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 December 3rd, 2012, 06:21 AM #1 Senior Member   Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0 A Challenging Integral Prove that: $\int_0^\infty \dfrac{\sin x \ln x}{x}dx=-\gamma \frac{\pi}{2}$ $\gamma$ is Euler's Constant.
December 3rd, 2012, 06:25 AM   #2
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Re: A Challenging Integral

Quote:
 Originally Posted by Etyucan Prove that: $\int_0^\infty \dfrac{\sin x \ln x}{x}dx=-\gamma \frac{\pi}{2}$ $\gamma$ is Euler's Constant.
You can post it here if you wish .

viewtopic.php?f=15&t=36221

December 3rd, 2012, 11:51 PM   #3
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Re: A Challenging Integral

Quote:
Originally Posted by zaidalyafey
Quote:
 Originally Posted by Etyucan Prove that: $\int_0^\infty \dfrac{\sin x \ln x}{x}dx=-\gamma \frac{\pi}{2}$ $\gamma$ is Euler's Constant.
You can post it here if you wish .

viewtopic.php?f=15&t=36221
No, thanks.

December 4th, 2012, 05:55 AM   #4
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Re: A Challenging Integral

Quote:
 Originally Posted by Etyucan Prove that: $\int_0^\infty \dfrac{\sin x \ln x}{x}dx=-\gamma \frac{\pi}{2}$ $\gamma$ is Euler's Constant.
Without parentheses this is ambiguous. Do you mean $\int_0^\infty sin(x) ln(x)dx$ or $\int_0\infty sin(x ln(x)) dx$? And, of course, there is a problem at x= 0. ln(0) is not defined so you really mean the limit as x goes to 0. Ignoring that for the moment, and assuming you meant the first of the two integrals, an obvious first step is to let u= ln(x) so that du= dx/x and $x= e^u$. Then the integral becomes $\int_{-\infty}^\infty sin(e^u)du$.

Now, what do you know about Euler's constant? In particular is there an integral representation for it?
http://en.wikipedia.org/wiki/Euler%E2%8 ... i_constant

 December 4th, 2012, 07:14 AM #5 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: A Challenging Integral I think you made a mistake in the substitution .
 December 4th, 2012, 10:48 AM #6 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: A Challenging Integral We can start with the identity: $\int_{0}^{\infty}x^{a-1}\sin(x)dx=\frac{2^{a-1}\sqrt{\pi}\Gamma(\frac{a}{2}+\frac{1}{2})}{\Gamm a(1-a/2)}$ Now, differentiate w.r.t a. Note the digamma functions involved due to the derivative of Gamma. $\int_{0}^{\infty}x^{a-1}\ln(x)\sin(x)dx=\frac{2^{a}\sqrt{\pi}\left(2\ln( 2)+\psi(a/2+1/2)+\psi(1-a/2)\right)\Gamma(a/2+1/2)}{4\Gamma(1-a/2)}$ Now, let $a=0$ and this simplifies to: $\int_{0}^{\infty}x^{-1}\ln(x)\sin(x)dx=\frac{\sqrt{\pi}\left(2\ln(2)+\p si(1/2)+\psi(1)\right)\Gamma(1/2)}{\Gamma(1)}$ But, $\Gamma(1)=1, \;\ \Gamma(1/2)=\sqrt{\pi}, \;\ \psi(1/2)=-\gamma-2\ln(2), \;\ \psi(1)=-\gamma$ So, we get: $\frac{-\pi}{2}\gamma$ The identity at the top I used at the start would be a good one to prove as well. I have to go now, though.
December 4th, 2012, 11:59 PM   #7
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Re: A Challenging Integral

Quote:
 Originally Posted by galactus We can start with the identity: $\int_{0}^{\infty}x^{a-1}\sin(x)dx=\frac{2^{a-1}\sqrt{\pi}\Gamma(\frac{a}{2}+\frac{1}{2})}{\Gamm a(1-a/2)}$ Now, differentiate w.r.t a. Note the digamma functions involved due to the derivative of Gamma. $\int_{0}^{\infty}x^{a-1}\ln(x)\sin(x)dx=\frac{2^{a}\sqrt{\pi}\left(2\ln( 2)+\psi(a/2+1/2)+\psi(1-a/2)\right)\Gamma(a/2+1/2)}{4\Gamma(1-a/2)}$ Now, let $a=0$ and this simplifies to: $\int_{0}^{\infty}x^{-1}\ln(x)\sin(x)dx=\frac{\sqrt{\pi}\left(2\ln(2)+\p si(1/2)+\psi(1)\right)\Gamma(1/2)}{\Gamma(1)}$ But, $\Gamma(1)=1, \;\ \Gamma(1/2)=\sqrt{\pi}, \;\ \psi(1/2)=-\gamma-2\ln(2), \;\ \psi(1)=-\gamma$ So, we get: $\frac{-\pi}{2}\gamma$ The identity at the top I used at the start would be a good one to prove as well. I have to go now, though.
Thank You!!!

 December 5th, 2012, 01:29 AM #8 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: A Challenging Integral I would know it involves the digamma or psi function , but I don't quite get the differentiating part.
 December 5th, 2012, 05:56 AM #9 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: A Challenging Integral The derivative of Gamma is $\Gamma'(x)=\Gamma(x)\psi(x)$ So, it involves that. Those solutions may be able to be reduced saomewhat if we look at them.
 December 5th, 2012, 06:18 AM #10 Senior Member   Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0 Re: A Challenging Integral I found out a way to calculate the integral $\int_0^\infty x^{a-1}\sin(x)dx=-\text{Im}\left[ \int_{0}^{\infty}x^{a-1}e^{-ix}dx\right]=-\text{Im}\left[ \frac{\Gamma (a)}{i^a}\right]$ $= -\Gamma (a)\text{Im}\left[ e^{-\frac{i\pi a}{2}}\right]=\Gamma (a) \sin \left( \frac{\pi a}{2}\right)$ Now, we can use Euler's Reflection Formula to express sin in terms of gamma function $\sin\left( \frac{\pi a}{2}\right)=\frac{\pi}{\Gamma (1-\frac{a}{2})\Gamma (\frac{a}{2})}$ $\int_0^\infty x^{a-1}\sin(x)dx= \frac{\pi \Gamma (a)}{\Gamma (1-\frac{a}{2})\Gamma (\frac{a}{2})}$ and finally from the duplication rule, $\int_0^\infty x^{a-1}\sin(x)dx= \frac{2^{a-1} \sqrt{\pi}\Gamma(\frac{a}{2}+\frac{1}{2})}{\Gamma (1-\frac{a}{2})$

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