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

  1. #1
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    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?
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  2. #2
    MHF Contributor matheagle's Avatar
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    \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}
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  3. #3
    Super Member redsoxfan325's Avatar
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    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.
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  4. #4
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    Smile Thanks RedSoxFan!

    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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