Brilliant Integration Contest - Season 2 (Part-1)

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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**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":

View Part 2

Note by Aditya Kumar
3 years, 10 months ago

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Problem 1:

Evaluate: 01xln(x)1x2dx\int _{ 0 }^{ 1 }{\frac { x \ln\left( x \right) }{ \sqrt { 1-{ x }^{ 2 } } } \, dx }

This problem has been solved by Surya Prakash.

Aditya Kumar - 3 years, 10 months ago

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Another solution: Let (1x2)=t\sqrt{(1-x^2)}=t

x(1x2)dx=dt\frac{-x}{\sqrt{(1-x^2)}}dx=dt

So integration becomes

10ln(1t2)dt- \int_1^0 ln\sqrt{(1-t^2)}dt

1210ln(1t2)dt-\frac{1}{2} \int_1^0 ln(1-t^2)dt

1201ln(1t2)dt\frac{1}{2} \int_0^1 ln(1-t^2)dt

1201(ln(1t)+ln(1+t))dt\frac{1}{2} \int_0^1 (ln(1-t)+ln(1+t))dt

This can be easily evaluated and comes out to be ln21ln2-1

Gautam Sharma - 3 years, 10 months ago

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Solution to Problem 1:\Large \text{Solution to Problem 1:}

01xln(x)1x2dx\int_{0}^{1} \dfrac{x \ln(x)}{\sqrt{1-x^2}} dx Take t=arcsin(x)    dt=dx1x2t = \arcsin(x) \iff dt = \dfrac{dx}{\sqrt{1-x^2}} and the limits changes to t=0t=0 to t=π2t=\dfrac{\pi}{2}. 0π2sin(t)ln(sin(t))dt\int_{0}^{\dfrac{\pi}{2}} \sin (t) \ln(\sin(t)) dt

Consider,

B(x,y)=01tx1(1t)y1dtB(x,y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt

Taking the transformation t=sin2(z)    dt=2sin(z)cos(z)dzt=\sin^2 (z) \iff dt = 2\sin(z) \cos (z)dz, the integral transforms into

B(x,y)=20π/2sin2x1(z)cos2y1(z)dzB(x,y) = 2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz

But B(x,y)=Γ(x)Γ(y)Γ(x+y)B(x,y) = \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}

So, 0π/2sin2x1(z)cos2y1(z)dz=12Γ(x)Γ(y)Γ(x+y)\int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz = \dfrac{1}{2} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}

Differentiate it w.r.t xx

20π/2sin2x1(z)cos2y1(z)ln(sin(z))dz=12xΓ(x)Γ(y)Γ(x+y)0π/2sin2x1(z)cos2y1(z)ln(sin(z))dz=14Γ(x)Γ(y)Γ(x+y)(ψ(x)ψ(x+y))2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{2} \dfrac{\partial}{\partial x} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} \\ \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{4} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} (\psi (x) - \psi(x+y))

Take x=1x= 1 and y=12y= \dfrac{1}{2}.

0π/2sin(z)ln(sin(z))dz=14Γ(1)Γ(1/2)Γ(3/2)(ψ(1)ψ(3/2))=ln(2)1\int_{0}^{\pi / 2} \sin(z) \ln (\sin (z)) dz = \dfrac{1}{4} \dfrac{\Gamma (1) \Gamma(1/2) }{\Gamma (3/2)} (\psi (1) - \psi(3/2)) = \ln (2) -1

Surya Prakash - 3 years, 10 months ago

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Problem 2:

Evaluate:

0xcos(x3)exp(x3)dx\Large \int_{0} ^{\infty} \dfrac{x \cos (x^3)}{\exp (x^3)} \, dx

This problem has been solved by Tanishq Varshney.

