=1π0sinπt0e(k+t)ydydt \displaystyle I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ { e }^{ -ky }dy } \int _{ 0 }^{ \infty }{ { e }^{ -ty }sin\pi tdt } =\frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ sin\pi t\int _{ 0 }^{ \infty }{ { e }^{ -(k+t)y }dy } dt }

I=1π0sinπtt+kdt \displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ \frac { sin\pi t }{ t+k } dt }

I=1πksin(πtπk)tdt \displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ k }^{ \infty }{ \frac { sin(\pi t-\pi k) }{ t } dt }

I=(1)kksin(πt)tdt\displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k }^{ \infty }{ \frac { sin(\pi t) }{ t } dt }

I=(1)kkπsin(t)tdt \displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k\pi }^{ \infty }{ \frac { sin(t) }{ t } dt }

Hence proved

Solution :

We start with x=eyx={e}^{-y} to get out integral as :

I=eyln(1+ey)y2+π2dy \displaystyle I = \int _{ -\infty }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

Break in into two parts :

I=0eyln(1+ey)y2+π2dy+0eyln(1+ey)y2+π2dy \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ -\infty }^{ 0 }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

In the second part put y=xy=-x to get :

I=0eyln(1+ey)y2+π2dy+0exln(1+ex)x2+π2dy\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { e }^{ -x }\ln { (1+{ e }^{ x }) } }{ { x }^{ 2 }+{ \pi }^{ 2 } } dy }

Manipulating it we have :

I=0eyln(1+ey)y2+π2dy+0eyln(1+ey)y2+π2dy+0yeyy2+π2dy \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { e }^{ -y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

I=0(ey+ey)ln(1+ey)y2+π2dy+0yeyy2+π2dy\displaystyle \Rightarrow I= \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ y }+{ e }^{ -y })\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }

We write I=J+K I = J+K

J=0(ex+ex)ln(1+ex)x2+π2dx \displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x })\ln { (1+{ e }^{ -x }) } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx

Wrting ln(1+ex) \ln(1+{e}^{-x}) in it's taylor series we have :

J=0(ex+ex)x2+π2r=1(1)r1rerxdx\displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x }) }{ { x }^{ 2 }+{ \pi }^{ 2 } } \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } { e }^{ -rx } } } dx

We interchange the integral and sum :

J=r=1(1)r1r(0e(r1)xx2+π2dx+0e(r+1)xx2+π2dx) \displaystyle J = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } (\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r-1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx+\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r+1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx } )

Using the lemma we have :

J=1πr=11r((r1)πsin(x)xdx+(r+1)πsin(x)xdx) \displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ (r-1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } +\int _{ (r+1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } } dx) }

Now it is worthy noting that :

rπsin(x)xdx=1sin(rπx)xdx \displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx } , for all positive rr, what if it was applied for r=0r=0, then the left integral is π2\frac{\pi}{2}, while the write one is 0, so for r=0r=0, we can write it as :

rπsin(x)xdx=1sin(rπx)xdx+π2 \displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx } +\frac { \pi }{ 2 } , for r=0 r=0, having said that we proceed further :

J=1πr=11r(12sin(2(r1)πx)+sin(2(r+1)πx)xdx)+12 \displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { \sin { (2(r-1)\pi x) } +\sin { (2(r+1)\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }

J=2πr=11r(12cos(2πx)sin(2rπx)xdx)+12 \displaystyle \Rightarrow J = \frac { 2 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { cos(2\pi x)\sin { (2r\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }

Again changing sum and integral we have :

J=2π12cos(2πx)r=1sin(2rπx)rxdx+12 \displaystyle \Rightarrow J = \frac { 2 }{ \pi } \int _{ \frac { 1 }{ 2 } }^{ \infty }{ \cos { (2\pi x) } \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ rx } } dx } +\frac { 1 }{ 2 }

It is a well known result that :

r=1sin(2rπx)rπ=x+12x \displaystyle \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ r\pi } } =\left\lfloor x \right\rfloor +\frac { 1 }{ 2 }-x

Using this we have :

J=212cos(2πx)x(x+x+12)dx+12 \displaystyle J = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 }

Leaving this let's evaluate KK first :

Using the same techniques as used in proving the lemma, we can show that :

