# Thread: Improper Integral

1. ## Improper Integral

here's the problem:
$\int_{7}^{\infty} \frac{\ln(x)}{x} dx$

I used substitution, letting $u = ln(x)$ which gave me $du=\frac{1}{x}dx$ and the resulting:
$\lim_{b\rightarrow \infty}\int_{ln(7)}^{ln(b)}{u} du$

This gave me $\lim_{b\rightarrow \infty}\frac{ln(b)^2}{2} - \frac{ln(7)^2}{2}$

Since infinity ends up in a numerator, I would think the answer is that the solution DNE, and it diverges, but webwork won't accept that. Where did I go wrong? Thanks MHF!!

2. ## Re: Improper Integral

Originally Posted by jjtjp
here's the problem:
$\int_{7}^{\infty} \frac{\ln(x)}{x} dx$
I used substitution, letting $u = ln(x)$ which gave me $du=\frac{1}{x}dx$ and the resulting:
$\lim_{b\rightarrow \infty}\int_{ln(7)}^{ln(b)}{u} du$
This gave me $\lim_{b\rightarrow \infty}\frac{ln(b)^2}{2} - \frac{ln(7)^2}{2}$
Since infinity ends up in a numerator, I would think the answer is that the solution DNE, and it diverges, but webwork won't accept that. Where did I go wrong? Thanks MHF!!
I think that your problem is webwork.

That may well be the very worst online course delivery system for mathematics there is.

3. ## Re: Improper Integral

You are correct, the integral diverges.

4. ## Re: Improper Integral

Is my math correct, though? Because when I graph $\frac{ln(x)}{x}$ it definitely approaches zero making me think I messed up somewhere.

5. ## Re: Improper Integral

okay, nvm. It was a webwork issue. I do hate how if you are wrong in webwork, it just tells you you're wrong, no explanation or anything. Thanks for verifying guys. I was thinking I was crazy after about 15 min.

6. ## Re: Improper Integral

Originally Posted by jjtjp
Is my math correct, though? Because when I graph $\frac{ln(x)}{x}$ it definitely approaches zero making me think I messed up somewhere.
An integral is really an infinite sum, and just because terms in a series tend to 0 is no guarantee that the series will still converge.