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

Question

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 $s\in(0,1)$.

Answer 1

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

Answer 2

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