Oloa Olivier's type integral π/4

[Tony1]

$$\int_{0}^{\pi\over 2}{\ln (2\cos x)\over x^2+\ln^2(2\cos x)}\mathrm dx={\pi\over 4}$$

I am very surprise it has a very simple closed form. I was expecting something else, probably involving a few other constants, like $\ln a$, $\pi$ ...

I don't know how to go about to show that it is $\pi\over 4$. I did try a substitution of $u=\ln(2\cos x)$, which make the integral looked more harder than it is...

$$\int_{0}^{\infty}{\mathrm du\over \tan x}\cdot{u\over x^2+u^2}$$

Can anyone help. Thank you!

 

[jvicens]

Hello there,

It is indeed surprising that the integral has a simple closed form, but it is true. Let me explain how to show that it is equal to $\pi\over 4$.

First, let's take a closer look at the integral:

$$\int_{0}^{\pi\over 2}{\ln (2\cos x)\over x^2+\ln^2(2\cos x)}\mathrm dx$$

We can see that the integrand involves both $\ln(2\cos x)$ and $x^2+\ln^2(2\cos x)$. This suggests that a substitution might be helpful. Let's try $u=\ln(2\cos x)$. This gives us $\mathrm du={\mathrm dx\over \tan x}$. Substituting this into the integral, we get:

$$\int_{0}^{\infty}{\mathrm du\over \tan x}\cdot{u\over x^2+u^2}$$

This is the same integral that you have in the forum post. Now, we can use the fact that $\tan x={x\over \sqrt{1+x^2}}$ to rewrite the integral as:

$$\int_{0}^{\infty}{\mathrm du\over \sqrt{1+x^2}}\cdot{u\over x^2+u^2}$$

Next, we can use the partial fraction decomposition to split the integrand into two parts:

$${\mathrm u\over x^2+u^2}={1\over 2}\cdot\left({1\over x+iu}+{1\over x-iu}\right)$$

Substituting this into the integral, we get:

$$\int_{0}^{\infty}{\mathrm du\over \sqrt{1+x^2}}\cdot{1\over 2}\cdot\left({1\over x+iu}+{1\over x-iu}\right)$$

Now, we can use the fact that $\int_0^{\infty}{1\over x+iy}\mathrm dx=\pi\over 2$ to evaluate the integral. This gives us:

$$\int_{0}^{\infty}{\mathrm du\over \sqrt{1+x^2}}\cdot{1\over