show that $\int_{0}^{\infty}e^{-x}\ln(x)dx=-\gamma=\Gamma'(1) $

Asked Aug 06 '13 at 12:55

Active Aug 06 '13 at 13:47

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show that $$\int_{0}^{\infty}e^{-x}\ln(x)dx= -\gamma=\Gamma'(1)$$

where $\gamma$ is Euler–Mascheroni constant

I started with $$\Gamma(z)=\left[ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)e^{-z/k}\right]^{-1}.$$

by take log then derivative with respect to z and put z=1 I got $\Gamma'(1)=-1-\gamma+\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+1})$

but I dont know how to compute $\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+1})$ I know the answer is 1 but how ?

this is my way . can any one show different way ? or give me how to compute $\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+1})$

edited Aug 06 '13 at 13:47

asked Aug 06 '13 at 12:55

mnsh

5,875

Is saying $\int_0^\infty e^{-x}\log x,dx = \Gamma'(1)$ legitimate? – Daniel Fischer Aug 06 '13 at 12:59
@DanielFischer yes – mnsh Aug 06 '13 at 13:04
Well, then you're done, since $\Gamma'(1) = -\gamma$. – Daniel Fischer Aug 06 '13 at 13:04
Hehe...@DanielFischer , that almost looked as a trap.+1 – DonAntonio Aug 06 '13 at 13:08
@DanielFischer hehe but why $\Gamma'(1)=-\gamma$ – mnsh Aug 06 '13 at 13:23
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This question is related, but doesn't answer all of the questions here. – Ayman Hourieh Aug 06 '13 at 13:28
@hmedan.mnsh For example, you get $\Gamma'(1) = -\gamma$ from the product representation $\Gamma(z) = e^{-\gamma z} \frac{1}{z}\prod\limits_{\nu = 1}^\infty \left(1 + \frac{z}{\nu}\right)^{-1}e^{z/\nu}$. – Daniel Fischer Aug 06 '13 at 13:35
@DanielFischer that is what i did but I dont know how to compute $\sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1})$ I know the answer is 1 but how ? – mnsh Aug 06 '13 at 13:40
@O.L. : thanks it is really has the answer but i make an edit which show where is the sum I dont know how to compute – mnsh Aug 06 '13 at 13:50
Telescopic sum is what you're looking for. – karakfa Aug 06 '13 at 14:05
@karakfa: no I say different method to show $\Gamma'(1)=-\gamma$ or the way to compute the sum which lead to show that $\Gamma'(1)=-\gamma$ – mnsh Aug 06 '13 at 14:11
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To compute $\sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1})$ Start by the following $\lim_{k\to \infty} \sum_{n=1}^{k}(\frac{1}{n}-\frac{1}{n+1}) =$ – Zaid Alyafeai Aug 06 '13 at 14:56
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$\sum_{n=1}^{k}(\frac{1}{n}-\frac{1}{n+1}) = 1-\frac{1}{2}+\frac{1}{2}+...+\frac{1}{k}-\frac{1}{k+1}$ – Zaid Alyafeai Aug 06 '13 at 14:58
@ZaidAlyafeai thanks – mnsh Aug 06 '13 at 15:02
that was telescopic sum. – karakfa Aug 14 '13 at 21:44

show that $\int_{0}^{\infty}e^{-x}\ln(x)dx=-\gamma=\Gamma'(1) $

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