I need some help with an indefinite integral problem (only the $2^{\textrm{nd}}$ part though). Problem is as follows. Consider the function $f(x) = \dfrac{\ln\!\left(x\right)}{x^{p}}$, where $p>1$ and $1 \le a \le b$.
1. Compute $\displaystyle \int_{a}^{b} f(x) dx$ in terms of $p, a$ and $b$.
2. Now compute $\displaystyle \int_{a}^{\infty} f(x) dx$ in terms of $p$ and $a$ by letting $b \to \infty$ in your first answer.
The correct answer to the first one is:
$\dfrac{-\left(b^{1-p}\!\left(1+\left(-1+p\right)\ln\!\left(b\right)\right)\right)}{\left(-1+p\right)^{2}}-\dfrac{-\left(a^{1-p}\!\left(1+\left(-1+p\right)\ln\!\left(a\right)\right)\right)}{\left(-1+p\right)^{2}}$
I only have difficulty in getting the $2^{\textrm{nd}}$ one right. Any ideas?
By letting $b \to \infty$ I get $\lim_{b\to\infty}\displaystyle \int_{a}^{b} f(x) dx$ and having calculated $\displaystyle \int_{a}^{b} f(x) dx$ I got:
$\lim_{b\to\infty}(\dfrac{-\left(b^{1-p}\!\left(1+\left(-1+p\right)\ln\!\left(b\right)\right)\right)}{\left(-1+p\right)^{2}}-\dfrac{-\left(a^{1-p}\!\left(1+\left(-1+p\right)\ln\!\left(a\right)\right)\right)}{\left(-1+p\right)^{2}})$
From here i am stuck.
I worked it out the long way. I must say that it is a bit long but still I was on the right track and the answer was accepted. What I did was I took:
$\lim_{b\to\infty}(\dfrac{-\left(b^{1-p}\!\left(1+\left(-1+p\right)\ln\!\left(b\right)\right)\right)}{\left(-1+p\right)^{2}}-\dfrac{-\left(a^{1-p}\!\left(1+\left(-1+p\right)\ln\!\left(a\right)\right)\right)}{\left(-1+p\right)^{2}})$
Then I continued on with all of the fraction inside of the limit. Expanded it, and got the form:
$\dfrac{a^{1-p}}{(p-1)^2}+\dfrac{p a^{1-p} \ln (a)}{(p-1)^2}-\dfrac{a^{1-p} \ln (a)}{(p-1)^2}-\dfrac{b^{1-p}}{(p-1)^2}-\dfrac{p b^{1-p} \ln (b)}{(p-1)^2}+\dfrac{b^{1-p} \ln (b)}{(p-1)^2}$
I took the limit of each one of them:
$\lim_{b\to\infty}(\dfrac{a^{1-p}}{(p-1)^2}+\dfrac{p a^{1-p} \ln (a)}{(p-1)^2}-\dfrac{a^{1-p} \ln (a)}{(p-1)^2}-\dfrac{b^{1-p}}{(p-1)^2}-\dfrac{p b^{1-p} \ln (b)}{(p-1)^2}+\dfrac{b^{1-p} \ln (b)}{(p-1)^2})$
We can see in the expression that only the last three have a $b$ in their expression and it is those that I evaluated. By evaluated each one of them I noticed that they clear out (meaning they $ = 0$). This leaves us with only the first 3 fractions expressed in terms of $a$ and $p$. That was the correct answer.
Evaluations:
1. $\lim_{b\to\infty}(-\dfrac{b^{1-p}}{(p-1)^2})$. In this case since $p>1$ the expression $b^{1-p}$ will have $1-p<0$, consequently, this will make $b^{1-p}$ be part of the denominator. As ${b\to\infty}$ the limit will be of the form $1/\infty$, which in limits is equal to $0$.
2. $\lim_{b\to\infty}(-\dfrac{p b^{1-p} \ln (b)}{(p-1)^2})$. Same principle as numer 1. The $b^{1-p}$ will go on bottom, in this case making the limit similar to $\lim_{x\to\infty}(\dfrac{ln(x)}{x})$ which is $=0$.
3. $\lim_{b\to\infty}(\dfrac{b^{1-p}\ln (b)}{(p-1)^2})$ Same analogy with #2. The $b^{1-p}$ will go on bottom, in this case making the limit similar to $\lim_{x\to\infty}(\dfrac{ln(x)}{x})$ which is $=0$.
Correct Answer: $\dfrac{a^{1-p}}{(p-1)^2}+\dfrac{p a^{1-p} \ln (a)}{(p-1)^2}-\dfrac{a^{1-p} \ln (a)}{(p-1)^2}$ This can can be factored.