# understanding this improper integral

• June 26th 2007, 05:31 PM
Possible actuary
understanding this improper integral
My teacher worked this problem one way and I think I have found an easier way. Also, I don't understand how he came to his conclusion.

Here is the problem: $\int_{0}^{e} ln {x}^2~dx$ This is how he worked the problem: $\lim_{b\to 0^+}\int_{b}^{e} 2~ln {x}~dx=2x~ln x -\int_{0}^{e} 2~dx=(2e-0)-2x (from~0~to~e)=2e-2e=0$. He also included $\lim_{x\to 0}\frac{2~lnx}{1/x}=\lim_{x\to 0^+}\frac{2/x}{-1/x^2}=\lim_{x\to 0^+}-2x=0$. This is what I did: $\lim_{b\to 0^+}2\int_{b}^{e} lnx~dx=\lim_{b\to 0^+}\frac{2}{x}~(from~b~to~e)=\frac{2}{e}-\frac{2}{b}=\frac{2}{e}$

Is my method correct? Any idea how he got what he did?
Thanks!
• June 26th 2007, 07:04 PM
curvature
• June 26th 2007, 07:05 PM
topsquark
Quote:

Originally Posted by Possible actuary
My teacher worked this problem one way and I think I have found an easier way. Also, I don't understand how he came to his conclusion.

Here is the problem: $\int_{0}^{e} ln {x}^2~dx$ This is how he worked the problem: $\lim_{b\to 0^+}\int_{b}^{e} 2~ln {x}~dx=2x~ln x -\int_{0}^{e} 2~dx=(2e-0)-2x (from~0~to~e)=2e-2e=0$. He also included $\lim_{x\to 0}\frac{2~lnx}{1/x}=\lim_{x\to 0^+}\frac{2/x}{-1/x^2}=\lim_{x\to 0^+}-2x=0$. This is what I did: $\lim_{b\to 0^+}2\int_{b}^{e} lnx~dx=\lim_{b\to 0^+}\frac{2}{x}~(from~b~to~e)=\frac{2}{e}-\frac{2}{b}=\frac{2}{e}$

Is my method correct? Any idea how he got what he did?
Thanks!

It looks to me like he integrated by parts to go from
$\lim_{b\to 0^+}\int_{b}^{e} 2~ln {x}~dx$
to
$2x~ln x -\int_{0}^{e} 2~dx$

In your work we have the problem that
$\lim_{b\to 0^+}\frac{2}{b} \to \infty$

-Dan
• June 26th 2007, 11:33 PM
curvature
Quote:

Originally Posted by Possible actuary
$\lim_{b\to 0^+}2\int_{b}^{e} lnx~dx=\lim_{b\to 0^+}\frac{2}{x}~(from~b~to~e)=\frac{2}{e}-\frac{2}{b}=\frac{2}{e}$

You might confuse the integral of lnx (which is x(lnx-1)) with the derivative of lnx (which is 1/x).