Question

How do you find the integral of `int sin(6x) dx` from negative infinity to infinity?

Answer 1

That integral does not converge (it diverges).

Explanation:

`int_-oo^oo sin(6x) dx = int_-oo^0 sin(6x) dx + int_0^oo sin(6x) dx`

Provided that both integrals on the right converge.

Let's look first at the integral on the positives.

`int_0^oo sin(6x) dx = lim_(brarroo)int_0^b sin(6x) dx`

`= lim_(brarroo) [-1/6cos(6x)]_0^b`

`= lim_(brarroo) [-1/6cos(6b)+1/6cos(0)]`

However `lim_(brarroo) [-1/6cos(6b)]` does not exist, so the integral on the positives diverges and the big integral (on `(-oo,oo)`) also diverges.