Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$ 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.

There have been similar integrals  such as this with a factor of $1/\log{t}$ in the integrand.  The way I have attacked these is to use the substitution $t=e^{-x}$; here, this produces
$$-\int_0^{\infty} dx \: \frac{e^{-x}}{x} \frac{1-e^{-2 x}}{1+e^{-2 x}} $$
$$ = -\int_0^{\infty} dx \: \frac{1}{x} (e^{-x} - e^{-3 x}) \sum_{k=0}^{\infty} (-1)^k e^{-2 k x} $$
Now, we reverse the order of the sum and integral.  This is justified by Abel's theorem, although I will leave the details for another time or another person to fill in:
$$ = -\sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} dx \: \frac{1}{x} (e^{-(2 k+1) x} - e^{-(2 k+3) x}) $$
The integrals have a simple closed form, and the result is now a sum:
$$ = \sum_{k=0}^{\infty} (-1)^{k+1} \log{ \left [ \frac{(2 k+1)}{(2 k+3)} \right ] } $$
You can form a log of a Wallis-type product from this sum:
$$ = \log{\left [ \prod_{k=0}^{\infty} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} \right ]} $$
Partial products of the above product may be evaluated:
$$ \prod_{k=0}^{n} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} = \frac{9 \Gamma{\left ( \frac{5}{4} \right )} \Gamma{\left ( \frac{9}{4} \right )} \Gamma{\left ( n + \frac{7}{4} \right )}^2}{5 \Gamma{\left ( \frac{7}{4} \right )}^2 \Gamma{\left (n+ \frac{5}{4} \right )} \Gamma{\left ( n + \frac{9}{4} \right )}}$$
You may show that the above expression converges as $n \rightarrow \infty$.  A little manipulation produces the stated result.
EDIT
@MikeSpivey points my attention to an equation out of Whittaker & Watson that applies here:
$$ \prod_{k=0}^{\infty} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} = \prod_{k=0}^{\infty} \frac{(k+\frac{3}{4})^2}{(k+\frac{1}{4})(k+\frac{5}{4})} = \frac{\Gamma{\left ( \frac{1}{4} \right )} \Gamma{\left ( \frac{5}{4} \right )}}{\Gamma{\left ( \frac{3}{4} \right )^2}} = \frac{2^2 \Gamma{\left ( \frac{5}{4} \right )^2}}{\Gamma{\left ( \frac{3}{4} \right )^2}} $$
The result immediately follows.
