I have proved that for [tex]\Re [z]>0,[/tex](adsbygoogle = window.adsbygoogle || []).push({});

[tex]\log \Gamma (z) = \int_{0}^{\infty} \left( z-1+\frac{1-e^{-(z-1)t}}{e^{-t}-1}\right) \frac{e^{-t}}{t}dt[/tex]

and I wish to justify differentiating under the integral sign (n+1)-times to give integral formulas for the polygamma functions. Let

[tex]f(t)=\left( z-1+\frac{1-e^{-(z-1)t}}{e^{-t}-1}\right) \frac{e^{-t}}{t}[/tex]

I have proved that if [tex]\Re [z]>0,\mbox{ and }|z|<\infty,[/tex] then

[tex]\lim_{t\rightarrow 0^+} f(t)=\frac{1}{2}z^2-\frac{3}{2}z+1,\mbox{ and }\lim_{t\rightarrow\infty} f(t)=0[/tex]

and if I recall correctly, it is sufficient that the integral in question be absolutely convergent to apply the differentiation rule.

EDIT: So how do I absolute convergence? Any hints?

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# Homework Help: Integrals for the LogGamma & polygamma fcns

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