# Thread: \int_{k}^{\infty} e^{-a^{2}} dy (gauss int. w. finite lim.)

1. ## \int_{k}^{\infty} e^{-a^{2}} dy (gauss int. w. finite lim.)

I am doing a calculation of overlap between wavefunctions in a physics thesis, however this integral puzzles me:

$\displaystyle \int_{-L/2+\kappa a^{2}}^{\infty} e^{-\frac{1}{2a^{2}}y^{2}} dy$

From a physic point of view it is very important that the integral is from $\displaystyle -L/2+\kappa a^{2}$ and not from $\displaystyle -\infty$. I am awere that is is impossible to write down an antiderivative of the function being integrated however, there must be some way to deal with the integral.

How do I calculate this integral (as a function of momentum $\displaystyle \kappa$)?

2. $\displaystyle D=-L/2+\kappa a^{2}$

$\displaystyle \int_{D}^{\infty} e^{-\frac{y^2}{2a^{2}}} dy$

$\displaystyle \int_{D}^{\infty} e^{-\left ( \frac{y}{a\sqrt{2}}\right )^2} dy$

Let $\displaystyle x =\frac{y}{a\sqrt{2}}$. Then $\displaystyle dy = a\sqrt{2}~dx$.

The bounds are now from $\displaystyle \frac{D}{a\sqrt{2}}$ to infinity.

It becomes $\displaystyle \int_{\frac{-L/2+\kappa a^{2}}{a\sqrt{2}}}^{\infty}e^{-x^2}~dx$.

$\displaystyle \int_{\frac{-L/2+\kappa a^{2}}{a\sqrt{2}}}^{\infty}e^{-x^2}~dx = \int_{0}^{\infty}e^{-x^2}~dx - \int_{0}^{\frac{-L/2+\kappa a^{2}}{a\sqrt{2}}}e^{-x^2}~dx$

$\displaystyle \int_{0}^{\infty}e^{-x^2}~dx$ is the Gaussian integral and $\displaystyle \frac{\sqrt{\pi}}{2}$

$\displaystyle \frac{\sqrt{\pi}}{2} - \int_{0}^{\frac{-L/2+\kappa a^{2}}{a\sqrt{2}}}e^{-x^2}~dx$

$\displaystyle \frac{\sqrt{\pi}}{2} - \frac{\sqrt{\pi}}{2} \mbox{Erf}(\frac{-L/2+\kappa a^{2}}{a\sqrt{2}})$

You have to use an approximation of the error function here.
An approximation of Erf is: $\displaystyle |\mbox{Erf}(x)| = \sqrt{1-e^{-(2x/\sqrt{\pi})^2}}$

Another one (and a better one, I think) is here: Error function - Wikipedia, the free encyclopedia

3. wingless, thanks a lot. I really appreciate it.

I've never come across error functions in this context if ever at all. Seems like I have only done easy -inf to inf gaussians in electro class.

4. Originally Posted by AA1983
wingless, thanks a lot. I really appreciate it.

I've never come across error functions in this context if ever at all. Seems like I have only done easy -inf to inf gaussians in electro class.
Not at all.. If you are looking for more about the error function, you can see the wikipedia thread