# Thread: understanding this improper integral

1. ## 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!

2. Your answer is correct. The answer is 0.

3. 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

4. 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).