11
$\begingroup$

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

$\endgroup$
8
  • 1
    $\begingroup$ Nothing elementary, I'm afraid. You'll need the Gaussian hypergeometric function for this... $\endgroup$ – J. M. isn't a mathematician Feb 9 '12 at 17:09
  • $\begingroup$ Please explain meaning of 'Nothing elementary'? Please provide me details regarding elementary and non elementary. $\endgroup$ – Inquisitive Feb 9 '12 at 17:15
  • $\begingroup$ You are aware of the concept of an "elementary function", no? The integral you have cannot be expressed in terms of those. $\endgroup$ – J. M. isn't a mathematician Feb 9 '12 at 17:17
  • 1
    $\begingroup$ Elementary Functions are the basic functions that we all know and all of the functions that we can make out of them, for example: $x$,$\sin x$, $e^x$, $\arctan (\ln x)$ and so.. J.M meant that the answer for this integral will not be any combination of these functions. $\endgroup$ – Jozef Feb 9 '12 at 17:40
  • 2
    $\begingroup$ @sree: Concerning the concept of "elementary function" see this Wikipedia's entry. $\endgroup$ – Américo Tavares Feb 9 '12 at 17:44
9
$\begingroup$

$$\tan(x) = -i \frac{1-e^{-2ix}}{1+e^{-2ix}} = -i - 2 i \sum_{k=1}^\infty (-1)^k e^{-2ikx}$$ (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:

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

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ – user856 Feb 9 '12 at 19:55
7
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Well, one possible route is to expand the integrand as a series, integrate termwise, and then apply any number of methods for recognizing hypergeometric series, like the ones embodied in the work of Gosper, Zeilberger, Koutschan, etc. $\endgroup$ – J. M. isn't a mathematician Feb 9 '12 at 17:46
  • $\begingroup$ 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...) $\endgroup$ – kcrisman Feb 10 '12 at 4:11
2
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ I couldn't find any method to solve this. :( $\endgroup$ – Inquisitive Feb 9 '12 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.