Integrate $\int_0^1 \ln{\left(\ln{\sqrt{1-x}}\right)} \mathop{dx}$ Problem says to integrate $$\int_0^1 \ln{\left(\ln{\sqrt{1-x}}\right)} \mathop{dx}$$
I try $u=1-x$ and got
$$\int_0^1 -\ln{2}+\ln{(\ln{u})} \mathop{du}$$
Then $t=\ln{u}$
$$-\ln{2}+\int_{-\infty}^0 e^t \ln{t}$$
Now what?
 A: HINT:
The Euler-Mascheroni constant can be expressed as
$$\gamma =-\int_0^\infty e^{-x}\log(x)\,dx$$
A: You are on the right track.  To evaluate $\int_{-\infty}^0 e^t \ln{t} \; dt$, I would do another substitution with $w=-t$:
$$\int_0^{\infty} e^{-w} \ln{(-w)} \; dw$$
$$=\int_0^{\infty} e^{-w} \ln{(-1)} \; dw + \int_0^{\infty} e^{-w} \ln{(w)} \; dw$$
Now, the left integral is easy and we will use only the principal value of $\ln{(-1)}=\pi i$:
$$\pi i \int_0^{\infty} e^{-w} \; dw = \pi i$$
To evaluate the right integral, you want to express $e^{-w}$ as its limit definition to eventually manipulate it into the Euler-Mascheroni constant:
$$\lim_{n \to \infty} \int_0^n {\left(1-\frac{w}{n}\right)}^{n-1} \ln{w} \; dw$$
Let $u=1-\frac{w}{n}$:
$$\lim_{n \to \infty} n\int_0^1u^{n-1} \ln{\left(n(1-u\right)} \; du$$
$$=\lim_{n \to \infty} n\ln{n}\int_0^1u^{n-1}\; du+n\int_0^1 u^{n-1} \ln{(1-u)} \; du$$
$$=\lim_{n \to \infty} \ln{n}-n\int_0^1 u^{n-1} \sum_{j=1}^{n} \frac{u^j}{j} \; du$$
And by the dominated convergence theorem we can interchange the summation and integral sign:
$$=\lim_{n \to \infty} \ln{n}-n\sum_{j=1}^{n}\int_0^1 \frac{u^{n+j-1}}{j} \; du$$
$$=\lim_{n \to \infty} \ln{n}-\sum_{j=1}^{n} \frac{1}{j}-\frac{1}{j+n}$$
$$=\lim_{n \to \infty} \ln{n}-\sum_{j=1}^{n} \frac{1}{j}$$
$$= -\gamma$$
Therefore, the original integral evaluates to $$\boxed{-\ln{2}-\gamma+ \pi i}$$
A: This is an interesting one. Note that if $x \in (0,1)$ then $1-x \in (0,1)$ and $\sqrt{1-x} \in (0,1)$, which makes $\ln\left(\sqrt{1-x}\right) \in (-\infty,0)$. What happens when you take another logarithm of that?
Hint: if you are working in the real numbers, this integral does not exist.
Are you looking for complex integration methods?
