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I am trying to compute this integral:

$$\int_{1}^2 \frac{e^{-x^2}}{x} dx$$

But I'm confused on how to do this since I'm aware that $e^{-x^2}$ has no integrand.

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But you know that you can easily find an antiderivative of $xe^{-x^2}$, right? –  Lubin Apr 11 '12 at 5:41
    
The indefinite integral can be expressed as $\frac{1}{2}\operatorname{Ei}(-x^2)$, where $\operatorname{Ei}$ is the exponential integral. –  Antonio Vargas Apr 11 '12 at 5:44
    
Yes, the antiderivative would be $\frac{-e^{-x^2}}{2}$, although I'm not sure how much that helps me ... but I may be forgetting something as it's been a while since I did some serious integration. Also, thanks for the link (and post edit) Vargas, but I've never seen that exponential integral, so I'm not sure how to use it. –  user16647 Apr 11 '12 at 5:48
    
Actually I just looked this up ... I don't think it can be solvable outside of using Ei .. –  user16647 Apr 11 '12 at 6:00
2  
$\ln(2)+\sum_{k=1}^{\infty} (-1)^k \frac{2^{2k} -1}{(2k)k!}$. –  copper.hat Apr 11 '12 at 6:11

1 Answer 1

up vote 1 down vote accepted

This integral is not elementary. Change variables. Let $z=x^2$. Then $$\begin{eqnarray} I &=& \int_1^2 dx\, x^{-1} e^{-x^2} \\ &=& \frac{1}{2} \int_1^4 dx\, z^{-1} e^{-z} \\ &=& \frac{1}{2} \left(\int_1^\infty dx\, z^{-1} e^{-z} - \int_4^{\infty} dx\, z^{-1} e^{-z}\right) \\ &=& \frac{1}{2}[\mathrm{E}_1(1) - \mathrm{E}_1(4)] \\ &=& 0.10780\cdots \end{eqnarray}$$ where the integral $\mathrm{E}_1(x)$ is closely related to the exponential integral $\mathrm{Ei}(x)$. (This can in fact be rewritten as $\frac{1}{2}\left[\mathrm{Ei}(-4) - \mathrm{Ei}(-1)\right]$.) An integral such as this cannot be written in terms of simple functions---this is as simple as it gets!

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Thanks for the answer. I'll keep this in mind. –  user16647 Apr 11 '12 at 6:15
    
@user16647: Glad to help. It looks like copper.hat has the correct infinite series in the comments above. –  user26872 Apr 11 '12 at 6:19

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