Complex integration: pole of order 2 at zero

Question

Fix $a>0$. The following integral seems to be related to a complex integration: $$ \int_{0}^{a}\frac{e^{-z}+z-1}{z^2}dz. $$ We see that $f(z):=\frac{e^{-z}+z-1}{z^2}$ has a pole of order 2 at $z=0$. However, the upper end is a fixed positive real number $a$ instead of +∞ in which case usually can be dealt with a Cauchy integral. Is it possible to study the integral by complex analysis?Please give me some help or hints. Thanks a lot.

Answer 1

Since the function is analytic (once the removable singularity is removed), all integrations along curves do not depend on the curve itself, only the endpoints.

Since $a$ is a real number, one curve connecting $0$ and $a$ is the real interval $[0,a]$. I.e., you can treat this as a function on the real line and perform the real integration. Any other curve connecting $0$ and $a$ would give you the same result.