Integrate $\int_0^1 \ln{\left(\ln{\sqrt{1-x}}\right)} \mathop{dx}$

Question

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?

Answer 1

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 $$\boxed{-\ln{2}-\gamma+ \pi i}$$

Answer 2

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?

Answer 3

HINT:

The Euler-Mascheroni constant can be expressed as

$$\gamma =-\int_0^\infty e^{-x}\log(x)\,dx$$