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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.

1 Answers 1

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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.

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    This would be perfect if the people that are taking the course at the moment learned Taylor series. They aren't that far yet. So far it's just been volumes, area between curves, separable/linear ODE's, convergence/divergence of a series and Integration techniques. – 2017-02-01
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    Never mind. I forgot I can use the comparison test for integrals for this. Thanks anyways. – 2017-02-01