Surya Prakash - 3 years, 10 months ago

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Solution to Problem 2\large{Solution~to~Problem~ 2}

Firstly I would add a proof of my solution

0eaxxn1dx=Γ(n)an\large{\displaystyle \int _{0}^{\infty} e^{-ax}x^{n-1} dx=\frac{\Gamma (n)}{a^{n}}} where Γ(n)\Gamma (n) is Gamma function

Replace aa+iba \rightarrow a+ib where i=1i =\sqrt{-1}

0eaxeibxxn1dx=Γ(n)(a+ib)n\large{\displaystyle \int _{0}^{\infty} e^{-ax}e^{-ibx}x^{n-1} dx=\frac{\Gamma (n)}{(a+ib)^{n}}}

Put a=rcosy and b=rsinya=r \cos y ~ and ~ b=r \sin y.

So that r2=a2+b2r^2=a^2+b^2 and y=arctan(ba)y=\arctan \left(\frac{b}{a}\right)

Using de moviers theorem

0eax(cosbxisinbx)xn1dx=Γ(n)rn(cosny+sinny)1\large{\displaystyle \int _{0}^{\infty} e^{-ax}(\cos bx-i \sin bx)x^{n-1} dx=\frac{\Gamma (n)}{r^{n}}(\cos ny+\sin ny)^{-1}}

Comparing the real and imaginary parts we finally get

0xn1eax(cosbx)dx=Γ(n)rn(cosny)\large{\displaystyle \int _{0}^{\infty}x^{n-1} e^{-ax}(\cos bx) dx=\frac{\Gamma (n)}{r^{n}}(\cos ny)}

now in the given integral put x3=tx^{3}=t

The integral now becomes

130t13etcos(t)dt\large{\frac{1}{3} \displaystyle \int^{\infty}_{0} t^{-\frac{1}{3}}e^{-t} \cos (t) dt}

here n=23n=\frac{2}{3};a=1a=1 and b=1b=1

Thus we get 13Γ(23)cos(π6)213\large{\frac{1}{3}\frac{\Gamma \left(\frac{2}{3}\right) \cos \left(\frac{\pi}{6}\right)}{2^{\frac{1}{3}}}}

3Γ(23)3.243\large{\boxed{\frac{\sqrt{3} \Gamma \left(\frac{2}{3}\right)}{3.2^{\frac{4}{3}}}}}

Tanishq Varshney - 3 years, 10 months ago

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Problem 23:

Evaluate:01log(1+x)log(1x3)dx\int _{ 0 }^{ 1 }{ \log\left( 1+x \right) \log\left( 1-{ x }^{ 3 } \right) dx }

Due to time constraint, Aditya Kumar decided to post a solution himself.

Aditya Kumar - 3 years, 10 months ago

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Solution to Problem 23:

Substitute: x1x1+xx\longrightarrow \frac { 1-x }{ 1+x }

I=01ln(1+x)ln(1x3)dx=201ln(2x(3+x2)(1+x)3)ln(21+x)dx(1+x)2I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}

On separating the integrands,

ln(2x(3+x2)(1+x)3)ln(21+x)=ln22ln2lnx+ln2ln(3+x2)4ln2ln(1+x)lnxln(1+x)+3ln2(1+x)ln(1+x)ln(3+x2)\ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2)

1st integral:

\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12

2nd integral:

01lnx(1+x)2dx=n1(1)nn=ln2\int_0^1 \frac{\ln x}{(1+x)^2}dx=\sum_{n\geq1} \frac{(-1)^n}{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=01xln(1+x)(1+x)(3+x2)dx=Re01ln(1+x)(1+x)(x+i3)dx=18ln22Re1i3101ln(1+x)x+i3dxJ=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx

Now on using the formula:

01ln(1+x)x+adx=ln2lna+1a1+Li2(21a)Li2(11a)\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right)

Putting a=i3a=i\sqrt{3} then gives, after taking the real part, and assuming the principle value of the logarithm,

J=18ln22+π312ln2+14ReLi2(eiπ/3)34ImLi2(eiπ/3)14ReLi2(12eiπ/3)+34ImLi2(12eiπ/3)\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})

Using:

ReLi2(eiπ/3)=π236\displaystyle \,\, \operatorname{Re}\text{Li}_2(e^{i\pi/3})=\frac{\pi^2}{36},

ImLi2(eiπ/3)=123ψ1(13)π233\displaystyle \,\, \operatorname{Im}\text{Li}_2(e^{i\pi/3})=\frac1{2\sqrt{3}}\psi_1\left(\frac13\right)-\frac{\pi^2}{3\sqrt{3}},

