I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. 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
Please post your solution and your proposed problem in a single new thread.
Format your post is as follows:
SOLUTION OF PROBLEM xxx (number of problem) :
[Post your solution here]
PROBLEM xxx (number of problem) :
[Post your problem here]
Please share this note so that lots of users here know this contest and take part in it. (>‿◠)✌
Okay, let the contest begin! Here is the first problem:
PROBLEM 1 :
For \(a>0\), show that
\[\begin{equation} \int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,dx=\pi\ln\left(\frac{a}{4}\right) \end{equation}\]
P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 2) and Brilliant Integration Contest - Season 1 (Part 3).
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Top NewestHere is the solution: \(\displaystyle I = \int \limits_0^a \frac{ \text{ln } x }{\sqrt{ax - x^2}} \text{d}x \)
\(\displaystyle I = \int \limits_0^a \frac{ \text{ln } (a-x) }{\sqrt{ax - x^2}} \text{d}x \)
\(\displaystyle \Rightarrow 2I = \int \limits_0^a \frac{ \text{ln } (ax - x^2) }{\sqrt{ax - x^2}} \text{d}x \)
Put \(\displaystyle x = \frac{a}{2} (1 - \sin t ) \Rightarrow ax - x^2 = \frac{a^2}{4} \cos ^2 t \)
Therefore, the integral becomes, \(\displaystyle 2I = \int \limits_{\frac{\pi}{2}}^{\frac{-\pi}{2}} \frac{ \text{ln } (\frac{a^2}{4} \cos ^2 t ) }{\sqrt{\frac{a^2}{4} \cos ^2 t }} (\frac{a}{2} \cos t )(-\text{d}t) \)
Rearranging, we obtain,
\(\displaystyle 2I = 4\int \limits_0^{\frac{\pi}{2}} \text{ln } (\frac{a}{2} \cos t ) \text{d}t \)
The value of \(\displaystyle \int \limits_0^{\frac{\pi}{2}} \text{ln } ( \cos \theta ) \text{d}\theta \) is \(\displaystyle \frac{ -\pi \text{ ln } 2 }{2} \) (which I have calculated separately and I can post if it is required).
Thus, we obtain, \(\displaystyle I = \pi \text{ ln } \bigg( \frac{a}{2} \bigg) - 2\frac{ \pi \text{ ln } 2 }{2} = \pi \text{ ln } \bigg(\frac{a}{4} \bigg)\)
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Find a closed form expression for the integral : \(\displaystyle I = \int \sin (2015 x) \sin ^{2013} x \text{ d}x \)
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\( Solution \quad of \quad problem \quad 2\)
Split \(sin(2015x)\) as \(sin(2014x+x)\) and our integral becomes :
\(I=\displaystyle \int { ({ sin }^{ 2014 }x.cos(2014x)+sin(2014x)cos(x){ sin }^{ 2013 }(x))\quad dx } \)
Multiply and divide it by \(2014\) to get :
\(\frac { 1 }{ 2014 } \displaystyle \int { ({ sin }^{ 2014 }x.(2014cos(2014x))+sin(2014x)(2014{ sin }^{ 2013 }(x)cos(x)))\quad dx } \)
\(\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { { sin }^{ 2014 }(x)dsin(2014x)+sin(2014x)(d{ sin }^{ 2014 }x) } \)
\(\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { d({ sin }^{ 2014 }(x)sin(2014x)) } \)
\(\large \displaystyle I=\frac { { sin }^{ 2014 }(x)sin(2014x) }{ 2014 } +C\)
\(\large Problem \quad 3\)
Evaluate \(I=\displaystyle \int _{ 0 }^{ \infty }{ { (\frac { sin(x) }{ x } })^{ 2 }dx } \)
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@Anastasiya Romanova I have posted my problem sorry for holding the contest.
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@Sudeep Salgia. See the rules & the format post. Thank you. :)
Nicely done, +1. Now you must post a new problem, but before that. I think it would be nice if you post this as a new post, no need to replyLog in to reply
Nicely done, +1! But you forget to post your problem anyway. You have a privilege for that so that this contest keeps going on. Post PROBLEM 2 in your thread solution. Thanks :)
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Well, there must be a rule for people like Ronak! Solving a problem doesn't give you the right to withhold the contest. In such a case, someone should be allowed to interfere, right?
