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Test the convergence of $$\int_0^\pi{\sqrt x\over \sin x}dx$$

I have to do it using comparison test. There is another test mentioned called the $\mu-test$ but its definition in the book doesn't seem right, cannot find it on the internet too. Is someone familiar with it ? It is a corollary of comparison test.

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Since $\dfrac{\sqrt{x}}{\sin x} \sim \dfrac{\sqrt{\pi}}{\pi-x}, x \to \pi$ the integral diverges. – njguliyev Oct 9 '13 at 17:46

HINT Close to zero the integrand behaves as

$$ \frac{\sqrt{x}}{x}$$

Since $\sin(x) \sim x$. Try to do similar analysis at the other end point.

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We have $$I = \underbrace{\int_0^{\pi} \dfrac{\sqrt{x}}{\sin(x)}dx = \int_0^{\pi} \dfrac{\sqrt{\pi-x}}{\sin(x)}dx}_{x \mapsto \pi-x}$$ Also, $\sin(x) \leq x$. Hence, $$I \geq \int_0^{\pi} \dfrac{\sqrt{\pi-x}}{x}dx \geq \int_0^{\pi/2} \dfrac{\sqrt{\pi-x}}{x}dx \geq \int_0^{\pi/2} \dfrac{\sqrt{\pi/2}}xdx = \infty$$

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