Li2(12eiπ/3)=Li2(i3)12ln2(34i34)\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right),

and

ReLi2(i3)=n=1(1)n(2n)23n=14Li2(13)\displaystyle \,\, \operatorname{Re}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)^2\, 3^n}=\frac14\text{Li}_2\left(-\frac13\right)

we have ReLi2(12eiπ/3)=π27218ln2(34)14Li2(13)\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)

and ImLi2(12eiπ/3)=π12ln(34)ImLi2(i3).\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi}{12}\ln\left(\frac34\right)-\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).

Finally, putting everything together gives: I=ln2218ln23+2ln2ln332ln36ln2+6π43(2+ln3)37π272+12ψ1(13)14Li2(13)+3Li2(i3)I=\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )

......

This was not original. I found it very awesome, so I thought to share it here.

Aditya Kumar - 3 years, 10 months ago

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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:

Julian Poon - 3 years, 10 months ago

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Problem 4:

Let P(x)P(x) be a polynomial such that P(x+1)P(x)=x2+x+1x2x+1\frac { P\left( x+1 \right) }{ P\left( x \right) } =\frac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 } and P(2)=3P(2)=3, evaluate01(tan1(P(x)).tan1(x1x))dx.\int _{ 0 }^{ 1 }{ \left( { \tan }^{ -1 }\left( P(x) \right) .{ \tan }^{ -1 }\left( \sqrt { \frac { x }{ 1-x } } \right) \right) \, dx }.

This problem has been solved by Tanishq Varshney.

Aditya Kumar - 3 years, 10 months ago

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Solution to Problem 4\text{Solution to Problem 4}

Clearly P(x)=x2x+1P(x)=x^2-x+1 and P(1x)=P(x)P(1-x)=P(x)

Also tan1(x)=cot1(1x)\tan^{-1} (x)=\cot^{-1} \left(\frac{1}{x}\right)

tan1(x)+cot1(x)=π2\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}

I=01(tan1(P(x))cot1(x1x))dx\large{I=\displaystyle \int^{1}_{0} \left( \tan^{-1}(P(x)) \cot^{-1} \left( \sqrt{\frac{x}{1-x}}\right) \right) dx}

2I=π201tan1(x2x+1)dx\large{2I=\frac{\pi}{2} \displaystyle \int^{1}_{0}\tan^{-1} (x^2-x+1) dx}

2I=π2(π201tan1(11+x(x1)))\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{1}{1+x(x-1)}\right)\right)}

2I=π2(π201tan1(x(x1)1+x(x1))dx)\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{x-(x-1)}{1+x(x-1)}\right) dx \right)}

2I=π2(π2(01tan1(x)dx01tan1(x1)dx))\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -( \displaystyle \int^{1}_{0} \tan^{-1} (x) dx- \displaystyle \int^{1}_{0} \tan^{-1} (x-1) dx)\right)}

2I=π2(π2(π22ln(2)))\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -\left( \frac{\pi}{2}-2 \ln (2)\right)\right)}

I=π2ln(2)\large{I=\frac{\pi}{2} \ln(2)}

Tanishq Varshney - 3 years, 10 months ago

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Whoops, I accidentally deleted Tanishq's comment:

Problem 5:

0π/2tan1(1729cos2x)dx \int_0^{\pi /2} \tan^{-1} (1729\cos^2 x) \, dx

This problem has been solved by Sudeep Salgia.

