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Let $a>0$, $b$ and $c$ be real constants. Evaluate the integral

$$ \int_{-\infty}^{\infty}e^{-ax^2}[ab\cos(bx)+bc\sin(bx)]dx. $$

Noting that $e^{-ax^2}bc\sin(bx)$ is odd, it would be nice if I could say that

$$ \int_{-\infty}^{\infty}e^{-ax^2}bc\sin(bx) = 0. $$

But is my understanding that for an improper integral of this form, the limit is taken with respect to one endpoint at a time, possibly giving me $\infty-\infty$. For the remaining part,

$$ \int_{-\infty}^{\infty}e^{-ax^2}ab\cos(bx), $$ I thought a technique similiar to the one found in Gaussian-like integral : $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$ might be useful, but I really have no clue.

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White Rice
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    You can indeed use the oddness for the sine part. And for the part with cosine, you can use the link you have. – mickep Jan 13 '15 at 20:04
  • Then why does the oddness work? Generally, it seems that you cannot use the oddness to show that an improper integral is equal to 0. Also, my integral is slightly different than the link I posted (b*cos(bx) vs cos(ax)). Finally I do not understand how you can differentiate past an improper integral, since you are differentiating past a limit(or two) as well as an integral. – White Rice Jan 13 '15 at 20:19
  • Since the factor $e^{-ax^2}$ makes this integral converge, you can integrate over $-R<x<R$ and let $R\to+\infty$. The problem of differentiating inside the integral is a bit to complicated to cover in a comment. What does your book say about it? – mickep Jan 13 '15 at 20:23
  • I do not have a book, I found this question on a practice exam. I have been referring to http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign. – White Rice Jan 13 '15 at 20:46

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