Question

How do you find the integral of `int (t^2)*sin 3t dt` from negative infinity to infinity?

Answer 1

`int_(-oo)^(oo) t^2 sin 3t dt` does not converge,

despite being the integral of an odd function.

Explanation:

`d/dt (-1/3t^2 cos 3t) = t^2 sin 3t - 2/3 t cos 3t`

`d/dt (2/9 t sin 3t) = 2/3 t cos 3t + 2/9 sin 3t`

`d/dt (2/27 cos 3t) = -2/9 sin 3t`

So

`int t^2 sin 3t dt = -1/3 t^2 cos 3t + 2/9 t sin 3t + 2/27 cos 3t + C`

Hence:

`int_0^x t^2 sin 3t dt = -1/3 x^2 cos 3x + 2/9 x sin 3x + 2/27 cos 3x - 2/27`

In particular:

`int_0^(2n pi)t^2 sin 3t dt = -1/3(2n pi)^2 -> -oo` as `n -> oo`

Also

`int_(-2n pi)^0 t^2 sin 3t dt = 1/3(2n pi)^2 -> oo` as `n -> oo`

So `int_(-oo)^(oo) t^2 sin 3t dt` does not converge.