K=0xexx2+π2dx=1/2cos(2πx)xdx \displaystyle K = \int _{ 0 }^{ \infty }{ \frac { x{ e }^{ -x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } dx } =-\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx }

Now I=J+K=212cos(2πx)x(x+x+12)dx+121/2cos(2πx)xdx=12+21/2cos(2πx)x{x}dx \displaystyle I = J+K = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 } -\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } =\frac { 1 }{ 2 } +2\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }

I=12+21cos(2πx)x{x}dx+2121cos(2πx)dx=12+21cos(2πx)x{x}dx \displaystyle I = \frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx } +2\int _{ \frac { 1 }{ 2 } }^{ 1 }{ \cos { (2\pi x) } dx } =\frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }

Now we will be evaluating :

M=1cos(2πx)x{x}dx \displaystyle M = \int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }

It can be written as :

M=n=1nn+1cos(2πx)x(xn)dx \displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } (x-n)dx } }

M=n=1nnn+1cos(2πx)xdx\displaystyle \Rightarrow M = -\sum _{ n=1 }^{ \infty }{ n\int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } dx } }

MM can also be written as :

M=n=1ncos(2πx)xdx\displaystyle M =-\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } }

This manipulation can easily be justified with the help of properties of double summations :

Using integration by parts we have :

M=n=1nsin(2πx)2πx2dxn=1sin(2πn)2πn \displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } } -\sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi n) } }{ 2\pi n } }

M=n=1nsin(2πx)2πx2dx\displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } }

M=n=11sin(2πnx)2πnx2dx \displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ 1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } dx } }

Again changing sum and integral we have :

M=1n=1sin(2πnx)2πnx2dx \displaystyle \Rightarrow M = \quad \int _{ 1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } } dx }

Using the result we have used above we have :

M=12112x2{x}x2dx \displaystyle M = \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }

M=12112x2{x}x2dx \displaystyle \Rightarrow M = \quad \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }

For solving H=1{x}x2dx \displaystyle H = \int _{ 1 }^{ \infty }{ \frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }

write it as a sum :

H=n=1nn+11xnx2dx \displaystyle H = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { 1 }{ x } -\frac { n }{ { x }^{ 2 } } dx } }

H=n=1ln(n+1)ln(n)1n+1 \displaystyle \Rightarrow H = \sum _{ n=1 }^{ \infty }{ \ln { (n+1) } -\ln { (n) } -\frac { 1 }{ n+1 } }

H=limn1+ln(n+1)Hn+1=1γ \displaystyle H = \lim _{ n\rightarrow \infty }{ 1+\ln { (n+1)-{ H }_{ n+1 } } } =1-\gamma

Hence M=γ214\displaystyle M = \frac { \gamma }{ 2 } -\frac { 1 }{ 4 }

Finally we got :

I=γ \Large \boxed { I=\gamma }

Ronak Agarwal - 3 years, 10 months ago

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Problem 26:

Show that 0πe2cosxsin2(sinx)dx=π2(J0(2i)1).\large\int _{ 0 }^{ \pi }{ { e }^{ 2\cos { x } }\sin ^{ 2 }{ \left( \sin { x } \right) } \, dx } =\frac { \pi }{ 2 } \left( J_{ 0 }\left( 2i \right) -1 \right).

Where Jn(z)J_n\left(z\right) denote the Bessel Function of First Kind.

This problem has been solved by Tanishq Varshney.

Julian Poon - 3 years, 10 months ago

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Solution to Problem 26\text{Solution to Problem 26}

Firstly some common things

2sin2x=1cos(2x)2 \sin ^{2} x= 1-\cos (2x)

πIn(z)=0πezcosθcos(nθ)dθ\large{\pi I_{n} (z)=\displaystyle \int^{\pi}_{0} e^{z \cos \theta} \cos (n \theta) d \theta}

where In(z)I_{n} (z) is modified Bessel function of first kind

Also In(z)=einπ2Jn(iz)\large{I_{n}(z) =e^{\frac{-in \pi}{2}} J_{n} (i z)} where i2=1i^2=-1 and Jn(z)J_{n} (z) is Bessel function of first kind.

