First, it would be good to know that the area bounded by the curve and the x-axis is convergent to begin with.

Let

, so that

, and the integral is equivalent to

The integrand is continuous everywhere except

, but that's okay because we have

. This means the integral is convergent - great! (Moreover, there's a special function designed to handle this sort of integral, aptly named the "sine integral function".)
Now, to compute the volume. Via the disk method, we have a volume given by the integral

By the same substitution as before, we can write this as

The half-angle identity for sine allows us to rewrite as

and replacing

,

, we have

Like the previous, this require a special function in order to express it in a closed form. You would find that its value is

where

is the Euler-Mascheroni constant and

denotes the cosine integral function.