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I have proved that for \Re [z]>0,
and I wish to justify differentiating under the integral sign (n+1)-times to give integral formulas for the polygamma functions. Let
I have proved that if \Re [z]>0,\mbox{ and }|z|<\infty, then
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?
\log \Gamma (z) = \int_{0}^{\infty} \left( z-1+\frac{1-e^{-(z-1)t}}{e^{-t}-1}\right) \frac{e^{-t}}{t}dt
and I wish to justify differentiating under the integral sign (n+1)-times to give integral formulas for the polygamma functions. Let
f(t)=\left( z-1+\frac{1-e^{-(z-1)t}}{e^{-t}-1}\right) \frac{e^{-t}}{t}
I have proved that if \Re [z]>0,\mbox{ and }|z|<\infty, then
\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
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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