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I know that $$ \int_0^\infty \frac{\log x}{\exp x} = -\gamma $$ where $ \gamma $ is the Euler-Mascheroni constant, but I have no idea how to prove this.

The series definition of $ \gamma $ leads me to believe that I should break the integrand into a series and interchange the summation and integration, but I can't think of a good series. The Maclaurin series of $ \ln x $ isn't applicable as the domain of $ x $ is not correct and I can't seem to manipulate the integrand so that such a Maclaurin series will work.

Another thing I thought of was using $ x \mapsto \log u $ to get $ \int\limits_{-\infty}^\infty \frac{\log \log u}{u^2} \ du $ and use some sort of contour integration, but I can't see how that would work out either.

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we easy show that $$\int_{0}^{\infty}\ln{x}e^{-x}dx=-\gamma$$ –  math110 Jun 13 '13 at 5:32
    
But for the sign, isn't this precisely what the OP's asking @math110? –  DonAntonio Jun 13 '13 at 10:43
    
That was a typo, my mistake. –  Jon Claus Jun 13 '13 at 16:29

4 Answers 4

up vote 8 down vote accepted

lemma1: $$\int_{0}^{1}\dfrac{1-e^{-x}}{x}dx-\int_{1}^{\infty}\dfrac{e^{-x}}{x}dx=\gamma$$

pf:use \begin{align}\sum_{i=1}^{n}\dfrac{1}{i}&=\int_{0}^{1}\dfrac{1-t^n}{1-t}dt\\ &=\int_{0}^{n}\dfrac{1-(1-\frac{x}{n})^n}{x}dx \end{align} and $$\ln{n}=\int_{1}^{n}\dfrac{1}{x}dx$$ so $$\gamma=\lim_{n\to\infty}(\sum_{i=1}^{n}\dfrac{1}{i}-\ln{n})=\lim_{n\to\infty}\left(\int_{0}^{1}\dfrac{1-(1-x/n)^n}{x}dx-\int_{1}^{n}\dfrac{(1-x/n)^n}{x}dx\right)$$

so $$\gamma=\int_{0}^{1}\dfrac{1-e^{-x}}{x}dx-\int_{1}^{\infty}\dfrac{e^{-x}}{x}dx=\int_{0}^{1}(1-e^{-x})d(\ln{x})-\int_{1}^{\infty}e^{-x}d(\ln{x})=\cdots=-\int_{0}^{\infty}e^{-x}\ln{x}dx$$

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For those who are wondering, I chose this one as it has the most elementary solution, relying only on the definition of $ \gamma $ as opposed to falling back on special functions. –  Jon Claus Jun 13 '13 at 16:30

Recalling the Mellin transform of a function $f$

$$ F(s)=\int_{0}^{\infty}x^{s-1}f(x)dx \implies F'(s) = \int_{0}^{\infty} x^{s-1}\ln(x)f(x) dx.$$

So, taking $f(x)=e^{-x}$ and finding its Mellin transform

$$ F(s)=\Gamma(s) \implies F'(s)=\Gamma'(s) $$

Taking the limit as $s\to 1$ yields the desired resuly

$$ \lim_{s\to 1}F'(s)=-\gamma. $$

Note: You can use the identity to find the limit

$$ \psi(x)=\frac{d}{dx}\ln\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}\implies \Gamma'(x)=\Gamma(x)\psi(x), $$

where $\psi(x)$ is the digamma function and $\psi(1)=-\gamma$.

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Here is one way, just derived it (by the way, the integral is equal to $-\gamma$). We know from Euler that you can write the logarithm as the following limit

$$\mathrm{ln}(x)=\lim_{n \to 0}\frac{x^n-1}{n}$$

The integral is now

$$\int_0^\infty \frac{\mathrm{ln} x}{\exp x}dx=\lim_{n\to 0}\frac{1}{n}\int_{0}^\infty\frac{x^n-1}{e^x}dx$$

This is just the Euler integral of second kind, thus you get

$$\int_0^\infty \frac{\mathrm{ln} x}{\exp x}dx=\lim_{n\to 0}\frac{\Gamma(n +1)-1}{n}=-\gamma$$

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+1. By the way, if you put your $\TeX$ in double dollar signs ($\$\$$ .... stuff .... $\$\$$), then the $\TeX$ gets centered and renders a bit better. –  JavaMan Jun 13 '13 at 5:41

This is just the definition of the gamma function: $$\Gamma(s)=\int_0^{\infty}x^{s-1}e^{-x}dx.$$ Differentiating it with respect to $s$ and then setting $s=1$, we get $$-\gamma=\psi(1)=\left[\frac{d}{ds}\ln\Gamma(s)\right]_{s=1}=\int_0^{\infty}\ln x\,e^{-x}dx.$$

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It's not completely clear to me how you get the log under the integral. –  NikolajK Jul 11 '13 at 20:19
    
@NickKidman $$\frac{d\Gamma(s)}{ds}=\frac{d}{ds}\int_0^{\infty }x^{s-1}e^{-x}dx=\int_0^{\infty}x^{s-1}\ln x\,e^{-x}dx.$$ Then it suffices to set $s=1$ in the final expression. –  O.L. Jul 11 '13 at 20:32
    
Ah okay, because the denominator in $\Gamma'(s)/\Gamma(s)$ becomes $1!$. –  NikolajK Jul 11 '13 at 20:35
    
@NickKidman Yes, exactly. –  O.L. Jul 11 '13 at 20:37

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