# Cosine evaluated at infinity

1. Sep 21, 2015

### maverick_76

okay so I have this integral:

2∫cos(kx)dk

The bounds are from zero to infinity, this doesn't converge but is there any other way to describe this?

2. Sep 21, 2015

### Staff: Mentor

Like this? $2\int_0^{\infty}\cos(kx) dk$
Because the upper limit of integration is ∞, this is an improper integral. You can't just "plug in" ∞ when you evaluate the antiderivative -- you need to use limits to evaluate it.

3. Sep 21, 2015

### maverick_76

okay so the limit would be lim_k->∞ sin(kx)
x

How would I go about evaluating it? Is there a substitution trick?

4. Sep 21, 2015

### Staff: Mentor

The limit doesn't exist because sin(kx) oscillates endlessly.

5. Sep 21, 2015

### maverick_76

okay gotcha. Thanks!