In my answer of the previous OP, I'm able to prove that
\begin{align}
I(a)&=\int_0^\infty e^{-(a-2)x}\cdot\frac{1-e^{-x}(1+x)}{x(1-e^{x})(e^{x}+e^{-x})}dx\tag1\\[10pt]
&=\int_0^1\frac{y^{a-1}}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy\tag2\\[10pt]
&=\log\Gamma\!\left(\frac{a+2}{4}\right)-\log\Gamma\!\left(\frac{a}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+1}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+2}{4}\right)\tag3
\end{align}
From the integral representations of $I(a)$ in $(1)$ and $(2)$, it's easy to show that
\begin{align}
\lim_{a\to\infty}I(a)&=\int_0^\infty \lim_{a\to\infty}e^{-(a-2)x}\cdot\frac{1-e^{-x}(1+x)}{x(1-e^{x})(e^{x}+e^{-x})}dx\\[10pt]
&=\int_0^1\lim_{a\to\infty}\frac{y^{a-1}}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy\quad,\quad 0<y<1\\[10pt]
&=0
\end{align}
But I'm having trouble proving
\begin{equation}
\lim_{a\to\infty}\left[\log\Gamma\!\left(\frac{a+2}{4}\right)-\log\Gamma\!\left(\frac{a}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+1}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+2}{4}\right)\right]=0
\end{equation}
Indeed the above result is confirmed by Wolfram Alpha. How does one prove the above limit?
Best Answer
$\log \Gamma$ is a convex function, hence $\psi$ is an increasing function and
$$\begin{eqnarray*}&&\int_{t}^{t+\frac{1}{2}}\psi(x)\,dx -\frac{\psi\left(t+\frac{1}{4}\right)+\psi\left(t+\frac{1}{2}\right)}{4}\\&=&\int_{t}^{t+\frac{1}{2}}\left(\psi(x)-\psi(t)\right)\,dx -\frac{\left(\psi\left(t+\frac{1}{4}\right)-\psi(t)\right)+\left(\psi\left(t+\frac{1}{2}\right)-\psi(t)\right)}{4}\tag{1}\end{eqnarray*}$$ is straightforward to handle through
that gives: $$ \psi(t+\theta)-\psi(t) = \theta\sum_{n\geq 0}\frac{1}{(n+t)(n+t+\theta)}\leq \frac{C\theta}{t} $$ proving that $(1)$ is $O\left(\frac{1}{t}\right)$. By using a sharpened version of the previous inequality, it is not difficult to check that the asymptotic expansion of $(1)$ is given by: