Diverge?

thanks (Rock)

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- September 28th 2008, 05:58 PMNachoImproper integral
Diverge?

thanks (Rock) - September 28th 2008, 06:17 PMKrizalid
It does converge. (Not absolutely, of course.)

- September 28th 2008, 06:23 PMNacho
- September 29th 2008, 11:59 AMMoo
Hello,

I'll try the proof I read a time ago...

Let

Integrate by parts with :

and . So we have and , and we'll take c=1. So

But we know by Taylor series that when x is near from 0.

So we can make a=0 since and have finite limits when

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See

By extension, the integrand is continuous for all x in [0,1] and finite. Thus the integral exists and it is finite value. So it doesn't intervene in the convergence or divergence of

(actually, you have to check this explanation...it's the sort of parts in which I am always wrong)

See

This obviously converges.

See

This converges by comparison with the Riemann integral, since :

. - September 29th 2008, 07:50 PMNacho
thanks!! (Clapping)