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

1 Answer
t0hierry
Apr 25, 2016

#\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

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