10
$\begingroup$

What is the integral of $e^x \tan(x)$? Using basic theorems it is difficult to get I think.

  • 2
    @sree: Concerning the concept of "elementary function" see [this](http://en.wikipedia.org/wiki/Elementary_function) Wikipedia's entry.2012-02-09

3 Answers 3

8

$\tan(x) = -i \frac{1-e^{-2ix}}{1+e^{-2ix}} = -i - 2 i \sum_{k=1}^\infty (-1)^k e^{-2ikx}$ (converging for $\text{Im}(x) < 0$)

$\begin{align} \int e^x \tan(x)\ dx &= -i e^x - 2 i \sum_{k=1}^\infty (-1)^k \int e^{(1-2ik)x}\ dx \\ &= -i e^x -2i \sum_{k=1}^\infty \frac{(-1)^k}{1-2ik} e^{(1-2ik)x} + C \\ &= -i{{ e}^{x}}-{\rm LerchPhi} \left( -{{ e}^{-2ix}},1,1+i/2 \right) {{ e}^{(1-2i)x}} + C \end{align}$

This Lerch Phi function can also be expressed in terms of a hypergeometric function:

${\rm LerchPhi} \left( z,1,1+\frac{i}{2} \right) = \frac{ 4-2i }{5} \ {\mbox{$_2$F$_1$}\left(1,1+\frac{i}{2};2+\frac{i}{2};z\right)}$

  • 1
    I took the liberty of reformatting your maths, as it was line breaking in an odd place in my browser. I hope that's OK.2012-02-09
7

I'm not sure sree was asking for an elementary answer, though it is possible. I can't comment yet, so I'm putting this in an answer area, though it doesn't answer; my apologies.

Comment: Are there good (and accessible) references for sree for how to utilize hypergeometric functions for doing indefinite integrals? For instance, would writing $e^x \tan(x)$ as a power series (or just the $\tan(x)$ part) and using some kind of uniform convergence and definitions of HG functions help?

  • 0
    Fair enough, though I think that "any number of methods" probably requires some elaboration. @Robert's answer is great - is there a _specific_ book which would introduce to these sorts of things by hand (as opposed to using an algorithm)? (Even if one has to pore through a massive table of such series...)2012-02-10
2

It is not expressible as an elementary function. integrals.com expresses it using hypergeometric functions:

$-i \left( -{{\rm e}^{x}} {\mbox{$_2$F$_1$}(1,-i/2;1-i/2;\,-{{\rm e}^{2ix}})}+ \left( 1/5-2i/5 \right) {{\rm e}^{ \left( 1+2i \right) x}} {\mbox{$_2$F$_1$}(1,1-i/2;2-i/2;\,-{{\rm e}^{2ix}})} \right)$

  • 0
    I couldn't find any method to solve this. :(2012-02-09