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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?

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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 $$\begin{align} &\frac{a+2}{4}\log\left(\frac{a+2}{4}\right)-\frac{a+2}{4}-\frac{1}{2}\log\left(\frac{a+2}{8\pi}\right)-\frac{a}{4}\log\left(\frac{a}{4}\right)+\frac{a}{4}+\frac{1}{2}\log\left(\frac{a}{8\pi}\right)\\ & -\frac{1}{4}\log\left(\frac{a+1}{4}\right)-\frac{1}{4}\log\left(\frac{a+2}{4}\right)+O\left(\frac{1}{a}\right)\\ &=\frac{1}{2}\log\left(\frac{a}{a+2}\right)+O\left(\frac{1}{a}\right)\rightarrow0. \end{align} $$

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  • $\begingroup$ Can you elaborate the series of $\log\Gamma(x)$ and $\psi(x)$? $\endgroup$ – Sophie Agnesi Jun 9 '16 at 8:03
  • $\begingroup$ @SophieAgnesi The first asymptotic follows from the Stirling's approximation. The second follows from the definition of Gamma as limit (see the wikipedia page, first formula in the "alternative definitions") and the definiton of Digamma function. For the second you can also observe that Digamma is the derivative of $\log(\Gamma(x)).$ $\endgroup$ – Marco Cantarini Jun 9 '16 at 8:24
  • $\begingroup$ OK, thanks for your answer. (+1) $\endgroup$ – Sophie Agnesi Jun 9 '16 at 9:08
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$\log \Gamma$ 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:

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

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  • $\begingroup$ Sorry Jack, I just don't get it this answer. Your answer seems to confuse me since it doesn't directly answer the given problem, but it was gratefully received nonetheless. (+1) $\endgroup$ – Sophie Agnesi Jun 9 '16 at 8:00
  • $\begingroup$ @SophieAgnesi: why do you say it does not answer your question? I simply wrote $\log\Gamma(a)-\log\Gamma(b)$ as $\int_{b}^{a}\psi(x)\,dx$. $\endgroup$ – Jack D'Aurizio Jun 9 '16 at 13:03

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