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Solution to problem 10
First integrate by parts -
\[I= \int_{0}^{\infty} \dfrac{\sin^3{x}}{x^2} dx =\int_{0}^{\infty} \dfrac{3\sin^2{x} \cos{x}}{x} dx\]
Then, use the property of the laplace transform that \(\mathcal{L} \left\{\dfrac{f(t)}{t}\right\}(s)=\displaystyle\int_{s}^{\infty} F(p) dp\) (in our case, \(s \rightarrow 0\)).
\[I=3\int_{0}^{\infty} \mathcal{L} \{\sin^2{t} \cos{t}\}(p) \ dp\]
\[I=3\int_{0}^{\infty} \dfrac{2s}{(s^2+1)(s^2+9)} ds=\boxed{\dfrac{3\log {3}}{4}}\] The last integral can be found by using partial fractions.
Note : \(\mathcal{L}\{f(t)\}(s)=\displaystyle\int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t\)
Problem 11
Find
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Solution of Problem 7:
Proposition :
Proof :
It can be proven by using Weierstrass substitution: \(t=\tan\left(\dfrac{x}{2}\right)\), then
\[\begin{align} \\ \int_0^{\pi} \frac{dx}{p+\cos x}\,dx &=\int_0^{\infty} \frac{2}{p+1+(p-1)t^2}\,dt\qquad\Rightarrow\qquad t=\sqrt{\frac{p+1}{p-1}}\tan y\\ &=\left.\frac{2}{\sqrt{p^2-1}}\arctan t\;\right|_0^{\infty}\\ &=\frac{\pi}{\sqrt{p^2-1}}\qquad\qquad\text{Q.E.D.} \end{align}\]
Differentiating the proposition w.r.t. \(p\) twice and setting \(p=\sqrt{10}\), we have
\[\begin{align} \\ \frac{\partial^2}{\partial p^2}\int_0^\pi\frac{1}{p+\cos x}\, dx&=\frac{\partial^2}{\partial p^2}\left[\frac{\pi}{\sqrt{p^2-1}}\right]\\ \int_0^\pi\frac{2}{\left(p+\cos x\right)^3}\, dx&=\frac{\pi\left(2p^2+1\right)}{\sqrt{\left(p^2-1\right)^5}}\\ \int_0^\pi\frac{1}{\left(\sqrt{10}+\cos x\right)^3}\, dx&=\frac{7\pi}{162} \end{align}\]
Problem 8 :
Prove
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Solution of Problem 8:
Substitute \(\tan x\mapsto x\), then the integral is:
\[\int_0^{\infty} \frac{dx}{(1+x^2)(1+8\sin^2x)}=\int_0^{\infty} \frac{dx}{(1+x^2)(5-4\cos(2x))}\]
\[=\int_0^{\infty} \frac{1}{1+x^2}\left(\frac{1}{3}\left(1+2\sum_{k=1}^{\infty} \frac{\cos(2kx)}{2^k}\right)\right)=\frac{\pi}{6}+\frac{2}{3}\sum_{k=1}^{\infty} \frac{1}{2^k}\int_0^{\infty} \frac{\cos(2kx)}{1+x^2}\,dx\]
\[=\frac{\pi}{6}+\frac{\pi}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2e^2}\right)^k=\frac{\pi}{6}+\frac{\pi}{3}\frac{1}{2e^2-1}=\boxed{\dfrac{\pi}{6}\left(\dfrac{2e^2+1}{2e^2-1}\right)}\]
I have used the following result:
\[\int_0^{\infty} \frac{\cos(mx)}{x^2+a^2}\,dx=\frac{\pi}{2a}e^{-am}\]
I cannot think of a challenging problem at the moment, I request somebody else to post one. Thanks!
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Ingenious! But I have a doubt - In your fourth step, how is it guaranteed that the function \(\dfrac{\cos (2kx)}{2^k}\) is positive on \((0,\infty)\)? PS : you forgot to write the \(\mathrm{d}x\) in the third step.
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@Anastasiya Romanova .
