# Math Help - Evaluating improper integral

1. ## Evaluating improper integral

$\int_e^{\infty} \frac{1}{x(\ln x)^3} dx$

First i made a u-substitution by making u = ln x.
After which i got $-\frac{1}{2(\ln x)^2}$
Then i went back to compute the improper integral again , but couldn't solve it.

Did i made any wrong move?
Thanks

2. Originally Posted by xcluded
$\int_e^{\infty} \frac{1}{x(\ln x)^3} dx$

First i made a u-substitution by making u = ln x.
After which i got $-\frac{1}{2(\ln x)^2}$
Then i went back to compute the improper integral again , but couldn't solve it.

Did i made any wrong move?
Thanks

That's the correct substitution, but you set it back up incorrectly. Once the substitution is made, $\ln x$ should not appear in the new integral.

It turns out that we should get $\int_e^{\infty}\frac{\,dx}{x\left(\ln x\right)^3}\xrightarrow{u=\ln x}{}\int_1^{\infty}\frac{\,du}{u^3}$

Does this make sense?

3. Originally Posted by xcluded
$\int_e^{\infty} \frac{1}{x(\ln x)^3} dx$

First i made a u-substitution by making u = ln x.
After which i got $-\frac{1}{2(\ln x)^2}$
Then i went back to compute the improper integral again , but couldn't solve it.

Did i made any wrong move?
Thanks

Since $\frac{d}{dx}\left[ -\frac{1}{2 (\ln(x))^2}\right]=\frac{1}{x(\ln(x))^3}$ by the fundamental theorem of calculus:

$\int_e^{\infty} \frac{1}{x(\ln x)^3}\; dx=$ $\lim_{\zeta \to \infty}\left[-\frac{1}{2(\ln(x))^2}\right]_e^{\zeta}=$ $

\frac{1}{2}-\lim_{\zeta \to \infty} \frac{1}{2(\ln(\zeta))^2}=\frac{1}{2}$

CB

4. Originally Posted by Chris L T521
That's the correct substitution, but you set it back up incorrectly. Once the substitution is made, $\ln x$ should not appear in the new integral.

It turns out that we should get $\int_e^{\infty}\frac{\,dx}{x\left(\ln x\right)^3}\xrightarrow{u=\ln x}{}\int_1^{\infty}\frac{\,du}{u^3}$

Does this make sense?
Hmm why does the lower limit becomes 1 ?

5. Originally Posted by xcluded
Hmm why does the lower limit becomes 1 ?
You substitute $u = ln(x)$ with the lower limit of x being $e$. So the lower limit of u is $ln(e)=1$

6. Thanks everyone !