Improper Integrals: Infinite Limits #2

[Silvanoshei]

Er?

\(\displaystyle \int\limits_{e}^{\infty}\frac{lnx}{x}dx\)

\(\displaystyle u=lnx \implies du = \frac{1}{x}dx \implies xdu=dx\)

\(\displaystyle udu \implies \frac{u^{2}}{2} \implies \left[\frac{lnx^{2}}{2}\right]_e^{\infty}\)

\(\displaystyle \infty - \frac{1}{2}\)

My books tells me I'm an idiot and the correct answer is \(\displaystyle \frac{1}{2}(ln2+1)\)

 

[lookagain]

Er?

\(\displaystyle \int\limits_{e}^{\infty}\frac{lnx}{x}dx\)

\(\displaystyle u=lnx \implies du = \frac{1}{x}dx \implies xdu=dx\)

\(\displaystyle udu \implies \frac{u^{2}}{2} \implies \left[\frac{lnx^{2}}{2}\right]_e^{\infty} \ \ \ \)That should be \(\displaystyle \ \ \dfrac{[\ln(x)]^2}{2}\bigg|_e^{\infty}.\) \(\displaystyle \ \ \ \ \ \ \)What you have in the numerator looks like the log of \(\displaystyle \ x^2.\)

\(\displaystyle \infty - \frac{1}{2} \ \ \ \ \) I don't get that. I get that the limit is a divergent one.

My books tells me I'm an idiot and the correct answer is \(\displaystyle \frac{1}{2}(ln2+1)\)

For the correct answer to be what the book is that you state it is, the problem would have to be different.

Look at the fraction for the integrand: \(\displaystyle \ \ \dfrac{ln(x)}{x}.\)

\(\displaystyle \dfrac{1}{x} \ < \ \dfrac{ln(x)}{x} \ \ for \ \ \ x \ge e.\)

Let k be any positive real number.


\(\displaystyle \ \ \displaystyle\sum_{x = k}^{\infty}\dfrac{1}{x} \ \ diverges.\)

And so the original integral doesn't have a finite area by comparison.