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I came across the following integral in a textbook without explanation. How can I prove it?

$$\int_0^\infty \frac{e^{-t}-1}{t^{s+1}}dt=\Gamma(-s)$$

Here $s\in(0,1)$.

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    $\begingroup$ Maybe use $$\Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx$$? $\endgroup$ – Andrew Li Jun 3 '18 at 17:39
  • $\begingroup$ See The gamma function $\endgroup$ – The Integrator Jun 3 '18 at 17:43
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Let $0<s<1$. Then $1-s>0$ and $$\Gamma(1-s)=\int_0^\infty \frac{e^{-t}}{t^s}\,dt.$$ Integrate by parts: $$\Gamma(1-s)=\left[\frac{1-e^{-t}}{t^s}\right]_0^\infty +\int_0^\infty\frac{s(1-e^{-t})}{t^{s+1}}\,dt =s\int_0^\infty\frac{1-e^{-t}}{t^{s+1}}\,dt.$$ But $\Gamma(1-s)=-s\Gamma(-s)$.

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  • $\begingroup$ I have a confusion here (probably a silly question). How does $\left[\dfrac{1 - e^{-t}}{t^s}\right]$ evaluate to $0$? At $t = 0$, doesn't it become undefined (division by $0$)? $\endgroup$ – an4s Jun 3 '18 at 17:56
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    $\begingroup$ @an4s Take it as a limit. $\endgroup$ – mickep Jun 3 '18 at 17:57
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    $\begingroup$ The numerator is $O(t)$ as $t\to0$ @an4s $\endgroup$ – Lord Shark the Unknown Jun 3 '18 at 17:58
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    $\begingroup$ @mickep Ah, that makes sense. For anybody else with the same question, I found this helpful as well. $\endgroup$ – an4s Jun 3 '18 at 18:11
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Since $s\in(0,1)$ and : $$ \int_0^\infty \frac{1}{t^{s+1}}dt=0 $$ we have: $$\int_0^\infty \frac{e^{-t}-1}{t^{s+1}}dt=\int_0^\infty \frac{e^{-t}}{t^{s+1}}dt-\int_0^\infty \frac{1}{t^{s+1}}dt= \int_0^\infty t^{-s-1}e^{-t} dt-0= \int_0^\infty t^{(-s)-1}e^{-t} dt =\Gamma (-s)$$

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  • $\begingroup$ Your first line is completely wrong. $\endgroup$ – Andrew Li Jun 3 '18 at 18:24
  • $\begingroup$ @AndrewLi What's wrong with it? $\endgroup$ – Anastassis Kapetanakis Jun 3 '18 at 18:25
  • $\begingroup$ It's divergent for the given $s$s because your lower bound is 0. $\endgroup$ – Andrew Li Jun 3 '18 at 18:25
  • $\begingroup$ Yes but there is 0<s<1 $\endgroup$ – Anastassis Kapetanakis Jun 3 '18 at 18:26
  • $\begingroup$ Have you tried $s=0.5$? $\endgroup$ – Andrew Li Jun 3 '18 at 18:27

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