# Convergence Problem... Problems

• Apr 21st 2010, 07:00 PM
scg780
Convergence Problem... Problems
Currently taking integral calculus and I have a problem that I have no idea how to approach.

Given the integral from 1 to infinity of dx/(x^p *(ln(x))^a)

- For what values of p and a does this converge

Even if you cant get the full solution, any thoughts on how to approach it would be greatly appreciated.
• Apr 22nd 2010, 12:19 AM
CaptainBlack
Quote:

Originally Posted by scg780
Currently taking integral calculus and I have a problem that I have no idea how to approach.

Given the integral from 1 to infinity of dx/(x^p *(ln(x))^a)

- For what values of p and a does this converge

Even if you cant get the full solution, any thoughts on how to approach it would be greatly appreciated.

For this integral to converge requires that the limit:

$\displaystyle \lim_{u\to \infty} \int_1^u\frac{1}{x^p (\ln(x))^a}\;dx$

exists. Now as $\displaystyle \ln(x)$ increases more slowly than any power of $\displaystyle x$ this is equivalent to asking for what values of $\displaystyle p$ does:

$\displaystyle \lim_{u\to \infty} \int_1^u\frac{1}{x^p}\;dx$

converge (with $\displaystyle p=1$ possibly being a special case).

CB
• Apr 22nd 2010, 06:44 AM
scg780
ok. Thank you. but i'm still confused on what the value of "a" would be
• Apr 22nd 2010, 12:46 PM
chisigma
Integrating by parts we obtain...

$\displaystyle \int \frac{dx}{x^{p}\cdot \ln^{a} x} = \frac{1}{1-p} \cdot \frac{1}{x^{p-1}\cdot \ln^{a} x} + \frac{a}{1-p} \int\frac{dx}{x^{p}\cdot \ln ^{a+1} x}$ (1)

... and then with another integration by parts...

$\displaystyle \int \frac{dx}{x^{p}\cdot \ln^{a} x} = \frac{1}{1-p} \cdot \frac{1}{x^{p-1}\cdot \ln^{a} x} + \frac{a}{(1-p)^{2}} \cdot \frac{1}{x^{p-1}\cdot \ln^{a+1} x} +$

$\displaystyle + \frac{a\cdot (a+1)}{(1-p)^{2}}\cdot \int \frac{dx}{x^{p}\cdot \ln^{a+2} x}$ (2)

If now we proceed indefinitely we obtain...

$\displaystyle \int \frac{dx}{x^{p}\cdot \ln^{a} x} = \frac{1}{1-p} \cdot \frac{1}{x^{p-1}\cdot \ln^{a} x} + \frac{a}{(1-p)^{2}} \cdot \frac{1}{x^{p-1}\cdot \ln^{a+1} x} + \dots$

$\displaystyle \dots + \frac{a\cdot (a+1) \dots (a+n-1)}{(1-p)^{n}}\cdot \frac{1}{x^{p-1}\cdot \ln^{a+n} x} + \dots$ (3)

At this point a careful investigation of (3) permits us to conclude that the [improper] integral...

$\displaystyle \int_{1}^{\infty} \frac{dx}{x^{p}\cdot \ln^{a} x}$ (4)

... converges if...

a) $\displaystyle p>1$

b) $\displaystyle a$ is a non positive integer...

If $\displaystyle a= -k, k\ge 0$ is ...

$\displaystyle \int_{1}^{\infty}\frac{\ln^{k} x}{x^{p}} = \frac{k!}{(p-1)^{k+1}}$ (5)

The convergence in the case $\displaystyle a<0$ but non integer can be easily verified by comparison...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$