In this question we have n=0n=0

Now in the integral

0πe2cosx(1cos(2sinx)2)dx\large{\displaystyle \int^{\pi}_{0} e^{2 \cos x} \left(\frac{1- \cos (2 \sin x)}{2}\right) dx}

On separating

=π2I0(2)120πe2cosxcos(2sinx)dx\large{= \frac{\pi}{2} I_{0} (2) -\frac{1}{2}\displaystyle \int^{\pi}_{0} e^{2 \cos x} \cos (2 \sin x) dx}

Let L=0πe2cosxcos(2sinx)dx\large{L=\displaystyle \int^{\pi}_{0} e^{2 \cos x} \cos (2 \sin x) dx}

L=(0πe2cosxe2isinxdx)\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2 \cos x}e^{2i \sin x} dx \right)}

L=(0πe2(cosx+isinx)dx)\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2 (\cos x+i \sin x)} dx \right)}

L=(0πe2eixdx)\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2e^{ix}} dx \right)}

L=(0πk=0(2eix)kk!dx)\large{L=\Re \left(\displaystyle \int^{\pi}_{0}\displaystyle \sum_{k=0}^{\infty} \frac{(2e^{ix})^{k}}{k!}dx \right)}

L=0πk=02kcos(kx)k!dx\large{L=\displaystyle \int^{\pi}_{0}\displaystyle \sum_{k=0}^{\infty} \frac{2^{k} \cos (kx)}{k!}dx}

Notice that 0πcos(kx)dx=0k1k is an integer\large{\displaystyle \int^{\pi}_{0} \cos (kx) dx=0 \quad \quad \forall k \geq 1 \\ \text{k is an integer}}

Thus only k=0k=0 case is considered and hence L=πL=\pi

On combining we have

π2(I0(2)1)\large{ \frac{\pi}{2} \left(I_{0}(2)-1 \right)}

π2(J0(2i)1)\large{ \frac{\pi}{2} \left(J_{0}(2i)-1 \right)}

Tanishq Varshney - 3 years, 10 months ago

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Problem 36:

Evaluate 0sinxex1dx. \displaystyle \int _{ 0 }^{ \infty }{ \dfrac { \sin x }{ { e }^{ x }-1 } \, dx } .

This problem has been solved by Deep Seth.

Ronak Agarwal - 3 years, 10 months ago

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Problem 27:

Evaluate

02dx(2x2x3)1/3.\large{\displaystyle \int^{2}_{0} \frac{dx}{(2x^2-x^3)^{{1} / {3}}}}.

This problem has been solved by Rajorshi Chaudhuri.

Tanishq Varshney - 3 years, 10 months ago

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Substitute x=2t So the equation becomes 01 t2/3(1t)1/3dt\large{\displaystyle \int^{1}_{0} \ { t }^{ -2/3 }{ (1-t) }^{ -1/3 } } dt This comes out as Γ(13)Γ(23)\large{\displaystyle \Gamma (\frac { 1 }{ 3 } )\Gamma (\frac { 2 }{ 3 } )} which is equal to 2π3\large{\displaystyle\frac { 2\pi }{ \sqrt { 3 }} }

Rajorshi Chaudhuri - 3 years, 10 months ago

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Problem 28:

Evaluate 0x5sin(x)(1+x2)3dx.\large{\displaystyle\int _{ 0 }^{ \infty }{ \frac { { x }^{ 5 }\sin { (x) } }{ { (1+{ x }^{ 2 }) }^{ 3 } } \, dx }}.

This problem has been solved by Sudeep Salgia.

Rajorshi Chaudhuri - 3 years, 10 months ago

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Solution to Problem 28:

The following result is directly used in the solution:0cos(mx)(a2+x2)2dx=π4a3eam(1+am) \displaystyle \int_0^{\infty} \frac{ \cos (mx)}{(a^2+x^2)^2} dx = \frac{\pi}{4a^3} e^{-am} ( 1 + am) .
This can be obtained easily by differentiating the integral w.r.t. aa evaluated in solution of problem 15 of this contest. Consider,
I(b)=0xsin(bx)(1+x2)3dx\displaystyle I(b) = \int_0^{\infty} \frac{x \sin(bx) }{ (1+ x^2)^3} dx . The required integral is 4b4I(b) \dfrac{\partial^4}{\partial b^4} I(b) . Integrating by parts, we get,
I(b)=sin(bx)4(1+x2)20+b40cos(bx)(1+x2)2dx=π16eb(b+b2) \displaystyle I(b) = - \frac{\sin(bx)}{4(1+x^2)^2} \bigg|_0^{\infty} + \frac{b}{4} \int_0^{\infty} \frac{ \cos (bx)}{(1+x^2)^2} dx = \frac{\pi}{16} e^{-b} ( b+ b^2)

Differentiate four times w.r.t bb and put b=1 b = 1 to get the value of the integral as π8e \dfrac{\pi}{8e} .

