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In Stein's Fourier Analysis, chapter 7, question 8(a), Stein hints that one can use the Fourier transform to verify the identity

$$ e^{-\beta} = \int_0^\infty \frac{e^{-u}}{\sqrt{\pi u}} e^{-\beta^2/4u}\; du $$

which holds for $\beta > 0$, by applying the Fourier transform in $\mathbf{R}^n$, writing $\beta = 2 \pi |x|$ for some $x \in \mathbf{R}^n$. All the proofs I have seen the identity have used the residue calculus in some form or another, and I can't seem to find a solution to this equation myself, or online. Does anyone have any idea how we might calculate that the Fourier transform of either side of the equation are equal, without using complex analysis?

Jacob Denson
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1 Answers1

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$$\int_0^\infty e^{-u - \frac{\beta^2}{4u}} \frac{du}{\sqrt{\pi u}} =\frac{2}{\sqrt{\pi}} \int_0^\infty e^{-u^2 - \frac{\beta}{4 u^2}} du =e^{-\beta}\frac{2}{\sqrt{\pi}} \int_0^\infty e^{-(u- \frac{\beta}u)^2} du.$$ Making the change of variable $u = \frac{\beta} v$, we have $$\int_0^\infty e^{-(u- \frac{\beta}u)^2} du = \beta \int_0^\infty e^{-(v- \frac{\beta}v)^2} \frac{dv}{v^2}.$$ Hence $$2 \int_0^\infty e^{-u - \frac{\beta^2}{4u}} \frac{du}{\sqrt{\pi u}}= e^{-\beta}\frac{2}{\sqrt{\pi}} \int_0^\infty e^{-(u -\frac\beta u)^2} (1 + \frac{\beta}{u^2})du.$$ Changing the variable $t = u -\frac{\beta}u$, we get $$2 \int_0^\infty e^{-u - \frac{\beta^2}{4u}} \frac{du}{\sqrt{\pi u}} = e^{-\beta}\frac{2}{\sqrt{\pi}} \int_{-\infty}^\infty e^{-t^2} dt = 2e^{-\beta}.$$ I hope that you understand why $$\int_{-\infty}^\infty e^{-t^2} dt = \sqrt{\pi}.$$

nguyen0610
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