Convergence test

**Wolvenmoon** · March 1st 2011, 05:10 PM

Hi,

I have to test the integral from 0 to pi of sin(x)/sqrt(pi-x) for convergence using the direct comparison test or the limit comparison test.

I know I can't integrate so I have to find a function that is either always greater than sin(x)/sqrt(pi-x) or do the limit comparison test on it.

I'm not quite sure how to approach this one.

**FernandoRevilla** · March 1st 2011, 07:17 PM

Is it compulsory to use those test?. Take into account that:

$\displaystyle\lim_{x \to \pi^-}{\dfrac{\sin x}{\sqrt{\pi -x}}}=\ldots=0$

so, we can expand by continuity the function taking $f(\pi)=0$ hence, the integral is convergent.

Alternatively:

$0\leq \dfrac{\sin x}{\sqrt{\pi -x}}\leq \dfrac{1}{\sqrt{\pi -x}}$

etc.

**Wolvenmoon** · March 1st 2011, 07:35 PM

Well, I think as long as we can prove it he won't care, but that's a guess.

Also, I didn't tackle the problem as a whole...sin(x)<1, always, and I forgot that.

Thanks!

**FernandoRevilla** · March 1st 2011, 07:59 PM

Originally Posted by Wolvenmoon

Well, I think as long as we can prove it he won't care, but that's a guess.

If so, the former is more elegant.

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