Two improper log integrals

Question

Evaluate $$\int_0^{\frac{\pi}{2}}\ln ^2(\tan x)\text{d}x$$ $$\int_0^{\frac{\pi}{2}}\ln ^2(\sin x)\text{d}x$$

Answer 1

Per @julien's comment: Let $u=\tan{x}$ and transform the integral into

$$\int_0^{\infty} du \frac{\log^2{u}}{1+u^2} $$

This can be evaluated via the residue theorem. Consider the integral

$$\oint_C dz \frac{\log^3{z}}{1+z^2}$$

where $C$ is a keyhole contour about the positive real axis. The integral about the contour arcs that go to infinity and zero, respectively, go to zero in those limits. We are left with

$$\begin{align}\oint_C dz \frac{\log^3{z}}{1+z^2} &= \int_0^{\infty} du \frac{\log^3{u}}{1+u^2} - \int_0^{\infty} du \frac{(\log{u}+ i 2 \pi)^3}{1+u^2}\\ &= -i 6 \pi \int_0^{\infty} du \frac{\log^2{u}}{1+u^2} + 12 \pi^2 \int_0^{\infty} du \frac{\log{u}}{1+u^2} + i 8 \pi^3 \int_0^{\infty} du \frac{1}{1+u^2} \end{align}$$

The value of this integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrands, which are $e^{i \pi/2}$ and $e^{i 3\pi/2}$. The residues at these poles are

$$\mathrm{Res}_{z=e^{i \pi/2}} = \frac{-i \pi^3/8}{2 i}$$ $$\mathrm{Res}_{z=e^{i 3\pi/2}} = \frac{-i 27\pi^3/8}{-2 i}$$

We may then write

$$i \left [- 6 \pi \int_0^{\infty} du \frac{\log^2{u}}{1+u^2} + 8 \pi^3 \int_0^{\infty} du \frac{1}{1+u^2} \right ] + 12 \pi^2 \int_0^{\infty} du \frac{\log{u}}{1+u^2} = i \frac{13 \pi^4}{4}$$

Now use the fact that

$$\int_0^{\infty} du \frac{1}{1+u^2} = \frac{\pi}{2}$$

and equate real and imaginary parts to get

$$\int_0^{\infty} du \frac{\log^2{u}}{1+u^2} = \frac{\pi^3}{8}$$ $$\int_0^{\infty} du \frac{\log{u}}{1+u^2} = 0$$

Therefore, the value of the stated integral is

$$\int_0^{\pi/2} dx \: \log^2{(\tan{x})} = \frac{\pi^3}{8}$$

Answer 2

For the first integral, making the change of variable $ \tan(x)=y $ yields

$$ \int_0^{\frac{\pi}{2}}\ln ^2(\tan x)\text{d}x = \int_0^{\frac{\pi}{2}}\frac{\ln ^2(x)}{1+x^2}\text{d}x = I. $$

to evaluate the last integral, we consider the integral

$$ \int_0^{\frac{\pi}{2}}\frac{x^s}{1+x^2}\text{d}x = \frac{\pi}{2} \,\sec \left( \frac{\pi\,s}{2} \right) = F(s). $$

which can be evaluated using $\beta$ function. Here is a related problem. Now, $I$ follows from $F$ as

$$ I = \lim_{s\to 0} \frac{d^2F(s)}{ds^2} = \frac{\pi^3}{8}. $$

For the second integral, you can use the same approach. Here is a related problem.

Answer 3

These integrals can be evaluated by using the following identities; $$\forall x\in \left(0, \frac{\pi}{2}\right);\phantom{;}\log(\sin x)=-\log 2-\sum_{n\ge 1}\frac{\cos(2nx)}{n} \quad\cdots (1)$$ $$\forall x\in \left(0, \frac{\pi}{2}\right);\phantom{;}\log(\cos x)=-\log 2-\sum_{n\ge 1}\frac{(-1)^n \cos(2nx)}{n} \quad\cdots (2)$$ For example, the second integral can be written like the following; $$\begin{aligned}A&:=\int_{0}^{\pi/2}\log^2 (\sin x)dx \\&=\int_{0}^{\pi/2}\left(-\log 2-\sum_{n\ge 1}\frac{\cos(2nx)}{n}\right)^2 dx\phantom{;}(\because (1)) \\&=\frac{\pi}{2}\log^2 2+\sum_{m, n\ge 1}\left(\frac{1}{mn}\int_{0}^{\pi/2}\cos(2nx)\cos(2mx)dx\right)+2\log 2 \sum_{n\ge 1}\left(\frac{1}{n}\int_{0}^{\pi/2}\cos(2nx) dx\right) \end{aligned}$$ (Note that I interchanged integral and summation in the third line.)
For positive integer $m$ and $n$, $$\int_{0}^{\pi/2}\cos(2nx)\cos(2mx)dx=\begin{cases}\frac{\pi}{4}\mbox{ (when }m=n\mbox{)}\\0\mbox{ (otherwise)} \end{cases}$$ $$\int_{0}^{\pi/2}\cos(2nx)dx=0$$ So $$A=\frac{\pi}{2}\log^2 2+\frac{\pi}{4}\sum_{n\ge 1}\frac{1}{n^2}=\color{blue}{\frac{\pi}{2}\log^2 2+\frac{\pi^3}{24}}$$
The first integral can be evaluated similarly; observing that $$\int_{0}^{\pi/2}\log^2 (\tan x) dx =\int_{0}^{\pi/2}(\log^2 (\sin x)+\log^2 (\cos x)-2\log(\sin x)\log(\cos x)) dx$$