TITLE: Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$
QUESTION [33 upvotes]: I am trying to prove that
$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$
I know how to deal with integrals involving cyclotomic polynomials and nested logarithms but I have no idea with this one.

REPLY [15 votes]: $$
\begin{align}
\int_0^1\frac{t+1}{t^2+1}\frac{t-1}{\log(t)}\,\mathrm{d}t
&=\int_0^1\frac{t+1}{t^2+1}\int_0^1t^x\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\int_0^1\frac{t^{x+1}+t^x}{t^2+1}\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^1\left(\frac1{x+1}+\frac1{x+2}-\frac1{x+3}-\frac1{x+4}+\dots\right)\,\mathrm{d}x\\
&=\left(\log\left(\frac21\right)+\log\left(\frac32\right)\right)
-\left(\log\left(\frac43\right)+\log\left(\frac54\right)\right)+\dots\\
&=\log\left(\frac31\right)-\log\left(\frac53\right)+\log\left(\frac75\right)-\log\left(\frac97\right)+\log\left(\frac{11}9\right)-\dots\\
&=\log\left(\frac31\cdot\frac35\cdot\frac75\cdot\frac79\cdot\frac{11}9\cdots\right)\\
&=\lim_{n\to\infty}\log\left(\frac{\Gamma\left(\frac54\right)^2}{\Gamma\left(\frac34\right)^2}
\frac{\Gamma\left(\frac{4n+3}4\right)^2}{\Gamma\left(\frac{4n+5}4\right)^2}(4n+3)\right)\\
&=2\log\left(2\frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac34\right)}\right)
\end{align}
$$
The last equality is due to Gautschi's inequality.