# find this integral with complex number

$$\int \frac{(z^{-3})e^{1/z}}1dz$$

I use tailor series but I cant solve it also I use common solution but I cant solve it

• Is the integral over some path, or do you just want an antiderivative? And why is there a $1$ in the denominator? – Arthur Jun 22 '17 at 12:34
• I just want its antiderivative, I cant write this function correctly because dz goes into denominator !! – Nima Gorjinezhad Jun 22 '17 at 12:43
• @NimaGorjinezhad You asked a similar question here. It seems you're having trouble using MathJax. Please go through this tutorial. – Sahiba Arora Jun 22 '17 at 12:46
• You mean $\int \frac{\exp(\frac{1}{z})}{z^3 dz}$ ? – user392395 Jun 22 '17 at 13:14

$\int z^{-3}e^{1/z}dz=- \int u(z)v'(z) dz$, where $u(z)=1/z$ and $v(z)=e^{1/z}$.
$$\int z^{-3}\, e^{1/z} dz=\frac{e^{\frac{1}{z}} (z-1)}{z}.$$