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I'm interested in the value of the integral: $$\ \int_{-\infty}^\infty e^{-x^2}\cos(ax) \ d x $$

One of the insightful ways I have seen this integral evaluated is by differentiating under the integral with respect to $\ a$ , and using a clever rearrangement to set up a differential equation that allows us to solve (see this post).

However, I am uncomfortable with moving our differential operator inside an improper integral. I'm pretty sure according to the Dominated Convergence Theorem that it should suffice to find some integrable $\ g(a,x)$ that bounds our sequence of functions: $$\ f_h(a,x)=\frac{f(a+h, x)-f(a,x)}{h}$$

for all h sufficiently small, and where $$\ \lim_{h \to 0}\strut f_h(a,x)= \frac{\partial}{\partial a}f(a,x)$$

for all a, x.

However, being somewhat of a novice at these things, I have no idea how to start on this track. Or, if I'm even on the right track to begin with. In particular, it has crossed my mind that showing uniform convergence might be sufficient? Any and all advice is welcome.

Math Enthusiast
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  • To address your discomfort with moving a differential operator inside an improper integral: you should not be uncomforatable at all. In this case, the integral is done over $x$, and thus you can write: $$f(a)=\int_{-\infty}^\infty e^{-x^2}\cos(ax)dx$$ as just that, a function of $a$. – Kraigolas Jan 22 '20 at 05:51

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For $|h| \leq 1$ a dominating integrable function is $(|x|+x^{2})e^{-x^{2}}$. To see this first write $\cos ((a+h)x) -\cos (ax)$ as $\cos (ax)\cos (hx)-\sin (ax)\sin (hx)-\cos (ax)$. Note that $|\sin (ax)\sin (hx)| \leq |h||x|$. Now observe that $1-\cos t \leq \frac {t^{2}} 2$ for all $t$ so $|\cos (ax) [\cos (hx)-1]| \leq \frac {|h|^{2}|x^{2}|} 2$. Can you show that $(|x|+x^{2})e^{-x^{2}}$ is integrable?

Kavi Rama Murthy
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  • Just what I was looking for! Unfortunately, I'm having trouble verifying that the integral of this bounding function does indeed converge on its infinite domain. Could you perhaps provide a further hint? – Math Enthusiast Jan 22 '20 at 16:12
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    @MathEnthusiast $e^{x^{2}/2} \geq x^{2}/2$ so $x^{2} e^{-x^{2}} \leq 2e^{-x^{2}/2}$. Using the fact that $2e^{-x^{2}/2}$ is integrable we see that $x^{2} e^{-x^{2}}$ is integrable. Can you show that $xe^{-x^{2}}$ is also integrable? – Kavi Rama Murthy Jan 22 '20 at 23:21
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Why not to consider$$I=\int e^{-x^2}e^{iax} \, dx=\int e^{-(x^2-iax)} \, dx$$ Complete the square and change variable to face the gaussian integral and, back to $x$ $$I=\frac{\sqrt{\pi }}{2} e^{-\frac{a^2}{4}} \text{erf}\left(x-i\frac{a}{2} \right)$$ and $$J=\int_{-\infty}^\infty e^{-x^2}e^{iax} \, dx=\sqrt{\pi }\, e^{-\frac{a^2}{4}}$$ which is your result (provided $a\in \mathbb{R}$).

Claude Leibovici
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