Sudeep Salgia - 3 years, 10 months ago

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Problem 29:

Show that 01ln(cos(πx2))x(x+1)dx=12(ln2)2(lnπ)(ln2). \large \int_0^1 \dfrac{\ln \left(\cos(\frac{\pi x}{2})\right)}{x(x+1)} \, dx = \dfrac{1}{2} (\ln2)^2 - (\ln \pi)( \ln 2 ) .

This problem has been solved by Tanishq Varshney.

Sudeep Salgia - 3 years, 10 months ago

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Solution to Problem 29

01ln(cos(πx2))xdx01ln(cos(πx2))x+1dx\large{\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x}dx - \displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}

=01ln(sin(πx)2sin(πx2))xdx01ln(cos(πx2))x+1dx\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2 \sin \left( \frac{\pi x}{2}\right)} \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}

=01ln(sin(πx)2)xdx01ln(sin(πx2))xdx01ln(cos(πx2))x+1dx\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\sin \left( \frac{\pi x}{2}\right) \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}

=01ln(sin(πx)2)xdx01ln(sin(π(1x)2))1xdx01ln(cos(πx2))x+1dx\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\sin \left( \frac{\pi(1- x)}{2}\right) \right)}{1-x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}

In the second integral xxx \rightarrow -x

=01ln(sin(πx)2)xdx10ln(cos(πx2))1+xdx01ln(cos(πx2))x+1dx\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{0}_{-1} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{1+x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}

=01ln(sin(πx)2)xdx11ln(cos(πx2))x+1dx\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{1}_{-1} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}

Now in the second integral x+1xx+1 \rightarrow x

=01ln(sin(πx)2)xdx02ln(sin(πx2))xdx\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{2}_{0} \frac{\ln \left(\sin \left( \frac{\pi x}{2}\right) \right)}{x}dx}

=lima0(a1ln(sin(πx)2)xdxa2ln(sin(πx2))xdx)\large{= \displaystyle \lim_{a \to 0} \left( \displaystyle \int^{1}_{a} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{2}_{a} \frac{\ln \left(\sin \left( \frac{\pi x}{2}\right) \right)}{x}dx \right)}

Again in the second integral x2x\large{\frac{x}{2} \rightarrow x}

=lima0(ln2a11xdx+a1ln(sin(πx))xdxa21ln(sin(πx))xdx)\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx+\displaystyle \int^{1}_{a} \frac{\ln (\sin (\pi x))}{x} dx-\displaystyle \int^{1}_{\frac{a}{2}} \frac{\ln (\sin (\pi x))}{x} dx \right)}

=lima0(ln2a11xdxa2aln(sin(πx))xdx)\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx -\displaystyle \int^{a}_{\frac{a}{2}} \frac{\ln (\sin (\pi x))}{x} dx \right)}

We know sinθθ\sin \theta \approx \theta when θ\theta approaches zero.

=lima0(ln2a11xdxa2aln(πx)xdx)\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx -\displaystyle \int^{a}_{\frac{a}{2}} \frac{\ln (\pi x)}{x} dx \right)}

=lima0(ln2lnalnπln212((lna)2(lna2)2))\large{= \displaystyle \lim_{a \to 0} \left( \ln 2 \ln a- \ln \pi \ln 2 -\frac{1}{2} \left((\ln a)^2-(\ln \frac{a}{2})^2 \right) \right)}

using a2b2=(ab)(a+b)a^2-b^2=(a-b)(a+b) and on simplifying the terms containing aa get cancelled and we finally have

12(ln2)2(lnπ)(ln2)\Large{\boxed{\frac{1}{2} (\ln 2)^2-(\ln \pi) (\ln 2)}}

Tanishq Varshney - 3 years, 10 months ago

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CHEATER!!!

Pi Han Goh - 3 years, 10 months ago

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Problem 30:

Prove that 0π/8ln(tan2x)dx=G2.\large{\displaystyle \int^{{\pi} / {8}}_{0} \ln (\tan 2x) \, dx=-\frac{G}{2}}.

