Integration by Parts help?

Question

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

Answer 1

I will ignore the first factor $\frac{1}{\sqrt{4\pi k t}}$ for the moment. You just need to split the integral in summands like this:

$$I= \int_{\mathbb{R}} \sigma^2 e^{-\sigma^2/(4kt)} d\sigma$$ and $$II= 2x \int_{\mathbb{R}} \sigma e^{-\sigma^2/(4kt)} d\sigma$$ and $$III= x^2\int_{\mathbb{R}}e^{-\sigma^2/(4kt)} d\sigma.$$ Then factor $I$ can be rewritten as:

$$I= \sqrt{2\pi (2kt)} \int_{\mathbb{R}} \sigma^2 \frac{1}{\sqrt{2\pi (2kt)}} e^{-\sigma^2/(4kt)} d\sigma = \sqrt{2\pi (2kt)} * (2kt) $$ where the last equality follows because the integral appearing above is the second order moment of a centered Gaussian random variable with variance $2kt$.

The second is zero since it is the integral of an odd function over a symmetric region (also the mean of a centered normal random variable) $$II=0$$ Finally, $$III=\sqrt{2\pi (2kt)} x^2 \int_{\mathbb{R}}\frac{1}{\sqrt{2\pi (2kt)}} e^{-\sigma^2/(4kt)} d\sigma = \sqrt{2\pi (2kt)} x^2. $$

Now if you put altogether you get $$\frac{1}{\sqrt{4\pi k t}}\int_{\mathbb{R}} e^{-\sigma^2/(4kt)}(\sigma^2+2\sigma + x^2) d\sigma = \frac{1}{\sqrt{4\pi k t}}(\sqrt{2\pi (2kt)} * (2kt) +0+x^2\sqrt{2\pi (2kt)})= 2kt + x^2.$$