Integral problem lnx/x^3: I think it diverges, but my book says it converges because lnx < x

[Guidoddd]

Hello! Can you help me with this integral?

[MATH]\int_{0}^{2} ln x /x^3 dx[/MATH]
I need to find if it converges or not.
The problem is in 0, and I think the function is asymptotic to [MATH]1/x^2[/MATH] so it diverges
But the solution of my book says: the integral converges because of the comparison [MATH]lnx < x [/MATH]Is this an error of the book or there us something I am missing ?

Thanks for help

 

[Win_odd Dhamnekar]

The integral is divergent

 

[ksdhart2]

While Winod's answer is completely correct, I'd like to expand a bit upon why it is correct. Using the Fundamental Theorem of Calculus, we know that:

\(\displaystyle \int\limits_{a}^{b} f(x) \: dx = F(b) - F(a)\)

where F(x) is the indefinite integral of f(x). Here we have \(\displaystyle f(x) = \dfrac{\ln(x)}{x^3} \implies F(x) = -\dfrac{2\ln(x) + 1}{4x^2}\). Clearly, F(x) is not defined at x = 0, both because of the denominator being 0 and because of ln(0) causing problems. However, what we can do is look at the limit as x approaches 0:

\(\displaystyle \lim\limits_{A \to 0} \int\limits_{A}^{2} \dfrac{\ln(x)}{x^3} \: dx = -\dfrac{2\ln(2) + 1}{16} + \lim\limits_{A \to 0} \dfrac{2\ln(A) + 1}{4A^2}\)

Looking at that limit, I can see it will be of the form \(\displaystyle \dfrac{-\infty}{\infty}\), so that suggests L'Hopital's Rule might be a good idea:

\(\displaystyle \lim\limits_{A \to 0} \dfrac{2\ln(A) + 1}{4A^2} = \lim\limits_{A \to 0} \dfrac{\frac{2}{A}}{8A} = \lim\limits_{A \to 0} \dfrac{1}{4A^2}\)

At this point, it should be obvious why the limit "blows up" to infinity.

 

[Dr.Peterson]

[MATH]\int_{0}^{2} ln x /x^3 dx[/MATH]
I need to find if it converges or not.
The problem is in 0, and I think the function is asymptotic to [MATH]1/x^2[/MATH] so it diverges
But the solution of my book says: the integral converges because of the comparison [MATH]lnx < x [/MATH]Is this an error of the book or there us something I am missing ?

It occurs to me that the book's comment might arise from thinking the limits of integration were 2 to infinity, because the comparison would be apt then. But the actual integral is improper at 0, and the comparison doesn't help there, as ln x is negative near 0.

 

[Guidoddd]

Thanks for your replies. I Think there is a print error of the book on the integration interval too, the previous exercises are from 2 to infinity and also the solution of this exercise makes sense changing the extremes