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Does $$\psi(z)=\int_{0}^{\infty}\left(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}\right) dt$$ converges for $\Re{z}>0$?

Bona
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    It is unclear what the "psi gamma function" is. The similarly named "digamma function" is frequently denoted using the symbol $\psi$ as $\psi(z)$. A generalization is the polygamma function also frequently denoted using $\psi$, as $\psi^{(m)}(z)$. – Eric Towers Jan 09 '20 at 00:06
  • Yes for this function I meant. So is the digamma function convergent for $Re(z)>0$? – Bona Jan 09 '20 at 08:56
  • Functions aren't "convergent". Sums, products, integrals, and other iterative/recursive operations are convergent. Are you asking whether the domain of the digamma function includes the halfplane $\Re(x) > 0$ or are you asking if a particular sum, product, integral, or other representation of digamma on part of its domain converges on the right halfplane? – Eric Towers Jan 09 '20 at 13:58
  • The integral representation. – Bona Jan 09 '20 at 14:11
  • Which integral representation. There is no integral in your Question. The Wikipedia lists several integral representations for digamma: https://en.wikipedia.org/wiki/Digamma_function#Integral_representations . MathWorld gives at least one more. http://mathworld.wolfram.com/DigammaFunction.html And the DLMF has a few as well. https://dlmf.nist.gov/5.9 – Eric Towers Jan 09 '20 at 14:18
  • $\psi(z)=\int_0^{\infty}(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}})dt$ – Bona Jan 09 '20 at 15:18
  • You should edit the correct name of the function and the integral into your Question so that it accurately captures your question. – Eric Towers Jan 09 '20 at 15:31
  • This answer might address your question. – robjohn Jan 11 '20 at 05:59

2 Answers2

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Yes, Gauss's integral representation of the digamma function is widely documented to converge on the (open) right halfplane.

Eric Towers
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Convergence at $0$: $$ \lim_{t \to 0}\left(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}\right) = z - \frac{3}{2} $$ so the apparent discontinuity at $t=0$ is removable.

Convergence at $\infty$:
Two parts, $$ \frac{e^{-t}}{t} < e^{-t}\quad \text{and}\quad \int^\infty e^{-t} dt\quad \text{converges, so}\quad \int^\infty \frac{e^{-t}}{t}\quad\text{converges.} $$ Also $$ \frac{e^{-zt}}{1-e^{-t}} \sim e^{-zt}\quad\text{as } t \to +\infty $$ But $$ \int e^{-zt} dt = \frac{e^{-zt}}{z} \tag1$$ Now $$ \left|e^{-zt}\right| = e^{\mathrm{Re}\;(-zt)} $$ So $$\text{if}\quad\mathrm{Re}\;z > 0, \quad\text{then}\quad |e^{-zt}| \to 0 \quad\text{so}\quad \int^\infty |e^{-zt}|dt \quad\text{converges and thus}\quad \int^\infty\frac{e^{-zt}}{1-e^{-t}}\;dt \quad\text{converges.} $$

On the other hand, $$ \text{if}\quad \mathrm{Re}\;z \le 1,\quad\text{then}\quad |e^{-zt}| \ge 1 $$

fortunately, the OP does not ask for this part. Absolute convergence fails, but conditional convergence may hold anyway.

GEdgar
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