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In Donald McQuarrie’s Mathematical Methods for Scientists and Engineers, he has a problem I would like to assign to my class, but I am having trouble solving it. It states

Show that $$\int_0^\infty e^{-x^2} \cos\alpha x \,dx = \frac{\sqrt{\pi}}{2} e^{-\alpha^2/4},$$ by expanding $\cos\alpha x$ in a Maclaurin series and integrating term by term.

Doing as suggested, you get $$ \int_0^\infty e^{-x^2} \left( 1 - \frac{\alpha^2}{2!} x^2 + \frac{\alpha^4}{4!} x^4 + \dots \right)dx. $$ I know that the first integral is $\sqrt{\pi}/2$ and that the remaining integrals can be evaluated using the following result from a table of integrals that I have $$ \int_0^\infty x^{2n} e^{-x^2} dx = \frac{1 \cdot 3 \cdot 5 \dots (2n-1) }{2^{n+1}} \sqrt{\pi}. $$ Doing this, I can complete the problem, the issue is that I do not know how to evaluate the above integral to obtain the result I found in the table of integrals.

Can anyone point me in the right direction?

Gary L. Gray
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  • Are you fine with $\int_0^{\infty} e^{-x^2} , \mathrm{d}x = \frac{\sqrt{\pi}}{2}$? If so, the remaining ones can be obtained by repeated integration by parts. – Klaus Aug 30 '22 at 13:29
  • You can also try differentiating with respect to $\alpha$ and then integrating by parts. This should give some identity that could lead to the result. – Ryszard Szwarc Aug 30 '22 at 13:29
  • https://en.wikipedia.org/wiki/Gamma_function This may help you, see also pp.192-pp.195 of baby rudin – Zhang Aug 30 '22 at 13:33
  • Also search “double facorial” for the $1\cdot2\cdots(2n-1)$ – Тyma Gaidash Aug 30 '22 at 13:47

3 Answers3

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Evaluate the table integral as follows $$ \int_0^\infty x^{2n} e^{-x^2} dx = (-1)^n \frac{d^n}{da^n} \bigg(\int_0^\infty e^{-ax^2} dx\bigg)_{a=1} \\= (-1)^n \frac{d^n}{da^n} \bigg(\frac{\sqrt{\pi}}2 a^{-1/2} \bigg)_{a=1} =\frac{1 \cdot 3 \cdot 5 \cdots (2n-1) }{2^{n+1}} \sqrt{\pi}. $$

Quanto
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As @Quanto already pointed out the Feynman method, let me add a more direct approach using integration by parts: $$\int_0^{\infty} x^{2n}e^{-x^2} \, \mathrm{d}x = \int_0^{\infty} x^{2n-1} \cdot xe^{-x^2} \, \mathrm{d}x = -\frac{1}{2}x^{2n-1}e^{-x^2}\Big|_0^{\infty} + \frac{2n-1}{2}\int_0^{\infty} x^{2n-2}e^{-x^2} \, \mathrm{d}x.$$ The first summand vanishes; the formula thus follows by induction.

Klaus
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The gamma function is defined as \begin{equation*} \Gamma(z) = \int_{0}^{\infty}t^{z-1} \, e^{-t} \, dt, \quad \Re(z) > 0. \end{equation*} Setting $t = x^{2}$ and $z=n + 1/2$, it follows that \begin{equation*} \frac{1}{2}\Gamma\left(n+\frac{1}{2}\right) = \int_{0}^{\infty}x^{2n} \, e^{-x^{2}} \, dx. \end{equation*} Repeated use of the recurrence $\Gamma(z+1) = z \, \Gamma(z)$, with $\Gamma(1/2) = \sqrt{\pi}$, gives the result \begin{equation*} \frac{1}{2}\Gamma\left(n+\frac{1}{2}\right) = \frac{1}{2}\left(\frac{2n-1}{2}\right)\left(\frac{2n-3}{2}\right)\cdots\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right) = \frac{(2n-1)(2n-3)\cdots 5 \cdot 3 \cdot 1}{2^{n+1}}\sqrt{\pi}. \end{equation*}

Barzane
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