Where G{G} denotes the Catalan's constant, G=n=0(1)n(2n+1)20.91596559\displaystyle G = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^2} \approx 0.915 965 59 .

This problem has been solved by Surya Prakesh (first) and Julian Poon (second) almost at the same time.

Tanishq Varshney - 3 years, 10 months ago

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Lemma : 01taln(t)dt=1(a+1)2\int_{0}^{1} t^{a} \ln(t) dt = -\dfrac{1}{(a+1)^2}

This lemma is already proved in many problems in this contest.

0π/8ln(tan(2x))dx=1/201ln(t)1+t2dt\int_{0}^{\pi/8} \ln(\tan(2x))dx =1/2 \int_{0}^{1} \dfrac{\ln(t)}{1+t^2} dt

I used the substitution t=tan(2x)t=\tan(2x).

01ln(t)1+t2dt=01i=0(1)it2iln(t)dt=i=0(1)i01t2iln(t)dt=i=0(1)i1(2i+1)2=β(2)=G\begin{aligned} \int_{0}^{1} \dfrac{\ln(t)}{1+t^2} dt &= \int_{0}^{1} \sum_{i=0}^{\infty}(-1)^{i} t^{2i} \ln(t)dt \\ &= \sum_{i=0}^{\infty}(-1)^{i} \int_{0}^{1} t^{2i} \ln(t) dt \\ &= -\sum_{i=0}^{\infty} (-1)^{i} \dfrac{1}{(2i+1)^2} \\ &= - \beta(2) \\ &= -G \end{aligned}

Therefore,

0π/8ln(tan(2x))dx=G2\int_{0}^{\pi/8} \ln(\tan(2x))dx = \dfrac{-G}{2}

Surya Prakash - 3 years, 10 months ago

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Not sure if this is considered but...

y=ln(tan(2x))y=\ln \left(\tan \left(2x\right)\right)

x=tan1(ey)2x=\frac{\tan ^{-1}\left(e^y\right)}{2}

0π8ln(tan(2x))dx=0tan1(ey)2dy=120tan1(ey)dy\int _{ 0 }^{ \frac { \pi }{ 8 } }{ \ln { \left( \tan \left( 2x \right) \right) } } dx=\int _{ -\infty }^{ 0 }{ \frac { \tan ^{ -1 } \left( e^{ y } \right) }{ 2 } } dy=-\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \tan ^{ -1 } \left( e^{ -y } \right) } dy

A known identity of the catalan constant is G=0tan1(ey)dyG=\int _{ 0 }^{ \infty }{ \tan ^{ -1 } \left( e^{ -y } \right) } dy

Hence the integral is G2-\frac{G}{2}

The identity above can be deduced through the Taylor Series of tan1(x)\tan^{-1}(x)

Julian Poon - 3 years, 10 months ago

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Problem 34:

Prove That 0(sin(x2)x2)(x2+x1) dx=π4π8. \large \int_{0}^{\infty} \left(\dfrac{\sin(x^2)}{x^2}\right)\cdot \left(x^2+x-1\right) \ \mathrm{d}x = \dfrac{\pi}{4} - \sqrt{\dfrac{\pi}{8}}.

This problem has been solved by Surya Prakesh.

Ishan Singh - 3 years, 10 months ago

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Lemma:

Consider the Gamma Function Γ(s)=0etts1dt\Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1} dt

Take the substitution, t=punt=pu^n.

Γ(s)=nps0epununs1du\Gamma(s) = n p^{s} \int_{0}^{\infty} e^{-pu^n} u^{ns-1} du

Taking p=a+ibp= a+ ib and comparing imaginary parts we get,

0eaununs1sin(bun)du=Γ(s)npssin(αs)\int_{0}^{\infty} e^{-au^n} u^{ns-1} \sin \left(bu^n \right) du = \dfrac{\Gamma(s)}{n |p|^s} \sin(\alpha s )

where p=a+ibp=a+ib and α\alpha is the argument of pp.


1st Integral:

0sin(x2)dx\int_{0}^{\infty} \sin \left(x^2 \right) dx

Comparing this with our lemma, we get a=0a=0, b=1b=1, n=2n=2, s=1/2s=1/2. Which gives us that p=1|p| = 1 and α=π/2\alpha = \pi /2.

