CONVERGENCE OF IMPROPER INTEGRAL

[Maryem Khalis]

The question is to test the convergence of this improper integral.



The integral is continuous so I calculated the left hand limit with a calculator an came to π so I know it's convergent. Calculating it by hand is possible but that wold be too long of a procedure so I tried the comparison test but I couldn't find any basic functions that would compare to this one and that I know are convergent/divergent.

I have the notes of my professor and she actually first used the p-Series test for convergence and because 2-ε>1 it converges.



Then she calculated the right hand limit on the integrand to see if it's continuous in t=0.



Then she concluded that the integral converges.

This is probably a dumb question but I can't seem to understand how she got the integrand to 1/t^2-ε or how it would behave like it. She didn't add the steps in between and I already tried multiple times and searched it on the internet but failed so this is my last my last resort.

Thank you in advance!

 

[ksdhart2]

I, too, am quite confused. What's written is here is simply not true. The supposition is that the given function f(t) "behaves like" \(\displaystyle \frac{1}{t^{2-\epsilon}}\) for every \( \epsilon > 0\). Immediately, I take some issue with this because "behaves like" is a somewhat ambiguous phrase and can mean multiple things. In this context, I suppose the most likely meaning would be "has the same limit at infinity and neither have any vertical asymptotes."

But one need only consider the case \(\epsilon = 2\) to see this is false, as the limit would be 1 (because \(\displaystyle \frac{1}{t^{2-2}} = \frac{1}{t^0} = 1\) everywhere). A deeper inspection shows that this is actually true for all \(\epsilon < 2\).

However, I think some of it can be salvaged. Have you learned about the comparison test for integrals yet? Essentially it says that if we have two functions \(0 \leq f(t) \leq g(t)\) on some interval \([a, b]\) then:

If \(\displaystyle \int\limits_{a}^{b} g(t) \: dt\) is convergent, then \(\displaystyle \int\limits_{a}^{b} f(t) \: dt\) is too.

If \(\displaystyle \int\limits_{a}^{b} f(t) \: dt\) is divergent, then \(\displaystyle \int\limits_{a}^{b} g(t) \: dt\) is too.

If we pick a specific value for \(\epsilon\), say \(\epsilon = 1\), then we can use the convergence test with \(\displaystyle f(t) = \frac{\ln(1+t^2)}{t^2}\) and \(\displaystyle g(t) = \frac{1}{t^{2-\epsilon}} = \frac{1}{t}\). First, check to see if \(0 \leq f(t) \leq g(t)\):

\(\displaystyle 0 \leq f(t) \implies \frac{\ln(1+t^2)}{t^2} > 0 \implies \ln(1+t^2) > 0\). Is this true for all \(t > 0\)? Why?

\(\displaystyle f(t) \leq g(t) \implies \frac{\ln(1+t^2)}{t^2} < \frac{1}{t} \implies t \cdot \ln(1+t^2) < t^2\). Is this true for all \(t > 0\)? Why?

Now that we know that condition is satisfied, it's only a matter of showing that \(\displaystyle \int\limits_{a}^{b} g(t) \: dt\) is convergent, but since \(\displaystyle g(t) = \frac{1}{t}\) you should be able to show that in a heartbeat.