# Orthogonal Integral

1. Jun 29, 2008

### nicksauce

On page 102 of Introduction to Quantum Mechanics, Griffiths writes that
$$\int_{-\infty}^{\infty}e^{i\lambda x}e^{-i \mu x}dx = 2\pi\delta(\lambda-\mu)$$

I don't see how this follows. If you replace lambda with mu, then you get a cancellation, and the integral of 1dx. Am I missing something?

2. Jun 30, 2008

### Fredrik

Staff Emeritus
I think you would have to know a bunch of stuff about distributions and integration theory to understand that. I can't say that I do, so I can only give you a very non-rigorous argument based on some ugly manipulations with Fourier transforms:

We know that if f is a nice enough function, it has a Fourier transform g that can be defined by

$$g(p)=\frac{1}{\sqrt{2\pi}}\int f(y)e^{-ipy}dy$$

The theorem that guarantees that also says that f can now be expressed as

$$f(x)=\frac{1}{\sqrt{2\pi}}\int g(p)e^{ipx}dp$$.

If we insert the first expression into the second and recklessly switch the order of the integrations, we get

$$f(x)=\int f(y)\bigg(\frac{1}{2\pi}\int e^{ip(x-y)}dp\bigg)dy$$

So if there's some way to justify switching the order of the integrations (which would have to include a new definition of what we mean by an "integral"), the expression in parentheses must be $\delta(x-y)$.

3. Jun 30, 2008

### nicksauce

Okay that (sort of) makes sense. Thanks!

4. Jun 30, 2008

### Marco_84

TO be more rigorous you should introduce delta as generalized function and then use a limiting procedure on test-function space D!