Yesterday I asked for the derivation of the Integral representation of the Digamma-Function: https://math.stackexchange.com/questions/3113119/integral-representation-of-digamma-function

Thanks again @Jair Taylor for the great and detailed explanation.

Today, however, I stumbled across the Integral representation of the Polygamma-Function on Wikipedia:

$\psi^{(n)}(x)=(-1)^{n+1}\int _{0}^{\infty }\left({\frac {t^{n}e^{-xt}}{1-e^{-t}}}\right)\,dt$

For $n=0$ this should represent an Integral representation for the Digamma-Function. So if I insert $n=0$ into the equation I end up with:

$\psi^{(0)}(x)=(-1)\int _{0}^{\infty }\left({\frac {e^{-xt}}{1-e^{-t}}}\right)\,dt$

This, however, is different from the Integral representation of the Digamma-Function I ended up with yesterday:

$\psi (x)=\int _{0}^{\infty }\left({\frac {e^{-t}}{t}}-{\frac {e^{-xt}}{1-e^{-t}}}\right)\,dt$

So how is it possible that the two Integral representation differ by the factor $\frac {e^{-t}}{t}$ within the Integral and still both are correct?

Please apologize if the solution is kind of obvious, I thought a lot about it and still don't see it.

Thank you so much for your help :)


closed as off-topic by Carlo Beenakker, Jan-Christoph Schlage-Puchta, Pace Nielsen, Johannes Hahn, Mark Wildon Feb 22 at 20:29

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  • 2
    $\begingroup$ it is all explained in the last small paragraph of en.wikipedia.org/wiki/…. :) $\endgroup$ – Wolfgang Feb 16 at 15:47
  • $\begingroup$ Yeah, I'm so sorry, sorry for the stupid question... $\endgroup$ – ansebene Feb 17 at 8:04

I tried $x=1$.
I get $$ \int _{0}^{\infty }\left({\frac {e^{-t}}{t}}-{\frac {e^{-t}}{1-e^{-t}}}\right)\,dt = -\gamma = \psi(1) $$ and $$ (-1)\int _{0}^{\infty }\left({\frac {e^{-t}}{1-e^{-t}}}\right)\,dt = -\infty \ne \psi(1) $$ even though Maple erroneously gets $-\gamma$ for this one also. The integral clearly diverges, since the integrand behaves like $1/t$ near $t=0$.

Where did you get $\psi^{(n)}(x)=(-1)^{n+1}\int _{0}^{\infty }\left({\frac {t^{n}e^{-xt}}{1-e^{-t}}}\right)\,dt$ ? Did it perhaps state $n > 0$ as a condition?

  • $\begingroup$ yes, the condition $n>0$ is stated, for example, in Wikipedia $\endgroup$ – Carlo Beenakker Feb 16 at 18:32
  • $\begingroup$ Yeah, I'm so sorry, I somehow overread the $n>0$. Sorry for the stupid question... $\endgroup$ – ansebene Feb 17 at 8:03

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