In my answer of the previous OP, I'm able to prove that
I(a)=∫∞0e−(a−2)x⋅1−e−x(1+x)x(1−ex)(ex+e−x)dx=∫10ya−11+y2(y1−y+1logy)dy=logΓ(a+24)−logΓ(a4)−14ψ(a+14)−14ψ(a+24)
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
\end{align}
But I'm having trouble proving
lima→∞[logΓ(a+24)−logΓ(a4)−14ψ(a+14)−14ψ(a+24)]=0
Indeed the above result is confirmed by Wolfram Alpha. How does one prove the above limit?
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