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.

Answer

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

Attribution
Source : Link , Question Author : Shobhit Bhatnagar , Answer Author : user 1591719

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