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I was recently asked to evaluate the following integral:

$$\int_0^x e^t \sqrt{2 + \sin(2t)} \, dt$$

It was beyond the ken of WolframAlpha, which I find quite discouraging.

Does anyone have an idea of how to tackle this problem?

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    Maple couldn't find the antiderivative either, and also couldn't integrate from $-\infty$ to $0$.2012-11-10
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    It means there is closed form for this integral?2012-11-10
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    It has a nice power series expression. Simply write-out the power series expansions of $e^t$, $\sqrt{1+t}$, re-scale, compose and multiply accordingly, then you take the anti-derivative.2012-11-14
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    Looks hard. Already, $\int\sqrt{2+\sin\,2u}\mathrm du$ is an elliptic integral; the additional exponential term makes me think a simple closed form is terribly unlikely.2012-11-17
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    Is this really what you were asked? The fact that you are integrating from 0 to $x$ makes me wonder if you were asked to compute the derivative of this expression with respect to $x$.2012-11-17
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    @JimConant That is really what I was asked.2012-11-18
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    @B.D. I just wanted to throw that out there just to make sure.2012-11-18
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    Not an elementary function for sure. What does the person who asked you about the integral want to know about it?2012-11-21
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    @fedja I highly suspect there was a transcription error somewhere along the way, because I am also quite convinced this is not an elementary function. Nonetheless, I made an attempt (somewhat feebly) at evaluating it, only to come up empty-handed. I would settle for a proof (probably using the well-known theorem of Liouville) that there is no such answer using only elementary functions. If there is another sneaky way of getting to the answer (through some unlikely manipulation of infinite series) that would be great too.2012-11-21
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    ---I would settle for a proof (probably using the well-known theorem of Liouville) that there is no such answer using only elementary functions--- The only little problem with that is that one needs a full day to carry out such an exercise by hand and to write a random formula to integrate takes under 20 seconds, so I'm not really inclined to enter this race on the "proving impossibility" side though I've done it a few times on AoPS. As to the series representation, mike4ty4 has already provided it :).2012-11-21
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    @fedja Okay; noted.2012-11-21

2 Answers 2

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If Mathematica (which is what Wolfram Alpha runs on) can't find a (closed-form) solution, it's likely it has none, neither in terms of "elementary" functions or even commonly used non-elementary special functions (of which Mathematica has quite a few). There are many integrals like that. It's not certain, but likely. There is a theorem of Liouville that can be applied to determine if elementary solutions are possible, not sure about solutions via more sophisticated functions.

An amazing fact is that even the seemingly very innocent-looking

$\int x^x dx$

has no solution not only in terms of elementary functions, but common non-elementary special functions as well (at least, I haven't seen any, and Mma doesn't seem to know of any).

Failing a closed form, you could try a series expansion, but I suspect it'll be a mess in this case.

If you need a value for a particular $x$, you can approximate it to arbitrary precision via a numerical integration ("quadrature") technique.

EDIT: NEW!!!

Following up on the "series expansion" idea, I found this infinite series expansion. It is based on the idea of representing sine by its complex exponential representation and using the Taylor series for the square root.

$$\int_{0}^{x} e^t \sqrt{2 + \sin(2t)} dt = \sqrt{2} \left(\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^{n+k} (2n)!}{(1 - 2n)(n!) 16^n i^n k! (n-k)! (2i(n - 2k) + 1)} e^{(2i(n - 2k) + 1)x}\right) - \sqrt{2} \left(\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^{n+k} (2n)!}{(1 - 2n)(n!) 16^n i^n k! (n-k)! (2i(n - 2k) + 1)}\right)$$.

The thing is, this series looks very hypergeometric. The only trouble I see is the presence of the factors $(n-k)!$ and $(2i(n - 2k) + 1)$ in the denominator, which depend on BOTH indices...

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    Yes, I am aware of this theorem of Liouville. See, e.g., http://mathoverflow.net/questions/108598/is-xtanx-integrable-in-elementary-functions/ where my musings were accepted before a *far* superior answer was provided.2012-11-10
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Ryan Budney's strategy can be expanded upon by WolframAlpha. It will also provide numerical approximations if asked. One can also see that the derivative gets twitchy at least twice, which may benefit an explanation of the lack of any representation of the anti-derivative in terms of elementary functions.