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.
 A: Let's introduce the parameter $\alpha$, and then differentiate with respect to $\alpha$ that yields
$$I(\alpha)=\int_0^1 \frac{t^\alpha-1}{(t^2+1)\ln t}dt  $$
$$I'(\alpha)=\int_0^1 \frac{t^{\alpha}}{(t^2+1)}dt=\frac{1}{4} \left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right)  $$
Then
$$I(\alpha)=\frac{1}{4} \int\left(-\psi_0\left(\frac{1 + \alpha}{4}\right) + \psi_0\left(\frac{3 + \alpha}{4}\right)\right) d\alpha= $$
$$I(\alpha)=\left(\ln \Gamma \left(\frac{3 + \alpha}{4}\right)- \ln \Gamma \left(\frac{1 + \alpha}{4}\right)\right)+C\tag1$$
If letting $\alpha=2$, then 
$$I(2)=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)+C$$
On the other hand, by letting $\alpha=0$ in $(1)$ we get
$$C=\ln \left(\frac{\Gamma \left(\frac{1}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}\right)$$
Thus
$$\int_0^1 \frac{t^2-1}{(t^2+1)\ln t}dt=\ln \left(\frac{\Gamma \left(\frac{5}{4}\right)\Gamma \left(\frac{1}{4}\right)}{\Gamma^2 \left(\frac{3}{4}\right)}\right)  $$
A: $$
\begin{align}
\int_0^1\frac{t+1}{t^2+1}\frac{t-1}{\log(t)}\,\mathrm{d}t
&=\int_0^1\frac{t+1}{t^2+1}\int_0^1t^x\,\mathrm{d}x\,\mathrm{d}t\\
&=\int_0^1\int_0^1\frac{t^{x+1}+t^x}{t^2+1}\,\mathrm{d}t\,\mathrm{d}x\\
&=\int_0^1\left(\frac1{x+1}+\frac1{x+2}-\frac1{x+3}-\frac1{x+4}+\dots\right)\,\mathrm{d}x\\
&=\left(\log\left(\frac21\right)+\log\left(\frac32\right)\right)
-\left(\log\left(\frac43\right)+\log\left(\frac54\right)\right)+\dots\\
&=\log\left(\frac31\right)-\log\left(\frac53\right)+\log\left(\frac75\right)-\log\left(\frac97\right)+\log\left(\frac{11}9\right)-\dots\\
&=\log\left(\frac31\cdot\frac35\cdot\frac75\cdot\frac79\cdot\frac{11}9\cdots\right)\\
&=\lim_{n\to\infty}\log\left(\frac{\Gamma\left(\frac54\right)^2}{\Gamma\left(\frac34\right)^2}
\frac{\Gamma\left(\frac{4n+3}4\right)^2}{\Gamma\left(\frac{4n+5}4\right)^2}(4n+3)\right)\\
&=2\log\left(2\frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac34\right)}\right)
\end{align}
$$
The last equality is due to Gautschi's inequality.
A: 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.