On substitution we get the integral evaluated to be I1=π22I_{1} = \dfrac{\sqrt{\pi}}{2 \sqrt{2}}.


2nd Integral:

0sin(x2)xdx\int_{0}^{\infty} \dfrac{\sin (x^2)}{x} dx

Take the substitution, x=t1/2x = t^{1/2}. We get,

120sin(t)tdt\dfrac{1}{2} \int_{0}^{\infty} \dfrac{\sin(t)}{t} dt

This is well known Integral (Dirichlet integral): 0sin(x)xdx=π2\int_{0}^{\infty} \dfrac{\sin(x)}{x} dx = \dfrac{\pi}{2}.

So, this integral evaluates to be I2=π4I_{2} = \dfrac{\pi}{4}.


3rd Integral:

This can evaluated in similar way as we obtained for first integral. And this is evaluated to be I3=π2I_{3} = \dfrac{\sqrt{\pi}}{\sqrt{2}}.

So, the overall integral I=I1+I2I3=π4π22I = I_{1} + I_{2} - I_{3} =\large{ \boxed{ \dfrac{\pi}{4} - \dfrac{\sqrt{\pi}}{2 \sqrt{2}}}}

Surya Prakash - 3 years, 10 months ago

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Problem 40:

Prove that 0π/2x(ln(sinx)2sinx)cotx dx=π8[Γ(14)]2\large \int_{0}^{{\pi} / {2}} x\left(\dfrac{\ln(\sin x)-2}{\sqrt{\sin x}}\right) \cot x \ \mathrm{d}x = - \sqrt{\frac{\pi}{8}} \left[\Gamma \left(\dfrac{1}{4}\right)\right]^{2}

This problem has been solved by Surya Prakesh.

Ishan Singh - 3 years, 10 months ago

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0π/2x(ln(sinx)2sinx)cot(x)dx=x((ln(sinx)2sinx)cot(x)dx)0π/20π/2((ln(sinx)2sinx)cot(x)dx)dx\int_{0}^{\pi /2} x \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \\ = x \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \right) \Biggl|_{0}^{\pi /2} - \int_{0}^{\pi /2} \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \right) dx

(ln(sinx)2sinx)cotxdx=(ln(t)2tt)dt=2lntd(t)td(ln(t))(t)2=2ln(t)t=2ln(sin(x))sin(x)\begin{aligned} \int \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx &= \int \left(\dfrac{\ln(t) - 2}{t \sqrt{t}} \right) dt \\ &=2 \int \dfrac{\ln t d(\sqrt{t} ) - \sqrt{t} d(\ln(t))}{(\sqrt{t}) ^2} \\ &= -2\dfrac{\ln(t)}{\sqrt{t}}\\ &= -2 \dfrac{\ln(\sin(x))}{\sqrt{\sin(x)}} \end{aligned}

So,

x((ln(sin(x))2sinx)cotxdx)0π/2=2xln(sin(x))sin(x)0=0lima02aln(sin(a))sin(a)=lima02aln(sin(a))sin(a)=0\begin{aligned} x \left(\int\left(\dfrac{\ln(\sin (x)) - 2}{\sqrt{\sin x}} \right) \cot x dx \right) \Biggl|_{0}^{\pi /2} &= \dfrac{-2x\ln(\sin(x))}{\sqrt{\sin(x)}} \Biggl|_{0}^{\infty}\\ &= 0 - \lim_{a \rightarrow 0} \dfrac{-2a \ln(\sin(a))}{\sqrt{\sin(a)}}\\ &= \lim_{a \rightarrow 0} \dfrac{2a \ln(\sin(a))}{\sqrt{\sin(a)}}\\ &= 0 \end{aligned}

Above limit is evaluated using L'Hospital rule.

