# Is the integral of arctan(x)/x^2 divergent?

• Mar 2nd 2009, 11:53 PM
pireskwared
Is the integral of arctan(x)/x^2 divergent?
By the following methodology I have determined that the integral of arctan(x)/x^2 dx with lower limit = 1 and upper limit = infinity is divergent. Can somebody verify this? If it is not divergent can you show me that the limit of the function as x-> infinity exists? By using substitution by parts and then partial fractions I found one of the terms to be infinity and concluded that it was, therefore, divergent. I don't have a math scripting program so it is difficult for me to show my work, but here goes:

solve by parts: with u = arctan(x) du = 1/(1+x^2)dx dv= 1/x^2 v= -1/x then limit as x -> inf [-arctan(x)/x + int[1/x(1+x^2)]. Then I use partial fractions on [1/x(1+x^2)] to get int(1/x) + int(-x/(x^2+1)). Integral of 1/x is ln(x). The limit of ln(x) as x-> infinity is infinity. I need go no further, then, because this term makes the limit of the integral indeterminate. So the function is divergent. Is that correct?
• Mar 2nd 2009, 11:56 PM
matheagle
$\displaystyle \arctan x <\pi/2$
thus

$\displaystyle \int_1^{\infty} {\arctan x\over x^2}dx < 100\int_1^{\infty} {dx\over x^2}=100$
where I'm being silly with the 100. Or, to be a little tighter...

$\displaystyle \int_1^{\infty} {\arctan x\over x^2}dx < {\pi\over 2}\int_1^{\infty} {dx\over x^2}={\pi\over 2}$
• Mar 3rd 2009, 12:25 AM
redsoxfan325
No, because if you find the whole indefinite integral, you get: $\displaystyle -\frac{arctan(x)}{x}-\frac{1}{2}ln(1+x^2)+ln(x)$.

Plugging in $\displaystyle \infty$:

$\displaystyle ln(x)-\frac{1}{2}ln(1+x^2)$ can be rewritten as $\displaystyle ln(\frac{x}{\sqrt{1+x^2}})$.

Since $\displaystyle \lim_{x\to\infty} \frac{x}{\sqrt{1+x^2}} = 1$, $\displaystyle \lim_{x\to\infty} ln(\frac{x}{\sqrt{1+x^2}}) = ln(1) = 0.$

Since $\displaystyle arctan(\infty) = \frac{\pi}{2}$, $\displaystyle \lim_{x\to\infty} -\frac{arctan(x)}{x} = -\frac{\frac{\pi}{2}}{\infty} = 0$.

Thus, the whole expression when $\displaystyle x=\infty$ is $\displaystyle 0$.

Plugging in $\displaystyle 1$:

This is a well-defined number.

$\displaystyle -\frac{arctan(1)}{1}-\frac{1}{2}ln(1+1^2)+ln(1) = -\frac{\frac{\pi}{4}}{1}-\frac{1}{2}ln(2)-0 = -\frac{\pi}{4}-\frac{1}{2}ln(2)$.

Subtracting:

Thus we have $\displaystyle 0-(-\frac{\pi}{4}-\frac{1}{2}ln(2)) = \frac{\pi}{4}+\frac{1}{2}ln(2)$.

Thus, $\displaystyle \int_{1}^{\infty} \frac{arctan(x)}{x^2}\,dx = \frac{\pi}{4}+\frac{1}{2}ln(2)$.

Maple confirms that this is the correct answer.
• Mar 3rd 2009, 10:48 AM
pireskwared
Thanks RedSoxFan!
I went back and worked out the entire integral as you suggested and came to the same answer you did.
Thanks again!