Computing $ \int_0^\infty \frac{\log x}{\exp x} \ dx $ 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. 
 A: 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
$$-\gamma=\psi(1)=\left[\frac{d}{ds}\ln\Gamma(s)\right]_{s=1}=\int_0^{\infty}\ln x\,e^{-x}dx.$$
A: We are all familiar with the famous limit $~\displaystyle\lim_{h\to0}\frac{a^h-1}h~=~\ln a,~$ which, by the simple 
substitution $h~=~\dfrac1n,~$ becomes $~\displaystyle\lim_{n\to\infty}~n\Big(\sqrt[^n]a-1\Big)~=~\ln a.~$ At the same time, using the 
Taylor series expansion of $~\ln(1-t)~=-\displaystyle\sum_{k~=~1}^\infty\frac{t^k}k,~$ we can show, by switching the order 
of summation and integration, and appealing to telescoping series, that $$\int_0^1\ln\Big(1-\sqrt[^n]x\Big)~dx~=-H_n,$$ the $n^{th}$ harmonic number. By combining the two formulas, we have $$\begin{align}\gamma&~=~\lim_{n\to\infty}\bigg[H_n-\ln n\bigg]~=~\lim_{n\to\infty}\bigg[-\int_0^1\ln\Big(1-\sqrt[^n]x\Big)~dx-\int_0^1\ln n~dx\bigg]~=~\\\\&~=-\lim_{n\to\infty}\int_0^1\ln\bigg[n\Big(1-\sqrt[^n]x\Big)\bigg]~dx~=-\int_0^1\ln\big(-\ln x\big)~dx~=-\Gamma'(1),
\end{align}$$ where the last identity has been obtained by differentiating under the integral sign Euler's 
first historical expression for the $\Gamma$ function, $~n!~=~\displaystyle\int_0^1\big(-\ln x\big)^n~dx,~$ at $n~=~0$. But the 
same function can also be written as $~n!~=~\displaystyle\int_0^\infty\frac{x^n}{e^x}~dx,~$ which, when differentiated under  the integral sign at $n~=~0$, yields the original integral. QED.
A: 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
$$\gamma=\int_{0}^{1}\dfrac{1-e^{-x}}{x}dx-\int_{1}^{\infty}\dfrac{e^{-x}}{x}dx=\int_{0}^{1}(1-e^{-x})d(\ln{x})-\int_{1}^{\infty}e^{-x}d(\ln{x})=\cdots=-\int_{0}^{\infty}e^{-x}\ln{x}dx$$
A: 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 
$$\int_0^\infty \frac{\mathrm{ln} x}{\exp x}dx=\lim_{n\to 0}\frac{\Gamma(n
+1)-1}{n}=-\gamma$$
A: 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 $\psi(1)=-\gamma$.
A: For $x\le n$, $\left(1-\frac{x}{n}\right)^n$ increases with $n$. By Bernoulli's Inequality,
$$
\begin{align}
\frac{\left(1-\frac{x}{n+1}\right)^{n+1}}{\left(1-\frac{x}{n}\right)^n}
&=\left(\frac{(n+1-x)\,n}{(n+1)(n-x)}\right)^{n+1}\left(1-\frac{x}{n}\right)\\
&=\left(1+\frac{x}{(n+1)(n-x)}\right)^{n+1}\left(1-\frac{x}{n}\right)\\[6pt]
&\ge\left(1+\frac{x}{n-x}\right)\left(1-\frac{x}{n}\right)\\[9pt]
&=\frac{n}{n-x}\frac{n-x}{n}\\[12pt]
&=1
\end{align}
$$
Therefore, by Monotone Convergence,
$$
\begin{align}
\int_0^\infty\log(x)e^{-x}\mathrm{d}x
&=\lim_{n\to\infty}\int_0^n\log(x)\left(1-\frac{x}{n}\right)^n\mathrm{d}x\\
&=\lim_{n\to\infty}n\int_0^1(\log(x)+\log(n))\,(1-x)^n\,\mathrm{d}x\\
&=\lim_{n\to\infty}n\int_0^1(\log(1-x)+\log(n))\,x^n\,\mathrm{d}x\\
&=\lim_{n\to\infty}\frac{n}{n+1}\log(n)-n\int_0^1\sum_{k=1}^\infty\frac{x^k}{k}x^n\,\mathrm{d}x\\
&=\lim_{n\to\infty}\frac{n}{n+1}\log(n)-n\sum_{k=1}^\infty\frac1{k(n+k+1)}\\
&=\lim_{n\to\infty}\frac{n}{n+1}\log(n)-\frac{n}{n+1}\sum_{k=1}^\infty\left(\frac1k-\frac1{n+k+1}\right)\\
&=\lim_{n\to\infty}\frac{n}{n+1}\left(\log(n)-\sum_{k=1}^{n+1}\frac1k\right)\\[6pt]
&=-\gamma
\end{align}
$$
