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 $s\in(0,1)$.
Let $0&lt;s&lt;1$. Then $1-s&gt;0$ 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 $\Gamma(1-s)=-s\Gamma(-s)$.
Since $s\in(0,1)$ and :
$$  \int_0^\infty \frac{1}{t^{s+1}}dt=0 $$
we have:
$$\int_0^\infty \frac{e^{-t}-1}{t^{s+1}}dt=\int_0^\infty \frac{e^{-t}}{t^{s+1}}dt-\int_0^\infty \frac{1}{t^{s+1}}dt= \int_0^\infty t^{-s-1}e^{-t} dt-0=  \int_0^\infty t^{(-s)-1}e^{-t} dt =\Gamma (-s)$$