Limit of gamma and digamma function

Question

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?

Answer 1

We have that $$\log\left(\Gamma\left(x\right)\right)\sim x\log\left(x\right)-x-\frac{1}{2}\log\left(\frac{x}{2\pi}\right)+O\left(\frac{1}{x}\right) $$ and $$\psi\left(x\right)\sim\log\left(x\right)+O\left(\frac{1}{x}\right) $$ as $x\rightarrow\infty$ (see here and here) hence we have to evaluate $$\begin{align} &\frac{a+2}{4}\log\left(\frac{a+2}{4}\right)-\frac{a+2}{4}-\frac{1}{2}\log\left(\frac{a+2}{8\pi}\right)-\frac{a}{4}\log\left(\frac{a}{4}\right)+\frac{a}{4}+\frac{1}{2}\log\left(\frac{a}{8\pi}\right)\\ & -\frac{1}{4}\log\left(\frac{a+1}{4}\right)-\frac{1}{4}\log\left(\frac{a+2}{4}\right)+O\left(\frac{1}{a}\right)\\ &=\frac{1}{2}\log\left(\frac{a}{a+2}\right)+O\left(\frac{1}{a}\right)\rightarrow0. \end{align} $$

Answer 2

$\log \Gamma$ 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

$$ \psi(a)-\psi(b) = (a-b)\sum_{n\geq 0}\frac{1}{(n+a)(n+b)} \tag{2}$$

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:

$$\int_{t}^{t+\frac{1}{2}}\psi(x)\,dx -\frac{\psi\left(t+\frac{1}{4}\right)+\psi\left(t+\frac{1}{2}\right)}{4}=\color{red}{-\frac{1}{192t^2}+O\left(\frac{1}{t^3}\right)}.\tag{3}$$