If pranav refuses to put up a question then who put will put it upLog in to reply
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Sorry for very late response. Using elementary techniques,-
Solution - Problem 7
\( \displaystyle \int_{0}^{\pi} \dfrac{dx}{p - cosx}\)
\( 2I = \displaystyle \int_{0}^{\pi} \dfrac{2p~dx}{p^2 - cos^2x}\)
\( I = \displaystyle \int_{0}^{\pi} \dfrac{p~dx}{p^2(1 + tan^2x) - 1}\)
\( I = p \displaystyle \int_{0}^{\pi} \dfrac{sec^2x~dx}{p^2tan^2x + \sqrt{(p^2 - 1)^2}}\)
tanx = t
\( I = 2p \displaystyle \int_{0}^{\infty} \dfrac{dt}{p^2t^2 + \sqrt{(p^2 - 1)^2}}\)
\( I = \dfrac{2p}{ \sqrt{p^2 - 1}} \times \dfrac{1}{p} \Big[_{0}^{\infty} tan^{-1} \dfrac{pt}{ \sqrt{p^2 - 1}}\)
\( I = \dfrac{\pi}{ \sqrt{p^2 - 1}}\)
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Solution of Problem 3 :
Use integration by parts by taking \(u=\sin^2x\) and \(dv=\dfrac{dx}{x^2}\), we get
\[\begin{align} \int_0^\infty\frac{\sin^2x}{x^2}\,dx&=-\left.\frac{\sin^2x}{x}\right|_0^\infty+\int_0^\infty\frac{2\sin x\cos x}{x}\,dx\\ &=0+\int_0^\infty\frac{\sin 2x}{x}\,dx\\ &=\int_0^\infty\frac{\sin t}{t}\,dt\qquad\Rightarrow\qquad t=2x\\ \end{align}\]
Now consider
\[\begin{equation} I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt\qquad a\ge0 \end{equation}\]
so that \(I(\infty)=0\) and our considered integral is \(I(0)\). Differentiating w.r.t. \(a\) and then integrating back, we get
\[\begin{align} I'(a)&=-\int_0^\infty e^{-at}\sin t\,dt\\ &=-\frac{1}{1+a^2}\qquad\Rightarrow\qquad\text{integration by parts twice}\\ I(a)&=-\int \frac{1}{1+a^2}\,da\\ &=-\arctan(a)+C \end{align}\]
For \(I(\infty)=0\), implying \(C=\dfrac{\pi}{2}\). Hence
\[\begin{equation} I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt=\frac{\pi}{2}-\arctan(a) \end{equation}\]
and
\[\begin{equation} I(0)=\int_0^\infty\frac{\sin t}{t}\,dt=\frac{\pi}{2} \end{equation}\]
Thus
Problem 4 :
Prove
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\(\large Solution\quad of\quad Problem\quad 4\)
Firstly we will prove a general result
Result
\(\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a }ln(x)dx } =\frac { -1 }{ { (1+a) }^{ 2 } }\)
Proof
\(I=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a }ln(x)dx }\)
Integrating by parts \(u=ln(x),dv={x}^{a}dx\)
\(I=\displaystyle ln(x)\frac { { x }^{ a+1 } }{ a+1 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 1 }{ a+1 } \int _{ 0 }^{ 1 }{ { x }^{ a }dx }\)
\(I=\frac { -1 }{ { (1+a) }^{ 2 } }\)
Hence proved
Now we have :
\(I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { ln(cos(x)) }{ sin(x) } dx }\)
Take \(cos(x)=y\) to get our integral as :
\(\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(y)dy }{ 1-{ y }^{ 2 } } } \)
Now \(\frac { 1 }{ 1-{ y }^{ 2 } } =\displaystyle \sum _{ n=0 }^{ \infty }{ { y }^{ 2n } } \)
Our integral becomes :
\(I=\displaystyle \sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ { y }^{ 2n }ln(y)dy } } \)
Using our proved result we get :
\(I=-\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { (2n+1) }^{ 2 } } } \)
Also the summation can we written as :
\(I=-(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2n) }^{ 2 } } } )\)
\(I=-(\zeta (2)-\frac { 1 }{ 4 } \zeta (2))=\frac { -3 }{ 4 } \zeta (2)=\large \frac{-{\pi}^{2}}{8}\)
Where \(\zeta(x)\) is the zeta function
\(\large Problem \quad 5\)
Find closed form of \(\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } }\)
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Keep going fella, +1! But please post your solution & proposed problem in a new thread like I did. No need to reply other threads to post a solution & a problem. Anyway, I've posted the solution of your problem & also proposed a new problem.