So,

0π/2x(ln(sinx)2sinx)cotxdx=0π/2((ln(sinx)2sinx)cotxdx)dx=20π/2ln(sin(x))sin(x)dx\int_{0}^{\pi /2} x \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx = - \int_{0}^{\pi /2} \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx \right) dx = 2 \int_{0}^{\pi /2} \dfrac{\ln(\sin(x))}{\sqrt{\sin(x)}} dx

Let I(a)=0π/2sin2a1(x)dxI(a) = \int_{0}^{\pi /2} \sin^{2a-1} (x) dx. So, our required integral is I(1/4)I'(1/4). By taking t=sin2(x)t = \sin ^2 (x), the integral transforms to I(a)=1201ta1(1t)1/2dt=12B(a,1/2)=12Γ(a)Γ(1/2)Γ(a+1/2)I(a) = \dfrac{1}{2} \int_{0}^{1} t^{a-1} (1-t)^{-1/2} dt = \dfrac{1}{2} B(a, 1/2) = \dfrac{1}{2} \dfrac{\Gamma(a) \Gamma(1/2)}{\Gamma(a+1/2)}

Differentiate it w.r.t aa,

I(a)=12Γ(a)Γ(1/2)Γ(a+1/2)(ψ(a)ψ(a+1/2))I'(a) = \dfrac{1}{2} \dfrac{\Gamma(a) \Gamma(1/2)}{\Gamma(a+1/2)} \left( \psi(a) - \psi(a+1/2) \right)

I(1/4)=12Γ(1/4)Γ(1/2)Γ(3/4)(ψ(1/4)ψ(3/4))I'(1/4) = \dfrac{1}{2} \dfrac{\Gamma(1/4) \Gamma(1/2)}{\Gamma(3/4)} \left( \psi(1/4) - \psi(3/4) \right)

Using Euler's reflection formula. We get

Γ(1/4)Γ(3/4)=πsin(π/4)=2π    Γ(3/4)=2πΓ(1/4)\Gamma(1/4) \Gamma(3/4) = \dfrac{\pi}{\sin(\pi/4)} = \sqrt{2} \pi \implies \Gamma(3/4) = \dfrac{\sqrt{2} \pi}{\Gamma(1/4)}

Also, ψ(1/4)ψ(3/4)=πcot(3π/4)=π\psi(1/4) - \psi(3/4) = \pi \cot(3 \pi /4) = - \pi.

Therefore the required integral evaluates to

π8(Γ(14))2-\sqrt{\dfrac{ \pi}{8}} \left( \Gamma \left(\dfrac{1}{4} \right) \right) ^2

Surya Prakash - 3 years, 10 months ago

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Problem 31:

Prove that limn01xnx2n1xdx=ln(2).\large \lim_{n \rightarrow \infty} \int_{0}^{1} \dfrac{x^n - x^{2n}}{1-x} \, dx = \ln (2).

This problem has been solved by Julian Poon.

Surya Prakash - 3 years, 10 months ago

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Using basic concepts in integration:

01xnx2n1xdx=01xn1xn1xdx=01xnk=0n1xk dx=k=0n101xn+k dx=k=0n11n+k+1=1nk=1n11+kn\quad\displaystyle\int_0^1\frac{x^n-x^{2n}}{1-x}\mathrm dx=\int_0^1x^n\frac{1-x^n}{1-x}\mathrm dx=\int_0^1x^n\sum_{k=0}^{n-1}x^k\ \mathrm dx=\sum_{k=0}^{n-1}\int_0^1x^{n+k}\ \mathrm dx=\sum_{k=0}^{n-1}\frac1{n+k+1}=\frac1n\sum_{k=1}^{n}\frac1{1+\frac kn}

Then:

limn01xnx2n1xdx=limn1nk=1n11+kn=0111+xdx=[ln(1+x)]01=ln2\displaystyle\lim_{n\to\infty}\int_0^1\frac{x^n-x^{2n}}{1-x}\mathrm dx=\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}\frac1{1+\frac kn}=\int_0^1\frac1{1+x}\mathrm dx=[\ln(1+x)]_0^1=\ln2

Kenny Lau - 3 years, 10 months ago

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Substituting k=0xk=11x\displaystyle \sum _{ k=0 }^{ \infty }{ x^{ k } } =\frac { 1 }{ 1-x }

01k=0xk+nk=0xk+2ndx=k=11k+nk=11k+2n\int _{ 0 }^{ 1 }{ \sum _{ k=0 }^{ \infty } x^{ k+n }-\sum _{ k=0 }^{ \infty } x^{ k+2n } } dx=\sum _{ k=1 }^{ \infty } \frac { 1 }{ k+n } -\sum _{ k=1 }^{ \infty } \frac { 1 }{ k+2n }

=H2nHn=ln(2)={ H }_{ 2n }-{ H }_{ n }=\ln(2)

The last step is from the approximation Hn=γ+ln(n)H_n=\gamma+\ln(n) as nn\to \infty