calculus - Limit of gamma and digamma function

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&=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?