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Solution of Problem 13 :
First we prove the following proposition:
Proposition :
Proof :
Note that \[ \int_0^1 x^a\ dx=\frac1{a+1}\qquad\text{for }\ \alpha>-1. \] Differentiating equation above \(k\) times w.r.t. \(a\) we have \[ \int_0^1 \frac{\partial^k}{\partial a^k}\left(x^a\right)\ dx=\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}}\qquad\text{Q.E.D.} \]
Now, we will evaluate the general case of
\[\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\int_0^\infty\frac{x^{a-1}\,e^{-bx}}{1-e^{-bx}}\,dx\qquad\text{for }\ a,b>0\]
Set \(y=e^{-bx}\), then
\[\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\frac{(-1)^{a-1}}{b^a}\int_0^1\frac{\ln^{a-1}y}{1-y}\,dy\]
Use a geometric series for \(\dfrac{1}{1-y}\) then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have
\[\begin{align} \int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx&=\frac{(-1)^{a-1}}{b^a}\int_0^1\sum_{k=0}^\infty y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty\int_0^1 y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty \frac{(-1)^{a-1}\, (a-1)!}{(k+1)^{a}}\\ &=\frac{(a-1)!}{b^a}\sum_{k=1}^\infty \frac{1}{k^{a}}\\ &=\frac{\Gamma(a)\,\zeta(a)}{b^a} \end{align}\]
where \(\Gamma(a)\) is the gamma function and \(\zeta(a)\) is the Riemann zeta function.
Hence, by setting \(a=n+1\) and \(b=1\), we have
\[\int_0^\infty\frac{x^{n}}{e^{x}-1}\,dx=\Gamma(n+1)\,\zeta(n+1)\]
Problem 14 :
Prove that
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\(Solution\quad of\quad problem\quad 9\)
\(\displaystyle I=\int _{ 0 }^{ 1 }{ ln(\frac { 1+x }{ 1-x } )\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } } \)
Put \(x=cos(\theta)\) to get our integral as :
\(\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln({ cot }^{ 2 }\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } } \)
\(\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(tan\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } } \)
Put \(tan(\theta/2)=x\) to get our integral as :
\(\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } \)
I proved in my solution to problem 4 that :
\(\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } =\frac { -{ \pi }^{ 2 } }{ 8 } \)
Using this I get :
\(I=\frac{{\pi}^{2}}{2}\)
\(Problem\quad 10\)
Find \(\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx } \)
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Solution of Problem 15 :
We have \[\begin{equation} \int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx \end{equation}\]
Proposition :
Proof :
Let
\[\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx \end{equation}\]
then
\[\begin{align} I'(s)&=\int_0^1 \frac{1-x^s}{1-x}\,dx\\ I''(s)&=-\int_0^1 \frac{x^s\ln x}{1-x} \,dx\\ &=-\int_0^1\sum_{n=0}^\infty x^{n+s}\ln x\,dx\\ &=-\sum_{n=0}^\infty\partial_s\int_0^1 x^{n+s}\,dx\\ &=-\sum_{n=0}^\infty\partial_s\left[\frac{1}{n+s+1}\right]\\ &=\sum_{n=0}^\infty\frac{1}{(n+s+1)^2}\\ &=\psi_1(s+1)\\ I'(s)&=\int\psi_1(s+1)\,ds\\ I'(s)&=\int\frac{\partial}{\partial s}\bigg[\psi(s+1)\bigg]\,ds\\ I'(s)&=\psi(s+1)+C\\ \end{align}\] For \(s=0\), we have \(I'(0)=0\). Implying \(C=-\psi(1)=\gamma\), then \[\begin{align} I'(s)&=\psi(s+1)+\gamma\\ I(s)&=\int \psi(s+1)\,ds +\gamma s+C\\ &=\int \frac{\partial}{\partial s}\bigg[\ln\Gamma(s+1)\bigg]\,ds +\gamma s+C\\ &=\ln\Gamma(s+1) +\gamma s+C\\ \end{align}\]
where \(\psi(z)\) is the digamma function. For \(s=0\), we have \(I(0)=0\). Implying \(C=0\), then
\[\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s\qquad\qquad\square \end{equation}\]
For \(s=1\), we have
\[\begin{equation} I(1)=\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx=\gamma \end{equation}\]
where \(\gamma\) is The Euler–Mascheroni constant.
Problem 16 :
Prove
where \(\operatorname{B}\left(x,y\right)\) is the beta function.
PS : POST YOUR SOLUTION BELOW EACH PROBLEM THREAD AND POST YOUR PROPOSED PROBLEM AS A NEW THREAD. PUT THEM IN SEPARATED THREAD. SO THAT THE POSTS LOOK MORE ORGANIZED. THANKS.
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Can you please post a all together new note of the integration contest it's getting messy and messier, Please move this contest to a new note @Anastasiya Romanova
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OK
The Brilliant Integration Contest - Season 1 (Part 1) moves to The Brilliant Integration Contest - Season 1 (Part 2). Enjoy it!! (>‿◠)✌
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_ Solution of problem 14 _
\(I = \displaystyle \int_{0}^{1} \ln \bigg(\dfrac{3+x}{3-x} \bigg) \dfrac{1}{\sqrt{x(1-x^2)}} {\mathrm dx} \)
= \(\displaystyle \int_{0}^{1} \ln(1 + x/3) \dfrac{1}{\sqrt{x(1-x)}} {\mathrm dx} \)
= \(\displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 + \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta - \displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 - \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta\)
Lemma \(\displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 \theta) = \pi \ln \bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)\)
Proof
Let \(I(a) = \displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 x) {\mathrm dx}\)
\(I'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{\sin^2 x}{1+ a \sin^2 x} {\mathrm dx}\)
\(I'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{1}{a}\bigg(1 - \dfrac{\sec^2 x}{1+(a+1) \tan^2x} {\mathrm dx}\bigg)\)
= \( \dfrac{\pi}{2a} - \dfrac{1}{a} \bigg(\displaystyle \int_{0}^{\infty} \dfrac{dz}{1+(a+1)z^2}\bigg)\)
=\(\dfrac{\pi}{2a} \bigg(1 - \dfrac{1}{\sqrt{a+1}}\bigg)\)
Hence, \(I(a) - I(0) = \displaystyle \int_{0}^{a} \dfrac{\pi}{2} \bigg(\dfrac{1}{x} - \dfrac{1}{x\sqrt{x+1}}\bigg) {\mathrm dx}\)
= \(\pi \ln(\sqrt{x+1} +1)\bigg|_{0}^{a} = \pi\ln\bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)\) (Note that \(I(0) = 0\))
Put \(a=1/3\) , and \(-1/3\), then substitute back in original integral to get:
\(I = 2 \pi \ln \bigg(\dfrac{1+\sqrt{4/3}}{1+\sqrt{2/3}} \bigg) = \boxed{\pi \ln \bigg(\dfrac{7 + 4\sqrt{3}}{5+2\sqrt{6}}\bigg)}\)
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Nice solution Jatin, +1! But you're a bit late about 10 min than Ronak, so the right to propose a new problem goes to Ronak.
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@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.
