Double Natural Logs

Set $u=e^{-t}$ then the integral becomes : $\int_0^{\infty} e^{-t} \ln t\ \mathrm{d}t = \frac{\mathrm{d}(\Gamma(x))}{\mathrm{d}x} \bigg|_{x=1} =-\gamma$ A Proof that $\Gamma'(1)=-\gamma$ : This proof can be found in many real analysis books.

Using dominated convergence theorem we can note that : $\int_0^{\infty} e^{-x} \ln x \ \mathrm{d}x= \lim_{n\to \infty} \int_0^{n}\left(1-\frac{x}{n} \right)^n \ln x \ \mathrm{d}x$ Set $x=n t$ you'll get $\lim_{n\to \infty} n \int_0^{1} \ln n (1-t)^n + (1-t)^n\ln t \ \mathrm{d}t=\lim_{n\to \infty} \frac{n}{n+1} \ln n + n\int_0^1 \ln t (1-t)^n \ \mathrm{d} t$ The latter integral can be evaluate by $y=1-t$ and then by parts easily : $\int_0^1y^n \ln(1-y)\ \mathrm{d}y =\frac{-1}{n+1} \int_0^1 \frac{y^{n+1}-1}{y-1} \ \mathrm{d} y = \frac{-H_{n+1}}{n+1}$ The last equality follows from the geometric sequence sum, then $\Gamma'(1)= \frac{n}{n+1}\left(\ln n - H_{n+1} \right) = -\gamma.$

Integrating by parts with $\displaystyle x=\ln (-\ln (u)) , dv=du$

$\displaystyle => dx=\frac{1}{u\ln u} du , v=u$

\displaystyle => I= \int_0^1 \mathrm \ln (-\ln u) \mathrm{d}u = \left. u\ln (-\ln u) \right|_0^1 - \int_0^1 \mathrm{ \frac{1}{\ln u} }\mathrm{d}u

Solving the second integral, let $\displaystyle u=e^{-t}, du = -e^{-t} dt$

$\displaystyle => \int_0^1 \mathrm{ \frac{1}{\ln u} }\mathrm{d}u= \int_{\infty }^0 \mathrm{ \frac{e^{-t}}{t} }\mathrm{d}t= -\int_0^{\infty} \mathrm{ \frac{e^{-t}}{t} }\mathrm{d}t$

Note the definition of the exponential integral: $\displaystyle -\int_{-x}^{\infty} \mathrm{ \frac{e^{-t}}{t} }\mathrm{d}t= Ei(x)$

$\displaystyle <=>\int_0^1 \mathrm{ \frac{1}{\ln u} }\mathrm{d}u= -\int_0^{\infty} \mathrm{ \frac{e^{-t}}{t} }\mathrm{d}t= \lim_{x \to 0} Ei(x)$

Hence, $\displaystyle I = \left. u\ln (-\ln u) \right|_0^1 - \lim_{x \to 0} Ei(x)$

we have: $\displaystyle \left. u\ln (-\ln u) \right|_0^1= \lim_{u \to 1} u\ln (-\ln u)- \lim_{u \to 0} u\ln (-\ln u)$

by direct substitution, $\displaystyle \lim_{u \to 0} u\ln (-\ln u)=\lim_{u \to 0} \frac{\ln (-\ln u)}{\frac{1}{u}} = \frac{\infty}{\infty}$

so using L'Hopital rule : $\displaystyle \lim_{u \to 0} \frac{\ln (-\ln u)}{\frac{1}{u}}=\lim_{u \to 0} \frac{\frac{1}{u\ln u}}{-\frac{1}{u^2}}=\lim_{u \to 0} -\frac{u}{\ln u} =0$

now we are left with $\displaystyle I= \lim_{u \to 1} u\ln (-\ln u) - \lim_{x \to 0} Ei(x)$ . With direct substitution, $\displaystyle I=-\infty +\infty$ , an indetermined form.

But $\displaystyle I= \lim_{u \to 1} u\ln (-\ln u) - \lim_{x \to 0} Ei(x) = \lim_{u \to 1} \ln (-\ln u) - \lim_{x \to 0} Ei(x)$

$\displaystyle = \lim_{x \to 0} \ln (x) - \lim_{x \to 0} Ei(x)$ (substitution $\displaystyle x=-\ln u$ )

$\displaystyle = \lim_{x \to 0} (\ln (x) - Ei(x))$

Note that $\displaystyle Ei(x) = \gamma + \ln x + \sum_{k=1}^{\infty} \frac{x^k}{k k!}$

$\displaystyle => I= \lim_{x \to 0} (\ln (x) - Ei(x))=\lim_{x \to 0} ( - \gamma - \sum_{k=1}^{\infty} \frac{x^k}{k k!})$

The terms of the sum are all zero for $\displaystyle x=0$ , therefore:

$\displaystyle \boxed{I=-\gamma}$