Evaluate $\int_{0}^{\infty}\frac{\log(2e^{x}-1)}{e^{x}-1}dx$?

Question

Evaluate $$\int_{0}^{\infty}\frac{\log(2e^{x}-1)}{e^{x}-1}dx$$ ?

This question is from MIT BEE Finals 2024

What I tried so far :

I wrote the integral as $$I=\int_{0}^{\infty}\frac{e^{x}\log(2e^{x}-1)}{e^{x}(e^{x}-1)}dx$$

That is I multiplied both the numerator and denominator by $e^{x}$

Now I substituted $e^{x}=t$ which implies $e^{x}dx=dt$

$\implies I=\int_{1}^{\infty}\frac{\log(2t-1)}{t(t-1)}dt$

Now I wrote the numerator as $[t-(t-1)]$

$\implies I=\int_{1}^{\infty}\frac{\log(2t-1)}{(t-1)}dt-\int_{1}^{\infty}\frac{\log(2t-1)}{t}dt$

Now I can't understand how to integrate these $2$ integrals.

$1^{st}$ :$\int_{1}^{\infty}\frac{\log(2t-1)}{(t-1)}dt$

$2^{nd}$ :$\int_{1}^{\infty}\frac{\log(2t-1)}{t}dt$

Please help me out with this integral.

Best Wishes

Mathematics

Answer 1

Substitute $2e^x-1=u^{-1}$ and use geometric series $$I=\int_0^\infty\frac{\ln(2e^x-1)}{e^x-1}dx=-2\int_0^1\frac{\ln u}{1-u^2}dx$$ $$=-2\sum_{n=0}^\infty\int_0^1u^{2n}\ln (u)du=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=2\left(\frac{\pi^2}{8}\right)=\frac{\pi^2}{4}$$

Answer 2

You can split the integrals because both diverge.

Starting from $I=\int_{1}^{\infty}\frac{\log(2t-1)}{t(t-1)}dt$ replace $t=u^{-1}$ to achieve $$ I { =\int_{0}^{1}\frac{\log(2/u-1)}{1-u}du \\=\int_{0}^{1}\frac{\log(2-u)}{1-u}-\frac{\log(u)}{1-u}du \\=\int_{0}^{1}\frac{\log(1+u)}{u}-\frac{\log(u)}{1-u}du \\=\text{Li}_2(1)-\text{Li}_2(-1) \\=\frac{\pi^2}{4}. } $$