Improper Integral/ Converge-Diverge.

**CHAMPAN** · April 6th 2010, 10:23 PM

Hello All,

I am trying to figure out how to solve the Improper Integral and then decide if it converges or diverges.

Upper limit 2, lower limit 0 (x-1)^-1/3

Any help you can give me would be greatly appreciated.

Thanks!

**CaptainBlack** · April 6th 2010, 10:38 PM

Originally Posted by CHAMPAN

Hello All,

I am trying to figure out how to solve the Improper Integral and then decide if it converges or diverges.

Upper limit 2, lower limit 0 (x-1)^-1/3

Any help you can give me would be greatly appreciated.

Thanks!

Put:

$I_{a,b}= \int_0^a (x-1)^{-1/3}\; dx+ \int_b^2 (x-1)^{-1/3}\; dx,\ a<1,\ b>1$

For the integral to converge requires that:

$I=\lim_{a\to 1_-, b \to 1_+}I_{a,b}$

exists, which is eqivalent to requiring that:

$\lim_{a \to 1_-}\int_0^a (x-1)^{-1/3}\; dx$

and

$\lim_{b \to 1_+}\int_b^2 (x-1)^{-1/3}\; dx$

both exist.

CB

**CHAMPAN** · April 6th 2010, 10:55 PM

Thanks for the explanation, do you happen to know how to tell where it converges?

Thank you again for your help! It is greatly appreciated.

**CaptainBlack** · April 7th 2010, 02:32 AM

Originally Posted by CHAMPAN

Thanks for the explanation, do you happen to know how to tell where it converges?

Thank you again for your help! It is greatly appreciated.

Where it converges? I'm not sure I know what you mean. Both the constituent limits exist so the improper integral converges (the only point that there is a problem is around x=1 where the integrand misbehaves, which is why that is the point about which we break up the integral)

CB

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