Pi Han Goh - 3 years, 10 months ago

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I'll post the outline of the solution as it is a bit cumbersome. Let
I(a)=0π2arctan(acos2x)dxI(a)=cos2x1+a2cos4xdx=sec2xsec4x+a2dx \displaystyle I(a) = \int_0^{\frac{\pi}{2}} \arctan (a \cos^2 x) dx \Rightarrow I'(a) = \frac{\cos^2 x}{1 + a^2 \cos^4 x } dx = \frac{\sec^2 x}{\sec^4 x + a^2 } dx Substitute tanx=t \tan x = t , to get I(a)=01t4+2t2+(a2+1)dt \displaystyle I'(a) = \int_0^{\infty} \frac{1}{t^4 + 2t^2 + (a^2 + 1)} dt . Put a2+1=k2a^2 + 1 = k^2 and with some simple rearrangement, we get,
2kI(a)=01+kt2(tkt)2+2(k+1)01kt2(t+kt)2+2(k+1)\displaystyle 2k I'(a) = \int_0^{\infty} \frac{ 1+ \frac{k}{t^2}}{\left( t - \frac{k}{t} \right)^2 + 2(k+1)} - \int_0^{\infty} \frac{ 1- \frac{k}{t^2}}{\left( t + \frac{k}{t} \right)^2 + 2(-k+1)} Substitute tkt=z \displaystyle t - \frac{k}{t} = z and t+kt=y \displaystyle t + \frac{k}{t} = y in the respective integrals and evaluate them easily (They come out to be arctan \arctan of something ). After all this we can write,
I(a)=π221a2+11a2+1+1da\displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{\sqrt{a^2 + 1}} \frac{1}{\sqrt{\sqrt{a^2 + 1} +1}} da
This can be rearranged to be written as
I(a)=π2211+a2+112×12a2+112×12×aa2+1da \displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{1+ \frac{\sqrt{a^2 + 1} - 1 }{2}} \times \frac{1}{2\sqrt{ \frac{\sqrt{a^2 + 1} - 1 }{2} }} \times \frac{1}{2} \times \frac{a}{\sqrt{a^2 +1}} da

This evaluates to π2arctan(a2+112) \displaystyle \frac{\pi}{2} \arctan \left( \sqrt{ \frac{\sqrt{a^2+1} - 1}{2}} \right)

The constant of integration is 0 0 from I(0)=0 I(0) = 0 . Put a=1729 a =1729 to get the answer.

Sudeep Salgia - 3 years, 10 months ago

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Problem 24:

Prove that

0π/21sin8x+cos8xdx=1022π. \large \int_0^{\pi /2} \dfrac1{\sin^8 x + \cos^8 x } \, dx = \dfrac{\sqrt{10-\sqrt2}}{2} \pi .

This problem has been solved by Tanishq Varshney.

Pi Han Goh - 3 years, 10 months ago

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Solution to Problem 24\text{Solution to Problem 24}

0xm11+xdx=πsinmπ0<m<1\large{\displaystyle \int ^{\infty}_{0} \frac{x^{m-1}}{1+x}dx=\frac{\pi}{\sin m \pi} \quad \quad 0<m<1}

Refer to problem 18 for this step.

Now in the integral tanxt\tan x \rightarrow t

0(1+t2)31+t8dt\large{\displaystyle \int ^{\infty}_{0} \frac{(1+t^2)^3}{1+t^{8}}dt}

t8z\large{t^8 \rightarrow z}

18(0z7811+z+3z5811+z+3z3811+z+z1811+zdz)\large{\frac{1}{8} \left(\displaystyle \int ^{\infty}_{0} \frac{z^{\frac{7}{8}-1}}{1+z}+ \frac{3z^{\frac{5}{8}-1}}{1+z}+ \frac{3z^{\frac{3}{8}-1}}{1+z}+ \frac{z^{\frac{1}{8}-1}}{1+z} dz \right)}

π4(1sin(π8)+3cos(π8))\large{\rightarrow \frac{\pi}{4} \left(\frac{1}{\sin \left(\frac{\pi}{8} \right)}+\frac{3}{\cos \left(\frac{\pi}{8} \right)} \right)}

=1022π\large{=\frac{\sqrt{10-\sqrt{2}}}{2} \pi}

Tanishq Varshney - 3 years, 10 months ago

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Problem 3\text{Problem 3}

Evaluate:

0π2tan35xdx\large {\displaystyle \int^{\frac{\pi}{2}}_{0} \tan^{\frac{3}{5}} x \, dx}

This problem has been solved by Aditya Kumar.

