Hard! Integrate $\int_0^{\frac{\pi}{2}}\frac{x\ln|\cot (x-\frac{\pi}{4})|}{\sin^2x}{d}x=\frac{\pi^{2}}{4}+\frac{\pi}{2}\ln2$

I am struggling with this integral, and I have tried to draw several contours but none did work. So proving: $$\int_0^{\frac{\pi}{2}}\frac{x}{\sin^2x}\ln\left|\cot\left(x-\frac{\pi}{4}\right)\right|\mathrm{d}x=\frac{\pi^{2}}{4}+\frac{\pi}{2}\ln2$$ So what I did is that I substitute $ t = \cot x$, which leads to the following integral: $$ \int_{0}^{\infty} \frac{\arctan{t}}{t^2} \cdot \ln\left|\frac{1+t}{1-t}\right| \mathrm{d}t$$ However the absolute value of the logarithm bothers me a lot. I did try to split it out and obtain an equivalent integral: $$ \int_{-\infty}^{\infty} \frac{\arctan{t}}{t^2} \cdot \ln\left| 1+t\right|\mathrm{d}t$$ It seems to me intuitively that one can construct a contour for this function but none of what I created works. The answer for this integral is considerably simple for me since there are various complicated integrals.

Solutions with the application of real analysis are also welcomed. Thanks in advance!

Solution 1:

You can indeed evaluate the integral $$\int_{0}^{\infty} \frac{\arctan x}{x^{2}} \, \log \left|\frac{1+x}{1-x} \right| \, \mathrm dx $$ using contour integration.

In case anyone is confused, you actually need two substitutions to get the integral in this form. After making the substitution $t = \cot x$, use the identity $\operatorname{arccot}(t) = \arctan \left(\frac{1}{t} \right)$, followed by the subtution $u = \frac{1}{t}$.

Let $R>1$, and consider the function $$f(z) = \frac{\log(1-iz)}{z^{2}} \, \log \left(\frac{1+z}{1-z} \right). $$

If we use the principal branch of the logarithm, then $\log \left(\frac{1+z}{1-z} \right) $ has a branch cut on $(-\infty, -1]$ and on $[1, \infty)$.

(Technically it's a single branch cut that connects the 2 branch points through the point at complex infinity.)

The function $\log(1-iz)$ also has a branch cut, but it's on the negative imaginary axis.

So let's integrate $f(z)$ around a contour that consists of the line segment just above the interval $[- R, -1]$, the interval $(-1,1)$, the line segment just above the interval $[1,R]$, and the upper half of the circle $|z|=R$.

Basically it looks just like the typical closed semicircular contour in the upper half-plane expect that we can't integrate on the branch cut itself. We need to integrate just above it. There should also be small quarter-circle indentations around the branch points at $z=-1$ and $z=1$, but their contributions vanish in the limit.

Now since $\log\left(\frac{1+z}{1-z} \right) \sim i \pi + \mathcal{O} \left(\frac{1}{z} \right)$ as $|z| \to \infty$, the integral along the semicircle vanishes as $R \to \infty$.

So letting $R \to \infty$, we get$$\int_{-\infty}^{-1} \frac{\log(1-ix)}{z^{2}} \left(\log \left|\frac{1+x}{1-x} \right| + i \pi \right)\, \mathrm dx+ \int_{-1}^{1} \frac{\log(1-ix)}{x^{2}} \, \log \left(\frac{1+x}{1-x} \right)\, \mathrm dx $$ $$+\int_{1}^{\infty} \frac{\log(1-ix)}{x^{2}} \, \left(\log \left|\frac{1+x}{1-x} \right| + i \pi \right) \, \mathrm dx=0. $$

And equating the imaginary parts on both sides of the equation and then rearranging, we get $$ \begin{align} \int_{-\infty}^{\infty} \frac{\arctan x}{x^{2}} \log \left|\frac{1+x}{1-x} \right| \, \mathrm dx &= \frac{\pi}{2} \int_{-\infty}^{-1} \frac{\log(1+x^{2})}{x^{2}} \, \mathrm dx + \frac{\pi}{2} \int_{1}^{\infty} \frac{\log(1+x^{2})}{x^{2}} \, \mathrm dx \\ &= \pi \int_{1}^{\infty} \frac{\log(1+x^{2})}{x^{2}} \, \mathrm dx \\ &= \pi \left( -\frac{\log(1+x^{2})}{x} \Bigg|^{\infty}_{1} + \int_{1}^{\infty} \frac{2}{1+x^{2}} \, \mathrm dx\right) \\ &= \pi \log(2) + \frac{\pi^{2}}{2}. \end{align}$$

The result then follows since the integrand is even.

EDIT: I added a hand-drawn picture of the contour.

Solution 2:

Here is to integrate with real method. Substitute $t= \tan(\frac\pi4-x)$

\begin{align} I=& \int_0^{\frac{\pi}{2}}\frac{x}{\sin^2x}\ln|\cot(x-\frac{\pi}{4})|{d}x=-2\int_{-1}^1 \frac{\frac\pi4 - \tan^{-1} t}{(1-t)^2}\ln|t|dt\\ =& -2\int_{0}^1 \frac{(\frac\pi4 - \tan^{-1} t)\ln t}{(1-t)^2} +\frac{(\frac\pi4 + \tan^{-1} t)\ln t}{(1+t)^2} \> dt\\ = &-\pi\int_0^1 \frac{\ln t}{(1+t)^2}dt- 8\int_0^1 \frac{(\frac\pi4 - \tan^{-1} t) t \ln t}{(1-t^2)^2}dt\tag1 \end{align} where $ \int_0^1 \frac{\ln t}{(1+t)^2}dt=-\ln2$ and \begin{align} & \int_0^1 \frac{(\frac\pi4 - \tan^{-1} t) t \ln t}{(1-t^2)^2}dt\\ =& \int_0^1 (\frac\pi4 - \tan^{-1} t)d\left( \frac14\ln(1-t^2) +\frac12 \frac{t^2\ln t}{1-t^2}\right)\\ =& \frac14 \int_0^1 \frac{\ln\frac{1-t^2}t}{1+t^2}dt+\frac14 \int_0^1 \frac{\ln t}{1-t^2}dt \end{align} Note $\int_0^1 \frac{\ln t}{1-t^2}dt =-\frac{\pi^2}8$ and $$J=\int_0^1 \frac{\ln\frac{1-t^2}t}{1+t^2}dt \overset{t\to\frac{1-t}{1+t}}= \int_0^1 \frac{2\ln2}{1+t^2}dt -J= \ln2 \int_0^1 \frac{1}{1+t^2}dt =\frac\pi4\ln2 $$ Substitute above results into (1) to obtain $$I= \frac{\pi^2}4+\frac\pi2\ln2$$