# How to compute $\int_0^\infty \frac{e^{-t}-1}{t^{s+1}}dt$

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 $s\in(0,1)$.

• Maybe use $$\Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx$$? – Andrew Li Jun 3 '18 at 17:39
• – The Integrator Jun 3 '18 at 17:43

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 $\Gamma(1-s)=-s\Gamma(-s)$.

• 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 $0$)? – an4s Jun 3 '18 at 17:56
• @an4s Take it as a limit. – mickep Jun 3 '18 at 17:57
• The numerator is $O(t)$ as $t\to0$ @an4s – Lord Shark the Unknown Jun 3 '18 at 17:58
• @mickep Ah, that makes sense. For anybody else with the same question, I found this helpful as well. – an4s Jun 3 '18 at 18:11

Since $s\in(0,1)$ and : $$\int_0^\infty \frac{1}{t^{s+1}}dt=0$$ we have: $$\int_0^\infty \frac{e^{-t}-1}{t^{s+1}}dt=\int_0^\infty \frac{e^{-t}}{t^{s+1}}dt-\int_0^\infty \frac{1}{t^{s+1}}dt= \int_0^\infty t^{-s-1}e^{-t} dt-0= \int_0^\infty t^{(-s)-1}e^{-t} dt =\Gamma (-s)$$

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