Tanishq Varshney - 3 years, 10 months ago

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Solution to Problem 3: 0π2(tanx)35dx=0π2(sinx)35(cosx)35dx=B(45,15)2...........{B(x,y)isbetafunction}=Γ(45)Γ(15)2Γ(1)=π2sin(π5)...............{byeulersreflectionformula}=2π1025\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( tanx \right) }^{ \frac { 3 }{ 5 } } } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( sinx \right) }^{ \frac { 3 }{ 5 } }{ \left( cosx \right) }^{ \frac { -3 }{ 5 } } } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { B\left( \frac { 4 }{ 5 } ,\frac { 1 }{ 5 } \right) }{ 2 } ...........\left\{ B(x,y)\quad is\quad beta\quad function \right\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \Gamma \left( \frac { 4 }{ 5 } \right) \Gamma \left( \frac { 1 }{ 5 } \right) }{ 2\Gamma \left( 1 \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \pi }{ 2sin\left( \frac { \pi }{ 5 } \right) } \quad ...............\{ by\quad euler's\quad reflection\quad formula\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 2\pi }{ \sqrt { 10-2\sqrt { 5 } } }

Aditya Kumar - 3 years, 10 months ago

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Problem 9:

Evaluate the indefinite integral: x2+20(xsinx+5cosx)2dx.\large{\displaystyle \int \frac{x^2+20}{(x \sin x+5 \cos x)^2} \, dx}.

This problem has been solved by Surya Prakesh.

Tanishq Varshney - 3 years, 10 months ago

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x2+20=(xsin(x)+5cos(x))2+(5sin(x)xcos(x))25=(xsin(x)+5cos(x))2+(5sin(x)xcos(x))25cos2(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))d(5sin(x)xcos(x))dx(5sin(x)xcos(x))d(xsin(x)+5cos(x))dx\begin{aligned} x^2 + 20 &= (x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \\ &=(x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \cos^2(x) - 5\sin^2 (x) -x \cos(x)\sin(x) + x\cos(x) \sin(x) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) + (5 \sin(x) - x \cos(x))(5 \sin(x) - x \cos(x) - \sin(x)) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) - (5 \sin(x) - x \cos(x))( x \cos(x) + \sin(x) - 5 \sin (x)) \\ &= (x \sin(x) + 5 \cos(x)) \dfrac{d(5 \sin(x) - x \cos(x))}{dx} - (5 \sin(x) - x \cos(x)) \dfrac{d(x \sin(x) + 5 \cos(x))}{dx} \end{aligned}

Let u=xsin(x)+5cos(x)u=x \sin(x) + 5 \cos(x) and v=5sin(x)xcos(x)v=5 \sin(x) - x \cos(x). So, x2+20=udvdxvdudx=u2d(vu)dxx^2 + 20 = u\dfrac{dv}{dx} - v\dfrac{du}{dx} = u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}.

So, x2+20(xsin(x)+5cos(x))2dx=u2d(vu)dxu2dx=vu=5sin(x)xcos(x)xsin(x)+5cos(x)\int \dfrac{x^2 + 20}{(x \sin(x) + 5 \cos(x))^2} dx = \int \dfrac{ u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}}{u^2} dx= \dfrac{v}{u} = \dfrac{5 \sin(x) - x \cos(x)}{x \sin(x) + 5 \cos(x)}

Surya Prakash - 3 years, 10 months ago

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Problem 13: (Originally posted by Tanishq)

Evaluate 01xln(1+x1x)dx. \int_0^1 x \ln \left( \dfrac{1+x}{1-x} \right) \, dx.

This problem has been solved by Surya Prakesh.

