I'm trying to evaluate the following indefinite integral:

$$ \int\frac{e^{-x^2}}{(1+2x^2)^2}dx $$

According to Wolfram|Alpha, this integral evaluates to:

$$ \int \frac{e^{-x^2}}{(1+2 x^2)^2} dx = \frac{1}{4}\left(\sqrt{\pi} \mathrm{erf}(x)+\frac{2x e^{-x^2}}{(2 x^2+1)}\right)+\mathrm{constant} $$

Is there a quick way of evaluating this integral to get that answer? A fairly obvious-looking integration-by-parts approach didn't work but looked fairly promising:

\begin{align} \int \frac{e^{-x^2}}{(1+2 x^2)^2} dx &= \frac{1}{4}\int \frac{e^{-x^2}}{x}\times\frac{4x}{(1+2 x^2)^2}dx \\ &=\frac{1}{4}\left(\frac{-e^{-x^2}}{x(1+2x^2)}+\int\frac{1}{(1+2x^2)}\times\frac{-2x^2e^{-x^2}-e^{-x^2}}{x^2}dx\right) \\ &=\frac{1}{4}\left(-\frac{e^{-x^2}}{x(1+2x^2)}-\int\frac{e^{-x^2}}{x^2}dx\right) \end{align}

If both the boundary term and the integrand were multiplied by $-2x^2$, this would be exactly what we want, since the integral on the right would then evaluate to $\sqrt{\pi}\mathrm{erf}(x)$. Is there some other integration by parts that I've missed? Or is it only possible to do the integral by more complicated means?

  • $\begingroup$ if the regular methods don't work then my next port of call would be differentiating under the integral or contour integration $\endgroup$
    – user27182
    Apr 25 '13 at 23:05

Try doing integration by parts on this term as well with $u = \exp(-x^2)$ and $dv = \frac{1}{x^2}dx$. Combine some terms and I think you'll have your result.

  • $\begingroup$ Thanks for you help. I'll let you know how it goes! $\endgroup$ Apr 25 '13 at 23:07
  • $\begingroup$ I edited my answer because while you dropped a negative in your LaTeX code, the second term should be negative in the last equality. $\endgroup$ Apr 25 '13 at 23:27
  • $\begingroup$ Yes - thanks. I'll change the incorrect $-$ sign now. Also, this method did indeed work! $\endgroup$ Apr 26 '13 at 10:19

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.