# Mathmetics in Introductory Quantum Mechanics book

1. Oct 5, 2011

### rar0308

I'm reading Introductory quantum mechanics written by liboff.
When I solve problems, I stuck with calculation such as Intergral(-inf to +inf) e^(-x^2)dx, Intergral(-inf to +inf) (x^2) e^(-x^2)dx, and other many integrals.
I studied thomas' calculus but I think I haven't seen these in the book. So can't do it. Are there math books about these integrals? What math subject is related to these (high level?)integrals?

Last edited: Oct 5, 2011
2. Oct 5, 2011

### Staff: Mentor

These are examples of Gaussian integrals:

http://en.wikipedia.org/wiki/Gaussian_integral

I don't recall seeing them in any of my undergraduate math courses. I think most physics students (in the USA at least) learn about them in their physics courses.

Last edited: Oct 5, 2011
3. Oct 5, 2011

### dextercioby

<Intergral(-inf to +inf) e^(x^2)dx> and the other one. I think you meant them with a minus e^(-x^2).

It depends on the curricula. An advanced course of calculus is normally taken before quantum mechanics or statistical mechanics.

4. Oct 5, 2011

### HallsofIvy

The first one, Integral(-inf to +inf) e^(x^2)dx, is not in Thomas because it does not converge. But $\int_{-infty}^\infty e^{-x^2}dx$, as dextercioby suggests, certainly is in Thomas, and the second is a variation. You may be trying to find an anti-derivative formula and not finding that- neither integrand has an "elementary" anti-derivative. Both are used extensively in probability ($e^{-x^2}$ is the "bell shaped curve") and so in quantum mechanics.

Here is a simple way to get the first integral:
Let $I= \int_{-\infty}^\infty e^{-x^2}dx$. Since the integrand is symmetric about x=0, we also have $I/2= \int_0^\infty e^{-x^2}dx$.

And, we can write $I/2= \int_0^\infty e^{-y^2}dy$. Multiplying those together, $I^2/4= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right)$. By Fubini's theorem, we can write that product of integrals as a double integral:
$$I^2/4= \int_{x= 0}^{\infty}\int_{y=0}^\infty e^{-(x^2+ y^2}dydx$$

Now, change to polar coordinates: $x^2+ y^2= r^2cos^2(\theta)+ r^2sin^2(\theta)= r^2 and [itex]dydx= r drd\theta$. The area of integration, with both x going from 0 to infinity is the first quadrant. In polar coordinates, r goes from 0 to infinity while $\theta$ goes from 0 to $\pi/2$. The integral becomes
$$I^2/4= \int_{\theta= 0}^{\pi/2}\int_{r=0}^\infty e^{-r^2} rdrd\theta= \frac{\pi}{2}\int_0^\infty e^{-r^2} rdr$$

That extra 'r' in the integrand now allows us to make the change of variable $u= r^2$ so $du= 2r dr$ and the integral becomes
$$I^2/4= \frac{\pi}{4}\int_0^\infty e^{-u}du$$
which is easy.

It was necessary, to make that change to polar coordinates, that the integral be over the entire first quadrant. Again, that function, $e^{-x^2}$, has no elementary anti-derivative. In fact, its anti-derivative is typically written "erf(x)", the "error function", and it values are got by a numerical integration.

Last edited by a moderator: Oct 5, 2011
5. Oct 5, 2011

### rar0308

Thank you very much for your helps and I'm glad to finally locate it.
Trying to find that integral, I have flipped over pages of thomas so many times.
Even I borrowed from library advanced calculus written by bucks. I found gamma function and Integral(-inf to +inf) e^(-x^2)dx in this book. Now i'm about to read both of them. Thanks again.
P.S. I'm curious about how to input mathematical notations at the post.

6. Oct 5, 2011