# Improper Integral

• July 13th 2010, 12:36 PM
MechEng
Improper Integral
For the integral:

$
\int_{0}^{\frac{1}{2}}{\frac{1}{x\ln^2{x}}}dx
$

Using the substitution:

$
u=\ln{x}
$

$
du=\frac{1}{x}
$

I am getting the following for my new terminals:

$
-\ln{2}
$

$-\infty$

To get my new terminals I have simply plugged the values of my old terminals into my u substitution (u=lnx).

When I evaluate the integral I wind up with a limit that needs to be evaluated, as follows:

$\lim_{t\rightarrow-\infty}\frac{-1}{ln(-ln(2)}+\frac{1}{ln(t)}$

When I evaluate that I come up with:

$\frac{1}{ln(-ln(2))}$

Have I done something wrong here?
• July 13th 2010, 12:40 PM
Ackbeet
Your notation is a little confused for the limit. However, 1/ln(2) is the correct answer. Certainly, your setup looks fine.
• July 13th 2010, 01:33 PM
MechEng
I tried to correct that last line, but somehow wound up making creating a double thread. I am not coming up with 1/ln(2). I am coming up with 1/ln(-ln(2)). I know that isn't right.

Can you tell me where I am going wrong?

Oh... and I did mean to say "add up the fractions" not "add up the constants".
• July 13th 2010, 01:36 PM
Ackbeet
• July 13th 2010, 02:06 PM
MechEng

$\int{\frac{du}{u^2}}$

Which gave me:

$-\frac{1}{u}$

I try to find my anti-derivatives and verify them with my TI-89. In this case, the 89 said I was correct.
• July 13th 2010, 02:32 PM
chisigma
The derivative of $\ln x$ is $\frac{1}{x}$ so that is...

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

... and ...

$\displaystyle \int_{0}^{\frac{1}{2}} \frac{dx}{x\ \ln^{2} x} = \frac{1}{\ln 2}$ (2)

Kind regards

$\chi$ $\sigma$
• July 13th 2010, 03:13 PM
Jester
$\int \limits_{-\infty}^{-\ln 2} \dfrac{du}{u^2}$
$\lim \limits_{a \to - \infty} \int \limits_{a}^{-\ln 2} \dfrac{du}{u^2} = \lim \limits_{a \to - \infty} \dfrac{-1}{u}\left. \right| \limits_{a}^{-\ln 2} = \lim \limits_{a \to - \infty} \dfrac{1}{\ln 2}+ \dfrac{1}{a}$