What kind of substitution should I use to obtain the following integrals?
$$\begin{align}
\int_0^1 \ln \ln \left(\frac{1}{x}\right)\,dx
&=\int_0^\infty e^{-x} \ln x\,dx\tag1\\
&=\int_0^\infty \left(\frac{1}{xe^x} – \frac{1}{e^x-1} \right)\,dx\tag2\\
&=-\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx\tag3\\
&=\int_0^\infty \left( e^{-x} – \frac{1}{1+x^k} \right)\,\frac{dx}{x},\qquad k>0\tag4\\
\end{align}$$
This is not homework problems and I know that the above integrals equal to negative of the Euler–Mascheroni constant. I got these integrals while reading this Wikipedia page: The Euler–Mascheroni constant. According to Wikipedia, the Euler–Mascheroni constant is defined as the limiting difference between the harmonic series and the natural logarithm:
$$\gamma=\lim_{N\to\infty} \left(\sum_{k=1}^N \frac{1}{k} – \ln N\right)$$
but I don't know why can this definition be associated to the above integrals?
I can obtain the equation $(1)$ using substitution $t=\ln \left(\frac{1}{x}\right)\rightarrow x=e^{-t} \rightarrow dx=-e^{-t}\,dt$ and I know that
$$\int_0^\infty e^{-x} \ln x\,dx=\Gamma'(1)=\Gamma(1)\psi(1)=-\gamma$$
but I can't obtain the rest. Any idea? Any help would be appreciated. Thanks in advance.
Best Answer
I am not sure if you can obtain the other three from the first integral through some substitution.
The $(3)$ one can be obtained from the definition of constant..
Notice that $$\sum_{i=1}^N\frac{1}{i}=\int_0^1 \frac{1-t^N}{1-t}\,dt$$ and $$\ln N=\int_0^1 \frac{t^{N-1}-1}{\ln t}\,dt$$ The above can be proved using Frullani's integral. Therefore $$\begin{aligned} \lim_{N\rightarrow \infty} \left(\sum_{i=1}^N\frac{1}{i}-\ln N\right) &=\lim_{N\rightarrow \infty}\int_0^1\left(\frac{1-t^N}{1-t}-\frac{t^{N-1}-1}{\ln t}\,dt\right)\,dt \\ &=\int_0^1 \left(\frac{1}{1-t}+\frac{1}{\ln t}\right)\,dt \end{aligned}$$ Make the substitution $\ln x=-t$ to obtain the $(2)$.