Answer to Question #95685 in Calculus for Rachel

Question

Let φ be the function on R^n defined by
φ(x) = e^(-|x|^(2)/2, x∈R
Compute (φ*φ)(x) for all x in R^n.

Answer 1

$(\varphi*\varphi)(x)=\int\limits_{\mathbb R^n}\varphi(y)\cdot\varphi(x-y)dy=\int\limits_{\mathbb R^n}e^{-\frac{|y|^2}{2}}e^{-\frac{|x-y|^2}{2}}dy$

Let $y_0=(y_1,\ldots,y_{n-1})$ and $x_0=(x_1,\ldots,x_{n-1})$ , then $\int\limits_{\mathbb R^n}e^{-\frac{|y|^2}{2}}e^{-\frac{|x-y|^2}{2}}dy=\int\limits_{\mathbb R}\int\limits_{\mathbb R^{n-1}}e^{-\frac{|y_0|^2+y_n^2}{2}}e^{-\frac{|x_0-y_0|^2+(x_n-y_n)^2}{2}}dy_0dy_n=$

$=\int\limits_{\mathbb R}\Bigl(\int\limits_{\mathbb R^{n-1}}e^{-\frac{|y_0|^2}{2}}e^{-\frac{|x_0-y_0|^2}{2}}dy_0\Bigr)e^{-\frac{y_n^2}{2}}e^{-\frac{(x_n-y_n)^2}{2}}dy_n=$

$=\int\limits_{\mathbb R^{n-1}}e^{-\frac{|y_0|^2}{2}}e^{-\frac{|x_0-y_0|^2}{2}}dy_0\cdot\int\limits_{\mathbb R}e^{-\frac{y_n^2}{2}}e^{-\frac{(x_n-y_n)^2}{2}}dy_n$

Calculate $\int\limits_{\mathbb R}e^{-\frac{y_n^2}{2}}e^{-\frac{(x_n-y_n)^2}{2}}dy_n$ :

$\int\limits_{\mathbb R}e^{-\frac{y_n^2}{2}}e^{-\frac{(x_n-y_n)^2}{2}}dy_n=\int\limits_{\mathbb R}e^{-\frac{2y_n^2+x_n^2-2x_ny_n}{2}}dy_n=$

$=\int\limits_{\mathbb R}e^{-\frac{\Bigl(y_n\sqrt{2}-\frac{x_n}{\sqrt{2}}\Bigr)^2+\frac{x_n^2}{2}}{2}}dy_n$

Let $z_n=y_n\sqrt{2}-\frac{x_n}{\sqrt{2}}$ , then $dy_n=\frac{1}{\sqrt{2}}dz_n$ and

$\int\limits_{\mathbb R}e^{-\frac{\Bigl(y_n\sqrt{2}-\frac{x_n}{\sqrt{2}}\Bigr)^2+\frac{x_n^2}{2}}{2}}dy_n=\int\limits_{\mathbb R}e^{-\frac{z_n^2+\frac{x_n^2}{2}}{2}}\cdot\frac{1}{\sqrt{2}}dz_n=$

$=\frac{1}{\sqrt{2}}e^{-\frac{x_n^2}{4}}\int\limits_{\mathbb R}e^{-\frac{z_n^2}{2}}dz_n$

We know that the probability density function of a normal disribution $N(0,1)$ is $\rho(u)=\frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}$ , so $\int\limits_{\mathbb R}\frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du=1$ , that is $\int\limits_{\mathbb R}e^{-\frac{u^2}{2}}du=\sqrt{2\pi}$ . We obtain $\frac{1}{\sqrt{2}}e^{-\frac{x_n^2}{4}}\int\limits_{\mathbb R}e^{-\frac{z_n^2}{2}}dz_n=\frac{1}{\sqrt{2}}e^{-\frac{x_n^2}{4}}\sqrt{2\pi}=e^{-\frac{x_n^2}{4}}\sqrt{\pi}$

So $\int\limits_{\mathbb R^n}e^{-\frac{|y|^2}{2}}e^{-\frac{|x-y|^2}{2}}dy=e^{-\frac{x_n^2}{4}}\sqrt{\pi}\int\limits_{\mathbb R^{n-1}}e^{-\frac{|y_0|^2}{2}}e^{-\frac{|x_0-y_0|^2}{2}}dy$

Therefore we can prove by induction on $n$ that $(\varphi*\varphi)(x)=\pi^{\frac{n}{2}}e^{-\frac{|x|^2}{4}}$

It is true for $n=1$ : we proved $(\varphi*\varphi)(x)=\int\limits_{\mathbb R}e^{-\frac{y^2}{2}}e^{-\frac{(x-y)^2}{2}}dy=e^{-\frac{x^2}{4}}\sqrt{\pi}$

Suppose that it is true for $n=k$, where $k\ge 1$ . Then $\int\limits_{\mathbb R^{k+1}}e^{-\frac{|y|^2}{2}}e^{-\frac{|x-y|^2}{2}}dy=e^{-\frac{x_{k+1}^2}{4}}\sqrt{\pi}\int\limits_{\mathbb R^k}e^{-\frac{|y_0|^2}{2}}e^{-\frac{|x_0-y_0|^2}{2}}dy$ , where $y_0=(y_1,\ldots,y_k)$ and $x_0=(x_1,\ldots,x_k)$ .

By induction hypohesis $\int\limits_{\mathbb R^k}e^{-\frac{|y_0|^2}{2}}e^{-\frac{|x_0-y_0|^2}{2}}dy=\pi^{\frac{k}{2}}e^{-\frac{|x_0|^2}{4}}$ , so $e^{-\frac{x_{k+1}^2}{4}}\sqrt{\pi}\int\limits_{\mathbb R^k}e^{-\frac{|y_0|^2}{2}}e^{-\frac{|x_0-y_0|^2}{2}}dy=e^{-\frac{x_{k+1}^2}{4}}\sqrt{\pi}\cdot\pi^{\frac{k}{2}}e^{-\frac{|x_0|^2}{4}}=$

$=\pi^{\frac{k+1}{2}}e^{-\frac{|x|^2}{4}}$ . We proved that $\int\limits_{\mathbb R^{k+1}}e^{-\frac{|y|^2}{2}}e^{-\frac{|x-y|^2}{2}}dy=\pi^{\frac{k+1}{2}}e^{-\frac{|x|^2}{4}}$.

By principle of mathematical induction we obtain $(\varphi*\varphi)(x)=\int\limits_{\mathbb R^n}e^{-\frac{|y|^2}{2}}e^{-\frac{|x-y|^2}{2}}dy=\pi^{\frac{n}{2}}e^{-\frac{|x|^2}{4}}$

Answer: $(\varphi*\varphi)(x)=\pi^{\frac{n}{2}}e^{-\frac{|x|^2}{4}}$