Integral of cos(ax)e^(-x^2)

Problem 1. Show that

i) $\displaystyle \int_0^{\infty}x^{2n}e^{-x^2}dx=\frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!}$ for all integers $n \ge 0.$

ii) $\displaystyle \int_0^{\infty}\cos(ax) e^{-x^2} dx = \frac{\sqrt{\pi}}{2}e^{-\frac{a^2}{4}}$ for all real numbers $a.$

Solution. i) Let $\displaystyle I_n:=\int_0^{\infty}x^{2n}e^{-x^2}dx.$ We showed here that $\displaystyle I_0=\int_0^{\infty} e^{-x^2}dx=\frac{\sqrt{\pi}}{2}.$ Now, in $I_n,$ we apply integration by parts with $\displaystyle x^{2n-1}=u$ and $\displaystyle xe^{-x^2}dx=dv$ to get $\displaystyle I_n=\frac{2n-1}{2}I_{n-1}$ and thus

$\displaystyle \begin{aligned} I_n=\frac{2n-1}{2} \cdot \frac{2n-3}{2}I_{n-2} = \cdots = \frac{2n-1}{2} \cdot \frac{2n-3}{2} \cdots \frac{1}{2} \cdot I_0 =\frac{(2n)!}{2^n(2n) (2n-2) \cdots 2}I_0 =\frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!}.\end{aligned}$

ii) Using i) and the Maclaurin series of $\cos(ax),$ we have

$\displaystyle \begin{aligned} \int_0^{\infty}\cos(ax) e^{-x^2}\ dx= \int_0^{\infty}\sum_{n=0}^{\infty}(-1)^n\frac{(ax)^{2n}}{(2n)!}e^{-x^2}dx=\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n)!} \int_0^{\infty}x^{2n}e^{-x^2}dx\end{aligned}$

$\displaystyle \begin{aligned}=\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n)!} \cdot \frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!} =\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{(\frac{-a^2}{4})^n}{n!}=\frac{\sqrt{\pi}}{2}e^{\frac{-a^2}{4}}. \ \Box \end{aligned}$

Unlike $\displaystyle \int_0^{\infty}\cos(ax) e^{-x^2} dx,$ there’s no closed form for $\displaystyle \int_0^{\infty}\sin(ax) e^{-x^2} dx$ but still there’s something nice and non-trivial that we can prove about the integral, as the following problem shows.

Problem 2. i) Show that

i) $\displaystyle \int_0^{\infty}x^{2n+1}e^{-x^2}dx=\frac{n!}{2}$ for all integers $n \ge 0.$

ii) $\displaystyle \int_0^{\infty}\sin(ax) e^{-x^2} dx = e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx$ for all real numbers $a.$

Solution. i) Let $\displaystyle J_n:=\int_0^{\infty}x^{2n+1}e^{-x^2}dx.$ See that $\displaystyle J_0=\int_0^{\infty}xe^{-x^2}dx=\frac{1}{2}.$ Integration by parts with $x^{2n}=u$ and $xe^{-x^2}dx=dv$ gives

$\displaystyle J_n=nJ_{n-1}=n(n-1)J_{n-2}= \cdots = n(n-1) \cdots 2 \cdot 1 \cdot J_0=\frac{n!}{2}.$

ii) Using i) and the Maclaurin series of $\sin(ax),$ we have

$\displaystyle \begin{aligned} \int_0^{\infty}\sin(ax) e^{-x^2} dx= \int_0^{\infty}\sum_{n=0}^{\infty}(-1)^n\frac{(ax)^{2n+1}}{(2n+1)!}e^{-x^2}dx=a\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n+1)!} \int_0^{\infty}x^{2n+1}e^{-x^2}dx\end{aligned}$

$\displaystyle \begin{aligned}=\frac{a}{2}\sum_{n=0}^{\infty} \frac{(-a^2)^{n}n!}{(2n+1)!}= \frac{a}{2}\sum_{n=0}^{\infty} \frac{(\frac{-a^2}{4})^n4^n}{(2n+1)\binom{2n}{n}n!}. \end{aligned} \ \ \ \ \ \ \ \ \ \ \ \ (*)$

But we showed here that $\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1}x \ dx = \frac{4^n}{(2n+1)\binom{2n}{n}}.$ Thus the substitution $x=\cos t$ gives

$\displaystyle \int_0^1 (1-x^2)^ndx = \int_0^{\frac{\pi}{2}} \sin^{2n+1}t \ dx=\frac{4^n}{(2n+1)\binom{2n}{n}}.$

So, by $(*),$

$\displaystyle \begin{aligned} \int_0^{\infty}\sin(ax) e^{-x^2} dx= \frac{a}{2}\sum_{n=0}^{\infty} \frac{(\frac{-a^2}{4})^n}{n!}\int_0^1(1-x^2)^n dx=\frac{a}{2}\int_0^1 \sum_{n=0}^{\infty} \frac{(\frac{-a^2(1-x^2)}{4})^n}{n!}dx\end{aligned}$

$\displaystyle=\frac{a}{2} \int_0^1 e^{\frac{-a^2(1-x^2)}{4}}dx =\frac{a}{2}e^{\frac{-a^2}{4}}\int_0^1 e^{\frac{a^2x^2}{4}}dx=e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx. \ \Box$

Exercise 1. In Problem 2, ii), we showed that $\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}n!}{(2n+1)!}= e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx,$ for all real numbers $a.$ Here’s another way to prove it. Let

$\displaystyle f(a):=\frac{1}{2}e^{\frac{a^2}{4}}\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}n!}{(2n+1)!}- \int_0^{\frac{a}{2}}e^{x^2}dx.$

Show that $f'(a)=0$ and conclude that $f(a)=0.$

Exercise 2. Evaluate $\displaystyle \lim_{a\to\infty} \int_0^{\infty} \sin(ax)e^{-x^2}dx$ and $\displaystyle \lim_{a\to\infty} a \int_0^{\infty} \sin(ax)e^{-x^2}dx.$