Friday, 12 May 2017

calculus - Limit of gamma and digamma function

In my answer of the previous OP, I'm able to prove that



I(a)=0e(a2)x1ex(1+x)x(1ex)(ex+ex)dx=10ya11+y2(y1y+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&=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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