So I have to determine if this integral converges or diverges:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} {\frac{1}{x\sin(x)}} dx$
So I can break u the integral at the discontinuity as such...
$\int_{-\frac{\pi}{2}}^{0} {\frac{1}{x\sin(x)}} dx+\int_{0}^{\frac{\pi}{2}} {\frac{1}{x\sin(x)}} dx$
I then tried to use Weierstrass Substitution so...
$\sin(x)=\frac{2t}{1+t^2}$ and $dx=\frac{2}{1+t^2} dt$
But then I have a problem. I end up with...
$\int {\frac{1}{xt}} dt$ because I can't figure out how to get rid of that pesky x. I have no idea how to approach it from there...
Any guidance would be appreciated.
Remark: generally, if you are asked to prove if an integral converges/diverges, most of the times it won't have a nice analytic closed form (as otherwise the question would be trivial).
Basically: Note that the function is even, so your integral equals $$2\cdot\int_0^{\pi/2}\frac{1}{x\sin(x)}\,dx.$$ Now Taylor-expand sine around $0$ and get an integral of the form $$\sim\int_0^\epsilon\frac{1}{x^2}\, dx,$$ which diverges. As the remaining part of the integral is positive, the whole integral diverges.