Using Comparison Theorem to solve a problem

[calculus_guy4]

Homework Statement

Use the Comparison Theorem to evaluate whether Integral(dx/(x*sinx)) on 0->pi/2 converges or diverges.

The Attempt at a Solution

I don't understand what to do about 0. Am I allowed to compare it to 1/x even though 1/x is undefined at 0? Like-wise am I allowed to compare it to 1/sinx even though 1/sinx is undefined at 0?

A related question I had was how do I use comparison theorem for trig functions? Since they are periodic can you not use it?

Thanks.

 

[lurflurf]

You can compare it to either.
x~sin(x)
I would write x sin(x)=x^2 [sin(x)/x] and compare it to x^2.

 

[calculus_guy4]

I still don't quite understand. So I need a function that converges that is bigger than the original function in order to prove the original function converges or a function smaller than the original function that diverges to prove that the original function diverges. Is that correct? So is 1/x^2 bigger or smaller than 1/xsin(x) on 0->pi/2 -- how do I figure out which function is bigger on my own? Do I just take random values and check? Also it is alright to compare to 1/x^2 even though 1/x^2 is undefined at 0?

 

[lurflurf]

The actual value at zero (or some other point in other problems) does not matter, it is the behavior near the point. The function is clearly large near zero, but wha matter is how fast it is growing. We know since sin'(0)=1 that sin(x) acts like x for small x. In particular for example x cos(x)<sin(x)<x. So we can also say 1/x^2<1/(x sin(x)) and thus the integral of 1/(x sin(x)) can converge only if the integral of 1/x^2 does. So does the integral of 1/x^2 converge?

 

[calculus_guy4]

"the integral of 1/(x sin(x)) can converge only if the integral of 1/x^2 does. So does the integral of 1/x^2 converge? "

Since 1/x^2 is smaller than 1/xsinx, if you compare the two functions all you can say is that 1/xsinx diverges if 1/x^2 diverges, right? You can't say anything about convergence since you would need to compare 1/xsinx to a bigger function. But since 1/x^2 converges you still don't know anything about 1/xsinx. So shouldn't I compare 1/xsinx to something else? Sorry, I still don't quite understand.

 

[LCKurtz]

"the integral of 1/(x sin(x)) can converge only if the integral of 1/x^2 does. So does the integral of 1/x^2 converge? "

Since 1/x^2 is smaller than 1/xsinx, if you compare the two functions all you can say is that 1/xsinx diverges if 1/x^2 diverges, right?

Yes. And does the integral of 1/x² diverge?