# Fourier Series of Secant

1. Feb 15, 2013

### mathskier

So I know that sec(x) has period 2Pi, and it's even so I don't need to figure out coefficients for bn.

Let's take the limits of the integral to go from -3/2 Pi to 1/2 Pi. How do I integrate sec(x) sin(nx) dx?! Am I on the right path?

PS: I know that this doesn't satisfy the Dirichlet Theorem, but the textbook I'm using still says to compute it.

2. Feb 16, 2013

### haruspex

Seems to me that integral is zero for n even and indeterminate for n odd.

3. Feb 16, 2013

### rbj

well since $\sec(x) = 1/\cos(x)$ then i think you can express the cosine functions, both on the top and on the bottom as exponentials with imaginary argument. i think you can get to an integral that looks like

$$c_n = \frac{1}{2 \pi} \int_{-\pi}^{+\pi} \frac{2}{e^{i (n+1) x}+e^{i (n-1) x}} \ dx$$

i think you can find an antiderivative of that. might still come out as indeterminate. but that's the integral you have to solve.

Last edited: Feb 16, 2013
4. Feb 16, 2013

### vela

Staff Emeritus
I evaluated the first few odd n integrals using Mathematica, and it looks like $a_{2n+1} = (-1)^n 2\pi$.

5. Feb 16, 2013

### haruspex

The reason I thought it indeterminate is that integrating over a 2π range effectively cancels a +∞ with a -∞. E.g. with n=1, it's the integral of tan, which is log cos. An interval that spans π/2 will appear to give a sensible answer, but in reality it's undefined.

6. Feb 16, 2013

### vela

Staff Emeritus
mathskier indicated the wrong integrand. It should be sec x cos(nx), which is the integrand I used.

7. Feb 16, 2013

### haruspex

That certainly makes a difference. It's not hard to show that the integral of sec x cos(nx) over a 2pi range is the negative of that of sec x cos((n-2)x), for n > 2.