Integrals for the LogGamma & polygamma fcns

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SUMMARY

This discussion centers on the derivation of integral formulas for the polygamma functions from the logarithm of the Gamma function, specifically using the formula for \log \Gamma (z) as defined for \Re [z]>0. The user has established the function f(t) and its limits as t approaches 0 and infinity, confirming that the integral is absolutely convergent under these conditions. The main query revolves around the application of the differentiation under the integral sign and the criteria for absolute convergence.

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  • Understanding of complex analysis, particularly the properties of the Gamma function.
  • Familiarity with integral calculus and differentiation under the integral sign.
  • Knowledge of limits and convergence in the context of improper integrals.
  • Experience with polygamma functions and their definitions.
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benorin
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I have proved that for [tex]\Re [z]>0,[/tex]

[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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