1. ## Heat diffusion, Gaussian

Hi,
I need to evaluate intergral (-inf, inf) of S(x-y, t)dx where y element R and where s(x, t) = (1/2sqrt(pi*k*t))e^((-x^2)/4kt). You are supposed to use the fact that intergral (-inf, inf) e^-ax^2dx = sqrt(pi/a). I know there is some trick you need to do at the beginning but not sure what so that it can be simplified easily.
Thanks

2. Originally Posted by tbyou87
Hi,
I need to evaluate intergral (-inf, inf) of S(x-y, t)dx where y element R and where s(x, t) = (1/2sqrt(pi*k*t))e^((-x^2)/4kt). You are supposed to use the fact that intergral (-inf, inf) e^-ax^2dx = sqrt(pi/a). I know there is some trick you need to do at the beginning but not sure what so that it can be simplified easily.
Thanks
Let me attempt to translate this into LaTeX:
$\displaystyle \int_{-\infty}^{\infty}\frac{1}{2}\sqrt{\pi k t}~e^{-(x - y)^2/4kt}dx$

using
$\displaystyle \int_{\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}$

-Dan

3. Originally Posted by topsquark
Let me attempt to translate this into LaTeX:
$\displaystyle \int_{-\infty}^{\infty}\frac{1}{2}\sqrt{\pi k t}~e^{-(x - y)^2/4kt}dx$

using
$\displaystyle \int_{\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}$

-Dan
It looks to me like you can do the following:
$\displaystyle \int_{-\infty}^{\infty}\frac{1}{2}\sqrt{\pi k t}~e^{-(x - y)^2/4kt}dx = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-(x - y)^2/4kt}dx$
(since t is constant with respect to the integral.)

Now let $\displaystyle u = x - y \implies du = dx$
$\displaystyle = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-u^2/4kt}du$

and temporarily set $\displaystyle a = \frac{1}{4kt}$:
$\displaystyle = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-au^2}du$

$\displaystyle = \frac{1}{2}\sqrt{\pi k t} \int_{-\infty}^{\infty}e^{-au^2}du$

$\displaystyle = \frac{1}{2}\sqrt{\pi k t} \sqrt{4 \pi kt}$

$\displaystyle = \pi kt$

But it worries me slightly that this is no longer a function of y.

-Dan