Complex Integration by Parts help

Question

Solve $\frac{1}{\sqrt{4\pi t}}\int\limits_{\mathbb{R}}e^{-\sigma^2/(4t)}(\sigma^2 +2\sigma x+x^2-1)d\sigma$

I am told the integral of the heat kernel is 1

Answer 1

This one is veeery similar to the one you just asked about :)

First write $\sigma^2 + 2\sigma x + x^2 = (\sigma + x)^2$ then

$$\frac{1}{\sqrt{4\pi t}} \int_{\mathbb{R}} (\sigma + x)^2 e^{-\sigma^2 /(4t)}d\sigma - \frac{1}{\sqrt{4\pi t}} \int_{\mathbb{R}}e^{-\sigma^2 /(4t)}d\sigma$$

Use the variable change $y=\sigma + x$ for the first integral and multiply and divide by $\sqrt{4\pi t}$ to make the integrand be a density kernel:

$$ \int_{\mathbb{R}} (\sigma + x)^2 e^{-\sigma^2 /(4t)}d\sigma = \sqrt{4\pi t} \int_{\mathbb{R}} y^2 \frac{1}{\sqrt{4\pi t}} e^{-(y-x)^2 /(4t)}dy = \sqrt{4\pi t} (2t + x^2)$$ since it is the second order moment of a NON-centered Gaussian random variable (or heat kernel) with mean $x$ and variance $2t$, (i.e. if $X\sim N(x,2t)$ then $E(X^2) = Var(X) + E(X)^2 = 2t+x^2$.

Finally $$\int_{\mathbb{R}}e^{-\sigma^2 /(4t)}d\sigma = \sqrt{4\pi t} \int_{\mathbb{R}} \frac{1}{\sqrt{4\pi t}} e^{-\sigma^2 /(4t)}d\sigma = \sqrt{4\pi t}$$ since the integral is 1.

Altogether, as before, $$\frac{1}{\sqrt{4\pi t}} \int_{\mathbb{R}} (\sigma^2 + 2\sigma x + x^2-1) e^{-\sigma^2 /(4t)}d\sigma = 2t + x^2 - 1.$$