Question

How do you integrate `int e^x tan x dx` using integration by parts?

Answer 1

`\tan(x) = -i \frac{1-e^{-2ix}}{1+e^{-2ix}} = -i - 2 i \sum_{k=1}^\infty (-1)^k e^{-2ikx}`
# \int e^x \tan(x)\ dx = -i e^x - 2 i \sum_{k=1}^\infty (-1)^k \int e^{(1-2ik)x}\ dx#
`= -i e^x -2i \sum_{k=1}^\infty \frac{(-1)^k}{1-2ik} e^{(1-2ik)x} + C`
that involves the Lerch Phi function