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\(Solution\quad of\quad problem\quad 14\)
Lemma
\(\displaystyle\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx } =2\pi ln(\frac { a+b }{ 2 } )\)
Proof :
\(\displaystyle I(a)=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx }\)
Differentiating with respect to \(a\) we get :
\(\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 2 }x }{ { a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x } dx } \)
Put \(tan(x)=t\) to get our integral as :
\(\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } } \)
\(\displaystyle =2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } } =\frac { 2a }{ { b }^{ 2 }-{ a }^{ 2 } } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { b }^{ 2 }dt }{ { a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 } } } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ 1+{ t }^{ 2 } } } )\)
Solving it we get :
\(\displaystyle {I}^{'}(a)=\frac { 2\pi }{ a+b } \)
\(\Rightarrow \displaystyle I(a)=2\pi ln(a+b)+C\)
Put \(a=b=1\) to get \(C=-2\pi ln(2)\)
Hence \(\displaystyle I(a)=2\pi ln(\frac{a+b}{2})\)
In our integral in the given question put \(x=\frac{1-sin\theta }{2}\) to get the integral as :
\(I= \displaystyle \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7-sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5+sin(\theta))dx } \)
Using the property \(\displaystyle \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx } \) we get :
\(\displaystyle I=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7+sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5-sin(\theta))dx } \)
Adding these two we get :
\(\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49-{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25-{ sin }^{ 2 }(\theta))dx } )\)
\(\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49{ cos }^{ 2 }\theta+48{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25{ cos }^{ 2 }\theta+24{ sin }^{ 2 }\theta)dx } )\)
Using our given lemma we get :
\(\displaystyle I=\frac { 1 }{ 2 } (2\pi ln(\frac{7+4\sqrt { 3 }}{2} )-2\pi ln(\frac{5+2\sqrt { 6 }}{2} ))=\pi ln(\frac { 7+4\sqrt { 3 } }{ 5+2\sqrt { 6 } } )\)
\(Problem\quad 15\)
Evaluate \(\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ ln(x) } +\frac { 1 }{ 1-x } dx } \)
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According to timeline, you're faster than @jatin yadav to post your solution, then the right to propose a new problem is yours.
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\(Solution\quad of\quad Problem\quad 12\)
It's a very good disguised integral.
First note that the function is an even function hence our integral can be written as :
\(\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { cos(arctan(2x)) }{ (1+{ x }^{ 2 })\sqrt { 1+4{ x }^{ 2 } } } dx } \)
Now we know that \(\displaystyle cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}\). Using this our integral becomes :
\(\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+{ x }^{ 2 })(1+4{ x }^{ 2 }) } dx } \)
Splliting it by partial fractions we get :
\(\displaystyle I=\frac { 2 }{ 3 } (\int _{ 0 }^{ \infty }{ \frac { dx }{ (\frac { 1 }{ 4 } +{ x }^{ 2 }) } } -\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } } )\)
\(\displaystyle I=\frac { 2 }{ 3 } (2{ tan }^{ -1 }(2x)\overset { \infty }{ \underset { 0 }{ | } } -{ tan }^{ -1 }(x)\overset { \infty }{ \underset { 0 }{ | } } )\)
\(\displaystyle I=\frac { \pi }{ 3 }\)
\(Problem\quad 13\)
Find closed form of \(\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } \)
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Wait a sec!? Could you rectify your solution, because the original problem is written as \(\arctan(2x)\). And one thing, what did you mean by "Now we know that \(\cos(\arctan x)=\frac{1}{\sqrt{1+4x^2}}\)"? Could you elaborate?
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That was my typing mistake, also Let\( arctan(2x)=\theta\), hence \(2x=tan(\theta)\), hence \(cos(\theta)=\frac{1}{\sqrt{1+4{x}^{2}}}\)
\(\Rightarrow cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}\)
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Solution of Problem 11 :
Using substitution \(u=\dfrac{\alpha^2}{x}\;\Rightarrow\;x=\dfrac{\alpha^2}{u}\;\Rightarrow\;dx=-\dfrac{\alpha^2}{u^2}\ du\), then
