Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

2 Answers 2

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.

share|improve this answer

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, $$I \geq \int_0^{\pi} \dfrac{\sqrt{\pi-x}}{x}dx \geq \int_0^{\pi/2} \dfrac{\sqrt{\pi-x}}{x}dx \geq \int_0^{\pi/2} \dfrac{\sqrt{\pi/2}}xdx = \infty$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.