# Harder improper integral

• Oct 28th 2010, 05:15 AM
Glitch
Harder improper integral
The question:
Find the $\displaystyle \lim_{R \to \infty} \int_{-R}^{2R} \frac{x}{1 + x^2} dx$

My attempt:
Let $\displaystyle u = 1 + x^2$

$\displaystyle du = 2x dx$

$\displaystyle \frac{1}{2}\int_{-R}^{2R} \frac{du}{u}$

$\displaystyle \frac{1}{2}_{-R}^{2R}[log(u)]$

$\displaystyle \frac{1}{2}[log(1+(2R)^2) - log(1 + (-R)^2)]$

$\displaystyle \frac{1}{2}[log(1+4R^2) - log(1+R^2)]$

$\displaystyle \frac{1}{2}[log(\frac{1+4R^2}{1+R^2})]$

Limit of the function inside log is 4, so:
$\displaystyle \frac{1}{2}log(4)$

However, the answer is log(2). Where did I go wrong? Thanks!
• Oct 28th 2010, 05:23 AM
TheCoffeeMachine
Hint: $\displaystyle b\log{a} = \log{a^b}$

Spoiler:
$\displaystyle \frac{1}{2}\log{4} = \log{4^{\frac{1}{2}}} = \log{2}.$
• Oct 28th 2010, 05:25 AM
Glitch
Oh dear, I can't believe I got so far and tripped up on that! Thanks!
• Oct 28th 2010, 05:28 AM
TheCoffeeMachine
Quote:

Originally Posted by Glitch
Oh dear, I can't believe I got so far and tripped up on that! Thanks!

Haha! It happens to us all. (Wink)