# Guassian Intergral doesn't seem to sum to sqrt(pi)

1. Apr 9, 2012

### thomas49th

1. The problem statement, all variables and given/known data
http://gyazo.com/93ec8b11261e3a00da976a0e28855960

2. Relevant equations
According to
http://en.wikipedia.org/wiki/Gaussian_function

3. The attempt at a solution
I have got as far as the integral at:

$$e^{-\frac{w^{2}}{4} } \int e^{-(t+ \frac{jw}{2})^{2}}dt$$

Apparently this makes the correct answer but I don't see how the integral evaluates to sqrt(pi)

Thanks
Thomas

Last edited: Apr 9, 2012
2. Apr 9, 2012

### Ray Vickson

Your integral is wrong; you have written
$$e^{-\frac{w^{2}}{4}\int e^{-(t^{2}+j\frac{w}{2})^{2}}},$$
whiich is not what the cited article wrote.

RGV

3. Apr 9, 2012

4. Apr 9, 2012

### thomas49th

thanks*

Last edited: Apr 10, 2012
5. Apr 10, 2012

### thomas49th

Maybe i wasn't very clear. I cannot see how

$$\int e^{-(t+ \frac{jw}{2})^{2}}dt$$

Becomes sqrt(pi). I can't be just sqrt(pi), as we don't just have the integral of e^(t^2))

6. Apr 10, 2012

### HallsofIvy

Staff Emeritus
$$\int_{-\infty}^\infty e^{-(t+ j\omega/2)^2}dt$$

Let $u= t+ j\omega^2/2$. $du= dt$. The integral becomes
$$\int_{-\infty}^\infty e^{-u^2}{du}$$

That's a well known integral. If you haven't seen it before, let
$$I= \int_{-\infty}^\infty e^{-u^2}{du}$$
so that, because of the symmetry,
$$I/2= \int_0^\infty e^{-u^2}{du}$$
Then it is also true that
$$I/2= \int_0^\infty e^{-v^2}dv$$
and, multiplying them
$$I^2/4= \left(\int_0^\infty e^{-u^2}du\right)\left(\int_0^\infty e^{-v^2}dv\right)= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-u^2-v^2}dudv$$

Now change to polar coordinates, r and $\theta$, with $u= r cos(\theta)$ and $v= r sin(\theta)$. The uv-integral takes each variable from 0 to $\infty$, the first quadrant, which, in terms of polar coordinates, has r from 0 to $\infty$ and $\theta$ from 0 to $\pi/2$. Of course, $-(u^2+v^2)=-r^2$ and $dudv= r drd\theta$

The integral becomes
$$\int_{\theta= 0}^{\pi/2}\int_{r=0}^\infty e^{-r^2}rdrd\theta= \frac{\pi}{2}\int_{r=0}^\infty e^{-r^2}rdr$$

To do that last integral, let $x= r^2$ so that dx= 2rdr and rdr= (1/2)dx.
Now we have
$$I^2/4= \frac{\pi}{4}\int_0^\infty e^{-x}dx$$

That last integral is, of course, 1 so we have $I^2/4= \pi/4$ and $I= \sqrt{\pi}$.

7. Apr 14, 2012

### thomas49th

nice going HallOfIvy, very good :)