\[\begin{align} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\int_0^\infty\frac{\ln \left(\dfrac{\alpha^2}{u}\right)}{\left(\dfrac{\alpha^2}{u}\right)^2+\alpha^2}\cdot \dfrac{\alpha^2}{u^2}\ du\\ I(\alpha)&=\int_0^\infty\frac{2\ln \alpha-\ln u}{\alpha^2+u^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-\int_0^\infty\frac{\ln u}{u^2+\alpha^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-I(\alpha)\\ I(\alpha)&=\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du. \end{align}\]
The last integral can easily be evaluated by using substitution \(u=\alpha\tan\theta\), then
\[\begin{align} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\frac{\ln \alpha}{\alpha}\int_0^{\Large\frac\pi2} \ d\theta\\ &=\frac{\pi\ln \alpha}{2\alpha} \end{align}\]
Problem 12 :
Prove
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The last substitution should be \(u=\alpha \tan \theta\) :)
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This is a simple kind of NCERT problem. It can be easily done by substituting \( t = tan^{-1}\dfrac{x}{\alpha}\) and then applying by parts
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Fixed it. Sorry, I'm too hasty. Thanks... :)
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Since @Pranav Arora is unable to propose a problem (I hope it's only temporary) and to make this contest sustains, then according to rule 4, I, as the last solver, have a right to propose a new one. Here is the problem:
PROBLEM 9
Prove that
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Solution -Problem - 9 We substitute, \[ t = \frac{1+x}{1-x} \implies x = \frac{1-t}{1+t} \] Doing the substitution and , simplifying , gives, \[ I = - \int_{0}^{1} \frac{\ln(t)}{(\sqrt{t})(1-t)} dt\] Now we substitute, \( t=\sin^2(x) \) \[ I = -4\int_{0}^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cos(x)} dx \] Using, Problem -\(4 \) \[ I = -4 \times \frac{-\pi^2}{8} = \frac{\pi^2}{2} \]
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I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I did @Shivang Jindal
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@Shivang Jindal
You can try my posted question.Log in to reply
I am able to write the integral as a series - \[I=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}\]
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Solution 9: First, consider the following Lemma.
LEMMA : \[\int_{0}^{1} \frac{\ln{x}}{1-x^2} dx=-\frac{\pi^2}{8}\]
Proof: Since, \(\frac{1}{1-x^2}=\sum\limits_{n=0}^{\infty} {x^{2n}}\), we have,
\( \int_0^1 \frac{\ln{x}}{1-x^2}=\sum\limits_{n=0}^{\infty} \int_0^1 x^{2n}{\ln{x}} \: dx\)
\(=-\sum\limits_{n=0}^{\infty} {\frac{1}{(2n+1)^2}}\) (Using Integration By Parts)
\(=\sum\limits_{n=0}^{\infty} {\frac{1}{(2n)^2}} - \sum\limits_{n=0}^{\infty} {\frac{1}{(n)^2}}\)
\(=\frac{-3}{4}\zeta(2)= -\frac{\pi^2}{8}\)
Now,
\(\text{I}= \int_{0}^{1} \ln({\frac{1+x}{1-x}}) \frac{dx}{x\sqrt{1-x^2}}\)
Substituting \(x=\frac{1-y^2}{1+y^2}\), we have,
\(\text{I}=-4 \times \int_{0}^{1} \frac{\ln{y}}{1-y^2} dx\)
\(=\frac{\pi^2}{2}\) (Using the Lemma)
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Nice answer, +1! But I'm really sorry, the one who deserves to propose the next problem is Ronak since according to the timeline his answer comes first than the other answers. Maybe, you can try to answer his proposed problem. \(\ddot\smile\)
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@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.
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Solution of Problem 5 :
I will prove the general form of
\[\int_0^\infty\frac{x^{m-1}}{1+x^n}\,dx=\frac{\pi}{n\sin\frac{m\pi}{n}}\]
It can easily be proven by taking substitution \(\displaystyle y=\frac{1}{1+x^n}\) and the integral becomes Beta function
\[\frac{1}{n}\int_0^1 y^{\large 1-\frac{m}{n}-1}\ (1-y)^{\large \frac{m}{n}-1}\,dy=\frac{\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)}{n}=\frac{\pi}{n\sin\frac{m\pi}{n}}\]
where the last part comes from Euler's reflection formula for the gamma function. Hence, by setting \(m=1\), we have
Problem 6 :
Show that
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\(I = \displaystyle \int_{0}^{\pi/2} \dfrac{\sin^2 x}{\sin x + \cos x} {\mathrm dx}\)