Pi Han Goh - 3 years, 10 months ago

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By using integration by parts,

xln(1+x1x)dx=x22ln(1+x1x)+x12ln(1+x)+12ln(1x)\int x \ln \left(\dfrac{1+x}{1-x} \right) dx = \dfrac{x^2}{2} \ln \left(\dfrac{1+x}{1-x} \right) + x - \dfrac{1}{2} \ln \left(1+x \right) + \dfrac{1}{2} \ln \left(1-x \right)

So,

01xln(1+x1x)dx=lima001axln(1+x1x)dx=lima0(1a)22ln(2aa)+1a12ln(2a)+12lna=1\begin{aligned} \int_{0}^{1} x \ln \left(\dfrac{1+x}{1-x} \right) dx &= \lim_{a \rightarrow 0} \int_{0}^{1-a} x \ln \left(\dfrac{1+x}{1-x} \right) dx \\ &= \lim_{a \rightarrow 0} \dfrac{(1-a)^2}{2} \ln \left(\dfrac{2-a}{a} \right) + 1 - a - \dfrac{1}{2} \ln \left(2-a \right) + \dfrac{1}{2} \ln a \\ &= 1\end{aligned}

Surya Prakash - 3 years, 10 months ago

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Problem 6:
Show that for positive reals, a,b a , b where a<b a<b, abarccos(x(a+b)xab)dx=π4(ba)2a+b \int_a^b \arccos \left( \frac{x}{\sqrt{(a+b)x - ab}} \right) \, dx = \frac{\pi}{4} \frac{ (b-a)^2}{a+b}

Statutory warning: The substitutions need not be very intuitive.

This problem has been solved Aditya Kumar.

Sudeep Salgia - 3 years, 10 months ago

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Solution to Problem 6:

I=abarccos(x(a+b)xab)dxI=abarccos(xab((a+b)ab)x1)dx((a+b)ab)x1=tI=abba(2ta+barccos((t2+1)abt))dtI=abba(2ta+barccos((t+1t)ab))dtI=\int _{ a }^{ b }{ \text{arccos}\left( \frac { x }{ \sqrt { \left( a+b \right) x-ab } } \right) dx } \\ I=\int _{ a }^{ b }{ \text{arccos}\left( \frac { \frac { x }{ \sqrt { ab } } }{ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } } \right) dx } \\ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } =t\\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \frac { \left( { t }^{ 2 }+1 \right) \sqrt { ab } }{ t } \right) \right) dt } \\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \left( { t }+\frac { 1 }{ t } \right) \sqrt { ab } \right) \right) dt }

Now using Integration By Parts,

I=abba(t2+1(baab)2(t1t)2)dtt=p+1+p2I=12(ba+ab)12(baab)(2p2ab+ba+24p2)dpI=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { { t }^{ 2 }+1 }{ { \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }^{ 2 }-{ \left( { t }-\frac { 1 }{ t } \right) }^{ 2 } } \right) dt } \\ t=p+\sqrt { 1+{ p }^{ 2 } } \\ I=\int _{ \frac { 1 }{ 2 } \left( -\sqrt { \frac { b }{ a } } +\sqrt { \frac { a }{ b } } \right) }^{ \frac { 1 }{ 2 } \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }{ \left( \frac { { 2p }^{ 2 } }{ \sqrt { \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } -{ p }^{ 2 } } } \right) dp }

Now,

p2=(ab+ba+24)sinθ{ p }^{ 2 }=\left( \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } \right) sin\theta

This can be evaluated easily.

I=π4(ba)2(a+b)\therefore \quad I=\frac { \pi }{ 4 } \frac { { \left( b-a \right) }^{ 2 } }{ \left( a+b \right) }

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.

Aditya Kumar - 3 years, 10 months ago

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Problem 8:

Evaluate

f(ω)=12πea2t2eiωtdt\large f(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a^2t^2}e^{i\omega t}dt

This problem has been solved by both Tanishq Varshney (first) and Surya Prakash (second) almost at the same time.

Abhishek Bakshi - 3 years, 10 months ago

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Solution to problem 8\large{\text{Solution to problem 8}}

ttt \rightarrow -t

adding this with the original equation

2f(w)=22πea2t2coswtdt\Large{2f(w)=\frac{2}{\sqrt{2 \pi}} \displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2} \cos wt dt}

as eiwt+eiwt=2coswt=2(eiwt)e^{iwt}+e^{-iwt}=2 \cos wt=2 \Re(e^{iwt})

f(w)=12π(ea2t2eiwtdt)\Large{f(w)=\frac{1}{\sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2}e^{iwt} dt \right)}

