Integral convergence

[shakalandro]

My task is to use the ratio test for impoper integral to show that $\displaystyle \int_{0}^{1}{\frac{-logx}{\sqrt{x}}}dx$ is convergent. But I cannot find the g(x) that makes it work.

 

[Deleted member 4993]

My task is to use the ration test for impoper integral to show that $\displaystyle \int_{0}^{1}{\frac{-logx}{\sqrt{x}}}dx$. But I cannot find the g(x) that makes it work.

Show what???

 

[shakalandro]

oops sorry, show that it is convergent.

 

[BigGlenntheHeavy]

I'm not sure what you are implying skakalandro, but here is how I would prove that the improper integral converges.

$\displaystyle We \ have \ \int_{0}^{1}\frac{-ln|x|dx}{\sqrt x}$

This integral is improper as we have an infinite discontinuity at x = 0.

$\displaystyle Hence, \ \int_{0}^{1}\frac{-ln|x|dx}{\sqrt x} \ = \ lim_{t\to0^{+}} \int_{t}^{1}\frac{-ln|x|dx}{\sqrt x}$

$\displaystyle = \ \ lim_{t\to0^{+}} -4\int_{t}^{1}ln|u|du, \ letting \ u \ = \ x^{1/2}, \ u^{2} \ = \ x, \ and \ 2udu \ = \ dx.$

$\displaystyle Ergo \ lim_{t\to0^{+}}-4[uln|u|-u]_{t}^{1} \ = \ lim_{t\to0^{+}}-4[(0-1)-[(tln|t|)-t]]$

$\displaystyle Now, \ tln|t| \ gives \ the \ indeterminate \ form \ of \ 0 \ times \ -infinity, \ therefore \ L' Hopital.$

$\displaystyle tln|t| \ = \ \frac{ln|t|}{t^{-1}} \ =H= \ -\infty/\infty \ = \ \frac{1/t}{-t^{-2}} \ = \ lim_{t\to0^{+}}(-t) \ = \ 0.$

$\displaystyle Therefore, \ \int_{0}^{1}\frac{-ln|x|dx}{\sqrt x} \ = \ 4, \ hence \ converges.$

Note: Any logarithm will do as long as its base is greater than 1.

 

[shakalandro]

Firstly, I know that their are two problems with the integral it has an infinite discontinuity at 0 in the numerator and a singularity in the denominator. I need to find some g(x) that whose integral from 0 to 1 is convergent and f(x) < g(x) for all x. Alternatively g(x) could be such that $\displaystyle \lim_{x \rightarrow 0}{\frac{f(x)}{g(x)}} = L$ where l is nonzero and finite.