Solving an Integral In A Central Limit Theorem Proof

Question

I stumbled across an interesting proof of the Central Limit Theorem that uses only Fourier transforms and some slick limit interchanges, which someone has been kind enough to post on Wolfram MathWorld. I'm stuck on this relation which ends up being key to finishing the proof: $$\int_{-\infty}^{\infty}e^{iaf-bf^2}df=e\strut^{\frac{-a^2}{4b}}\sqrt{\frac{\pi}{b}}$$ The only justification that is given for this equivalence is " Abramowitz and Stegun 1972, p. 302, equation 7.4.6" which I looked up and it's just a table of integrals. Furthermore, the page states that the LHS can be immediately recognized as the Fourier Transform of a Gaussian, which I don't quite follow. The equation referred to in Abramowitz and Stegun doesn't even match the stated equivalence:

Does anyone have a nice proof of the equivalence I'm looking for, or a link to another resource that provides more detail?

Answer 1

First, $e^{iaf} = \cos(af) + i \sin(af)$. So $$ \int_{-\infty}^{\infty}e^{iaf-bf^2}df = \int_{-\infty}^{\infty}\cos(af) e^{-bf^2}df + i\int_{-\infty}^{\infty}\sin(af) e^{-bf^2}df $$ Now the second one has an odd function as integrand, so it is $0$. The first one is found from 7.4.6. It is an even function, so $\int_{-\infty}^{+\infty} = 2 \int_0^{+\infty}$.

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proof for 7.4.6
Begin with $$ \int_{-\infty}^{+\infty} e^{-v^2} dv = \sqrt{\pi} $$ Then do a translation, $v=u-c$, $c \in \mathbb R$, to get $$ \sqrt{\pi} = \int_{-\infty}^{+\infty} e^{-(u-c)^2} du =\int_{-\infty}^{+\infty} e^{-c^2+2uc-u^2} du =e^{-c^2}\int_{-\infty}^{+\infty} e^{2uc-u^2} du \\ \int_{-\infty}^{+\infty} e^{2uc-u^2} du=e^{c^2}\sqrt{\pi},\qquad c \in \mathbb R $$ Change variables $u=At$ where $A>0$ $$ \int_{-\infty}^{+\infty} e^{2Atc-A^2 t^2}dt = \frac{e^{c^2}\sqrt{\pi}}{A} $$ Now let $A = \sqrt{a},a > 0$. $$ \int_{-\infty}^{+\infty} e^{2\sqrt{a}\,tc-a t^2}dt = \frac{e^{c^2}\sqrt{\pi}}{\sqrt{a}},\qquad c \in\mathbb R, a > 0 \tag{1}$$ For a fixed value $a>0$, both sides of $(1)$ are entire functions of the complex variable $c$. The equation $(1)$ holds for a set of values of $c$ with a limit point (all real values), so it holds for all $c \in \mathbb C$. Thus: $$ \int_{-\infty}^{+\infty} e^{2\sqrt{a}\,tc-a t^2}dt = \frac{e^{c^2}\sqrt{\pi}}{\sqrt{a}},\qquad c \in\mathbb C, a > 0 $$ Finally, let $c =ix/\sqrt{a}$ to get $$ \int_{-\infty}^{+\infty} e^{2ixt-a t^2}dt = \frac{e^{-x^2/a}\sqrt{\pi}}{\sqrt{a}},\qquad x \in \mathbb R, a>0. $$ As noted above, the imaginary part is $0$ and the real part yields $$ \int_{-\infty}^{+\infty} \cos(2xt)e^{-a t^2}dt = \frac{e^{-x^2/a}\sqrt{\pi}}{\sqrt{a}},\qquad \int_{0}^{+\infty} \cos(2xt)e^{-a t^2}dt = \frac{e^{-x^2/a}\sqrt{\pi}}{2\sqrt{a}} $$

Answer 2

To show equivalence of a Fourier transform, $\int_0^\infty=\frac{1}{2}\int_{=\infty}^{\infty}$, since the integrand $(e^{-at^2}cos(2xt))$ is even. Further $e^{2ixt}=cos(2xt)+isin(2xt)$. Because the integrand $(e^{-at^2})$ is even, the integral using $sin(2xt)$ $=0$.

Net result $\int_0^\infty e^{-at^2}cos(2xt)dt=\frac{1}{2}\int_{-\infty}^\infty e^{-at^2}e^{2ixt}dt$ which is the Fourier transform.