Testing For Convergence Notation

I was wondering that if I was testing for the convergence of the integral, $\int _{x=a}^b\:f\left(x\right)dx$ and the problem point was that the integrand was undefined at x = a.

Then, after the Taylor series, I saw that $f(x) \geq g(x)$ near x = a.

So, I can roughly say that $f(x) \sim g(x)$ as $x \rightarrow a+$.

Would this notation then be correct:

$\int _{x=a}^b\:f\left(x\right)\:dx\: \sim \:\int _{x=a}^b\:g\left(x\right)\:dx$ as $x \rightarrow a+$ and so if the integral $\:\int _{x=a}^b\:g\left(x\right)\:dx$ converges or diverges, then so does $\:\int _{x=a}^b\:f\left(x\right)\:dx$?

Thank you!

Not necessarily.

For that, you would require $f(x) \sim g(x)$ over the entire interval $(a,b)$ and not *just* in asymptotically as $x \to a_+$.

Oh, so would I just say that as f(x) behaves like g(x) near x = a+, then the convergence/divergence of the integral of f(x) is the same as the integral of g(x) and then use the Limit Comparison Test? I was just wondering so I am writing it correctly

Also, just wondering for the Direct Comparison Test, we usually say if $f(x) \leq g(x)$ or $f(x) \geq g(x)$. Does it still apply if we have a strict inequality?

The idea is that as $x \to a_+$, then at some point (let's call it $a + \epsilon$) $f(x)$ will be very well approximated by $g(x)$.

Thus, the integral can be split as

$\displaystyle \int_a^b f(x) dx = \int_a^{a+ \epsilon} f(x) dx + \int_{a+ \epsilon}^b f(x) dx$

where the second integral converges (hopefully this is obvious), and the divergence/convergence of the first integral is obtained by comparing it with

$\displaystyle \int_a^{a+ \epsilon} g(x) dx$


since over the region $(a,a+\epsilon)$ we have $f(x) \sim g(x)$ hence $\displaystyle \int_a^{a+ \epsilon} f(x) dx \sim \int_a^{a+ \epsilon} g(x) dx$




What do you reckon?

How often do you end up with the equality case when comparing things?

Oh right, now I understand. Thank you so much, that makes a lot of sense! So, we assume epsilon > zero and very small right, to show a + epsilon is close to a?
Yes, I think if we have a strict inequality, it holds even more and so must be correct.

Yep.