A novelty integral for $\pi$

Question

My lab friends always play a mentally challenging brain game every month to keep our mind running on all four cylinders and the last month challenge was to find a novelty expression for $\pi$. In order to stick to the rule, of course we must avoid the good old Ramanujan and online available expressions, for instance: the coolest ways of expressing $\pi$ on Quora. The winner of the last month challenge is this integral

$${\large\int_0^\infty}\frac{(1+x)\log(1+x)(2+\log x)\log\left(\!\frac{1+x}{2}\!\right)-2x\log(1+x)\log x}{x^{3/2}(1+x)\log^2x}\ dx={\Large\pi}$$

The equality is precise to at least thousand decimal places. Unfortunately, my friend who proposes this integral keeping the mystery to himself. I tried to crack this integral while waiting for a solution to be offered by one of my friends, but failed to get any.

I have tried to break this integral into two part:

$${\large\int_0^\infty}\frac{\log(1+x)(2+\log x)\log\left(\!\frac{1+x}{2}\!\right)}{x^{3/2}\log^2x}\ dx-2{\large\int_0^\infty}\frac{\log(1+x)}{\sqrt{x}(1+x)\log x}\ dx$$

but each integrals diverges. I have tried many substitutions like $x=y-1$, $x=\frac{1}{y}$, or $x=\tan^2y$ hoping for familiar functions, but couldn't get one. I also tried the method of differentiation under integral sign by introducing

$$I(s)={\large\int_0^\infty}x^{s}\cdot\frac{(1+x)\log(1+x)(2+\log x)\log\left(\!\frac{1+x}{2}\!\right)-2x\log(1+x)\log x}{(1+x)\log^2x}\ dx$$

and differentiating twice with respect to $s$ to get rid of $\log^2x$ couldn't work either. I have a strong feeling that I miss something completely obvious in my calculation. I'm not having much success in evaluating this integral since two weeks ago, so I thought it's about time to ask you for help. Can you help me out to prove it, please?

Answer 1

The integrand can be broken up as $$I=\int_0^{\infty} \left(\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x} +\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln x}-\frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x}\right)dx.$$

But, by integration by parts, $$\int \frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x} dx= \int \frac{2\ln(1+x)}{x^{1/2} \ln x} d\left(\ln\left(\frac{1+x}{2}\right)\right) \\=\small\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{1/2} \ln x}-2\int \left( \frac{\ln\left(\frac{1+x}{2}\right)}{x^{1/2} (1+x) \ln x}-\frac{ \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x}-\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{2 x^{3/2} \ln x}\right)dx$$

That is, $$\int \left(\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln^2 x} +\frac{\ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{3/2} \ln x}-\frac{2\ln(1+x)}{x^{1/2} (1+x) \ln x}\right)dx \\= -\frac{2 \ln(1+x)\ln\left(\frac{1+x}{2}\right)}{x^{1/2} \ln x}+2 \int \frac{\ln\left(\frac{1+x}{2}\right)}{x^{1/2} (1+x) \ln x} dx$$

And the claim follows from Jack D'Aurizio's preliminary result.

Answer 2

A preliminary result.

$$I_1=\int_{0}^{+\infty}\frac{2\log\left(\frac{1+x}{2}\right)}{\sqrt{x}(1+x)\log(x)}=\color{red}{\pi}.\tag{1}$$

Proof: through the substitution $x=e^t$, $$\begin{eqnarray*}I_1=\int_{-\infty}^{+\infty}\frac{\log\left(\frac{e^t+1}{2}\right)}{\cosh\left(\frac{t}{2}\right)t}\,dt&=&\color{red}{\frac{1}{2}\int_{-\infty}^{+\infty}\frac{dt}{\cosh\left(\frac{t}{2}\right)}}+\color{blue}{\int_{-\infty}^{+\infty}\frac{\log\cosh\left(\frac{t}{2}\right)}{t\cosh\left(\frac{t}{2}\right)}\,dt}\end{eqnarray*}$$ where the red integral is easy to compute and the blue one vanishes by symmetry.

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Thanks to nospoon, this is ultimately everything we need to prove the OP's identity through integration by parts.