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

**pireskwared** · March 2nd 2009, 11:53 PM

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?

**matheagle** · March 2nd 2009, 11:56 PM

$\arctan x <\pi/2$
thus

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

$\int_1^{\infty} {\arctan x\over x^2}dx < {\pi\over 2}\int_1^{\infty} {dx\over x^2}={\pi\over 2}$

**redsoxfan325** · March 3rd 2009, 12:25 AM

No, because if you find the whole indefinite integral, you get: $-\frac{arctan(x)}{x}-\frac{1}{2}ln(1+x^2)+ln(x)$ .

Plugging in $\infty$ :

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

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

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

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

Plugging in $1$ :

This is a well-defined number.

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

Subtracting:

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

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

Maple confirms that this is the correct answer.

**pireskwared** · March 3rd 2009, 10:48 AM

I went back and worked out the entire integral as you suggested and came to the same answer you did.
Thanks again!

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

Thanks RedSoxFan!

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