The sum of $\sum_{n=1}^{\infty}\frac 1{2n^2-1}$ using the beta function I am trying to find the sum of the series
$$
\sum_{n=1}^{\infty}\frac 1{2n^2-1}
$$
using definite integrals. After several substitutions I get
$$
\sum_{n=1}^{\infty}\frac 1{2n^2-1}=\dots=\frac{\sqrt 2}{4}\int_0^1\frac{(1-t)^{-\frac{\sqrt 2}{2}}-(1-t)^{\frac{\sqrt 2}{2}}}{t}\,\mathrm dt
$$
Now I am in stuck. It's like the combination of two beta functions $B(0,1-\sqrt 2/2)$ and $B(0,1+\sqrt 2/2)$, however they are not defined at $0$ in the first input. Is there any way of handling with the integral? Thank you in advance.
Derivation of the integral:
\begin{align*}
\sum_{n=1}^{\infty}\frac 1{2n^2-1}
&=\frac 12\sum_{n=1}^{\infty}\left(\frac 1{\sqrt 2n-1}-\frac 1{\sqrt 2n+1}\right)
=\frac 12\sum_{n=1}^{\infty}\int_0^1\left(x^{\sqrt 2n-2}-x^{\sqrt 2n}\right)\mathrm dx\\[12pt]
&=\frac 12\int_0^1\sum_{n=1}^{\infty}\left(x^{\sqrt 2n-2}-x^{\sqrt 2n}\right)\mathrm dx
=\frac 12\int_0^1\frac{x^{\sqrt 2-2}-x^{\sqrt 2}}{1-x^{\sqrt 2}}\,\mathrm dx\\[12pt]
&=\left[\text{subst. $x=y^{\sqrt 2}$}\right]\ \frac {\sqrt 2}2\int_0^1\frac{y^{-\sqrt 2}-y^{\sqrt 2}}{1-y^2}\cdot y\,\mathrm dy\\[12pt]
&=\left[\text{subst. $t=1-y^2$}\right]\ \frac {\sqrt 2}4\int_0^1\frac{(1-t)^{-\frac{\sqrt 2}{2}}-(1-t)^{\frac{\sqrt 2}{2}}}{t}\,\mathrm dt
\end{align*}
 A: let's consider $I(\alpha)=\int_0^1\frac{(1-t)^{-\alpha}-(1-t)^{\alpha}}{t}dt$, where $\alpha\in(0;1)$.
We can apply the regularisation and consider $I(\alpha)=\lim_{\epsilon\to 0}I_\epsilon(\alpha)=\lim_{\epsilon\to 0}\int_0^1\frac{(1-t)^{-\alpha}-(1-t)^{\alpha}}{t^{1-\epsilon}}dt$.
Then
$$I_\epsilon(\alpha)=B(\epsilon;1-\alpha)-B(\epsilon;1+\alpha)=\Gamma(\epsilon)\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\alpha+\epsilon)}-\frac{\Gamma(1+\alpha)}{\Gamma(1+\alpha+\epsilon)}\Big)$$
Given that at $\epsilon\to0\,\,\,\epsilon\,\Gamma(\epsilon)=\Gamma(1+\epsilon)=1-\gamma\epsilon+O(\epsilon^2)\,\,\Rightarrow\,\,\Gamma(\epsilon)=\frac{1}{\epsilon}-\gamma+O(\epsilon)$,
and $\Gamma(1\pm\alpha+\epsilon)=\Gamma(1\pm\alpha)+\Gamma'(1\pm\alpha)\epsilon+O(\epsilon^2)=\Gamma(1\pm\alpha)\Big(1+\psi(1\pm\alpha)\epsilon+O(\epsilon^2)\Big)$
(where $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$ - digamma function)
$$I_\epsilon(\alpha)=\Big(\frac{1}{\epsilon}+O(1)\Big)\Big(\frac{1}{1+\psi(1-\alpha)\epsilon+O(\epsilon^2)}-\frac{1}{1+\psi(1+\alpha)\epsilon+O(\epsilon^2)}\Big)$$
$$=\frac{1}{\epsilon}\big(\psi(1+\alpha)\epsilon-\psi(1-\alpha)\epsilon\big)+O(\epsilon)$$
$$I(\alpha)=\lim_{\epsilon\to 0}I_\epsilon(\alpha)=\psi(1+\alpha)-\psi(1-\alpha)$$
Given that $\psi(1+\alpha)=\psi(\alpha)+\frac{1}{\alpha}$ and $\psi(1-\alpha)-\psi(\alpha)=\pi\cot\pi\alpha$ (digamma-function)
$$I(\alpha)=\frac{1}{\alpha}-\pi\cot\pi\alpha$$
