Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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
$\ln(2)+\sum_{k=1}^{\infty} (-1)^k \frac{2^{2k} -1}{(2k)k!}$. – 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!

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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