# Integral representation of the Digamma function vs. Integral representation of the Polygamma function [closed]

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 WildonFeb 22 at 20:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Carlo Beenakker, Johannes Hahn, Mark Wildon
If this question can be reworded to fit the rules in the help center, please edit the question.

• it is all explained in the last small paragraph of en.wikipedia.org/wiki/…. :) – Wolfgang Feb 16 at 15:47
• Yeah, I'm so sorry, sorry for the stupid question... – 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?
• yes, the condition $n>0$ is stated, for example, in Wikipedia – Carlo Beenakker Feb 16 at 18:32
• Yeah, I'm so sorry, I somehow overread the $n>0$. Sorry for the stupid question... – ansebene Feb 17 at 8:03