\(I = \displaystyle \int_{0}^{\pi/2} \dfrac{\cos^2 x}{\sin x + \cos x} {\mathrm dx}\) (As \(\displaystyle \int_{0}^{a} f(x) {\mathrm dx} = \int_{0}^{a} f(a-x) {\mathrm dx} \) )
Hence, \(2I = \displaystyle \int_{0}^{\pi/2} \dfrac{1}{\sin x + \cos x} {\mathrm dx}\)
Hence, \(I = \dfrac{1}{2} \displaystyle \int_{0}^{\pi/2} \dfrac{1 + \tan^2 (x/2)}{ 2 \tan(x/2) + 1 -\tan^2(x/2)} {\mathrm dx}\)
Putting \(\tan \dfrac{x}{2} = t\),
\(I = \displaystyle \int_{0}^{1} \dfrac{dt}{2 - (1-t)^2}\)
=\(\displaystyle \dfrac{1}{2 \sqrt{2}} \ln \bigg(\dfrac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} \bigg) \bigg|_{0}^{1} \)
= \(\displaystyle \dfrac{\ln(3 + 2 \sqrt{2})}{2 \sqrt{2}}\)
Problem 7
Prove that \(\displaystyle \int_{0}^{\pi} \dfrac{1}{(\sqrt{10}+ \cos x)^3} = \dfrac{7 \pi}{162}\)
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\[ I(a) = \int_{0}^{\pi} \frac{1}{(a-\cos(x))}, a>1 \] Use Weierstrass substitution, its not hard to show that , this integral \( =\frac{\pi}{\sqrt{a^2-1}} \) Now , we differentiate, both sides with respect to \( a \). \[ I'(a) = \int_{0}^{\pi} -\frac{1}{(a-\cos(x))^2} = \frac{d}{da} \frac{\pi}{\sqrt{a^2-1}} \] Again differentiating both sides, \[ I''(a) = \int_{0}^{\pi} \frac{2}{(a-\cos(x))^3} = \frac{d}{d^2a} \frac{\pi}{\sqrt{a^2-1}} \] Taking, 2 to RHS gives, \[ I''(a) = \int_{0}^{\pi} \frac{1}{(a-\cos(x))^3} = \frac{d}{d^2a} \frac{\pi}{2\sqrt{a^2-1}} \] Note that, we need \( I''(\sqrt{10}) \) . Plugging, \( a=\sqrt{10} \) , and evaluating the expression in RHS gives, \( \frac{7\pi}{162} \)
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It looks like there's someone here who is interesting in knowing the closed-form of your previous problem 7 and he posted it on Math S.E.. I am wondering, does that really have a closed-form? I tried to solve it for hours but no success & you really gave me lots of trouble cracking that integral.
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this problem and got struck at it.
Well ,actually I had to change the problem as I didn't know its solution. Actually, few days back, at the same website I tried answeringLog in to reply
his question but only gets 3 upvotes. It's not worth enough. Luckily, he accepts my answer. I also answer one of his question on Integrals and Series, but of course not for free. LOL
OMG! Why did you propose then here on this contest? Anyway, just ignore that user. He always ask two or three hard integrals in a single question on Math S.E. I can answer one ofLog in to reply
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this. You're very good at all math subjects & physics too. That's truly impressive!!
No worries. It's so-so in my country. I learned from my late grandpa, my big brother (I hate him!), and lots of people on internet since I was 7. You might be interested in seeingAnyway, I've just remembered that the same user (Samurai, the one who posts your previous problem 7 on Math S.E.) also posted one of my problem on Brilliant at Math S.E. too. What a cheater!
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Advanced System Of Equations and Geometry problem: All Three Concurrency. Answering Algebra problem makes me level up, so I must down my level again. I just want the other know me because I'm sorta good at Calculus, not the other subjects. Knowing Calculus in such young age also gives me lots of advantages on the other forums where I join in. Okay, I think it's enough. I hope I am not making any immodest statements or showing narcissism on my part. \(\ddot\smile\)
Because I wanna beat my big brother. Seriously, he always says to me that calculus is the highest level mathematics, so that's why I'm really interested in calculus specially integrals & series and almost every weekend we have a competition of solving integrals & series problems although he always beats me easily. Of course I love math, it means I also love the other fields of math but I'm just not too interested in them. Besides, my first crush on math because of calculus. I don't wanna tell the story because it's too personal. Special for you, I have answered two problems here that don't relate to Calculus to show my capability other than Calculus (It's also because I'm bored). Here they are Algebra problem:Log in to reply
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Ans:ln4a/2a^2
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