Let at=zat=z

f(w)=1a2π(ez2eiwazdz)\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-z^2}e^{i\frac{w}{a}z} dz \right)}

f(w)=1a2π(e(ziw2a)2ew24a2dz)\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-(z-\frac{iw}{2a})^2}e^{-\frac{w^2}{4a^2}} dz \right)}

Using Gaussian integral

f(w)=1a2ew24a2\Large{f(w)=\frac{1}{a \sqrt{2}} e^{-\frac{w^2}{4a^2}}}

Tanishq Varshney - 3 years, 10 months ago

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f(ω)=12πea2t2eiωtdtf(\omega) = \dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} e^{i\omega t}dt

It implies that f(ω)=[F(F)](ω)f(\omega) = \big[\mathcal{F}(F) \big](\omega), where F(t)=ea2t2F(t) = e^{-a^2 t^2}.

Now, differentiate F(t)F(t), we get F(t)=2a2tF(t)F'(t) = -2a^2 t F(t).

Applying fourier transformation on both sides gives iωf(ω)=2ia2f(ω)i \omega f(\omega) = -2i a^2 f'(\omega).

    ωf(ω)=2a2df(ω)dω    f(ω)=ceω24a2\implies \omega f(\omega) = -2a^2 \dfrac{d f(\omega)}{d \omega} \\ \implies f(\omega) = c e^{-\dfrac{\omega^{2}}{4a^{2}}}

Since c=f(0)=12πea2t2dt=212π0ea2t2dt=1a2πΓ(1/2)=1a2c = f(0) =\dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} dt = 2 \dfrac{1}{\sqrt{2 \pi}} \int_{0} ^{\infty} e^{-a^2 t^2} dt =\dfrac{1}{a \sqrt{2 \pi}} \Gamma(1/2) = \dfrac{1}{a\sqrt{2}}.

Therefore, f(ω)=1a2eω24a2f(\omega) =\dfrac{1}{a\sqrt{2}} e^{-\dfrac{\omega^{2}}{4a^{2}}}.

Surya Prakash - 3 years, 10 months ago

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Problem 10:

Evaluate

0π/21(9tan2(x)+16)3dx\large \int_{0}^{\pi/2} \dfrac{1}{(9 \tan^2 (x) + 16)^3} \, dx

This problem has been solved by Pi Han Goh.

Surya Prakash - 3 years, 10 months ago

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Let II denote the value of the integral. And let 3tan(x)=4tan(y)3\tan (x) = 4\tan(y) , differentiate with respect to xx: 3sec2(x)=4sec2(y)(dydx)dx=4sec2(y)3sec2(x)dy=12sec2(y)9sec2(x)dy=12sec2(y)9tan2(x)+9dy=12sec2(y)16tan2(y)+9dy3\sec^2(x) = 4\sec^2(y) \left( \frac{dy}{dx} \right) \Rightarrow dx = \dfrac{4\sec^2(y)}{3\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\tan^2(x) + 9} dy= \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} dy

And of course, 9tan2(x)+16=(3tan(x))2+16=16(tan2(y)+1)=16sec2(y)9\tan^2(x) + 16 = (3\tan(x))^2 + 16 = 16(\tan^2 (y) + 1) = 16\sec^2(y) .

When x=0x = 0, y=0y = 0; when xπ2x\to \frac\pi2^- , yπ2y \to \frac\pi2 ^- as well. The integral becomes

I=0π/21(16sec2(y))312sec2(y)16tan2(y)+9dy=121630π/21sec4(y)116tan2(y)+9dy=310240π/2cos6(y)16sin2(y)+9cos2(y)dy=310240π/2cos6(y)16(1cos2(y))+9cos2(y)dy=310240π/2cos6(y)167cos2(y)dy. \begin{aligned} I &=& \int_0^{\pi/2} \dfrac{1}{(16\sec^2(y))^3} \cdot \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} \, dy \\ &=& \dfrac{12}{16^3} \int_0^{\pi/2} \dfrac1{\sec^4(y)} \cdot \dfrac1{16\tan^2 (y) + 9} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16\sin^2 (y) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16(1-\cos^2(y)) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16 - 7\cos^2(y)} \, dy. \\ \end{aligned}