# Thread: Comparison Test for Improper Integrals

1. ## Comparison Test for Improper Integrals

Can someone help explain this problem to me? I'm not sure how i am able to find what P can be so that the integral converges.

2. ## Re: Comparison Test for Improper Integrals

$\ln x\geq 1$ for $x\geq 3$ so $\frac{1}{\ln x}\leq 1$

3. ## Re: Comparison Test for Improper Integrals

but, it is asking for a value of P? I understand what you mean since you found the comparison. But, im not understanding what P stands for

4. ## Re: Comparison Test for Improper Integrals

Idea showed that $\displaystyle \int_8^\infty { \dfrac{1}{x^p\ln x}dx } \le \int_8^\infty {\dfrac{1}{x^p}dx}$

You should be able to calculate the latter integral directly. It only converges for certain values of p.

5. ## Re: Comparison Test for Improper Integrals

Hi, thanks for the response. I just got around to check the page, but I was wondering where the x>=3 comes from? Also, I'm kinda new to the whole p-integral concept/comparison test. My professor kinda rush this part of the lecture so quick that i didn't fully grasp the idea. I only understood the improper integrals and how to evaluate them.

I also know the meaning of convergent and divergent. But, im not sure how the comparison test helps here

6. ## Re: Comparison Test for Improper Integrals

Okay, so i evaluated that latter integral. At the end, i got the lim as c goes to infinity of 1/ (c^(p-1)) - 1/(8^(p-1))

I think p must be 0 for the function to converge, since the limit would exist when p=0. But, is that all of the p values?

7. ## Re: Comparison Test for Improper Integrals

Originally Posted by lc99
Okay, so i evaluated that latter integral. At the end, i got the lim as c goes to infinity of 1/ (c^(p-1)) - 1/(8^(p-1))

I think p must be 0 for the function to converge, since the limit would exist when p=0. But, is that all of the p values?
Not quite...

If $p=0$, then you have $\dfrac{1}{c^{-1}} = c$ which approaches infinity (does not converge).

In fact, for $p<1$, you have the final exponent for $c$ is positive, so as $c$ approaches infinity, you get infinity. Suppose $p=1$. Then your integral becomes $\displaystyle \int_8^\infty \dfrac{dx}{x} = \lim_{c\to \infty} \ln c = \infty$. So, even when $p=1$, it still diverges.

You need $p>1$ or $p \in (1,\infty)$.

The question becomes, what about $\displaystyle \int_8^\infty \dfrac{dx}{x\ln x}$ (where p=1 for the original integral). That is a straightforward integration by substitution problem.

Let $u = \ln x$. Then $du = \dfrac{dx}{x}$. Your bounds of integration change to $u(8) = \ln 8$ and $\lim_{x \to \infty} u(x) = \infty$:

$\displaystyle \int_{\ln 8}^\infty \dfrac{du}{u} = \left. \ln u \right]_{\ln 8}^\infty = \infty$ (does not converge).

This proves that the original integral does not converge for $p \le 1$

8. ## Re: Comparison Test for Improper Integrals

I'm sorry... i keep typing the opposite of what's in my mind when im on here!! I meant to say p=1, which is still wrong. But, your explanation really simplifies things for me. Thank you.

Although i do have one question.. it seems to me that you plugged in a p-value before integrating. However, i kept the p-value and integrated. And, the limit i replied you with was what i got... But, when i plugged in p=1, i get lim c-> infinity of 1/c^0 -(1/8^0) which gives the value 0.
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