thus
where I'm being silly with the 100. Or, to be a little tighter...
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?
No, because if you find the whole indefinite integral, you get: .
Plugging in :
can be rewritten as .
Since ,
Since , .
Thus, the whole expression when is .
Plugging in :
This is a well-defined number.
.
Subtracting:
Thus we have .
Thus, .
Maple confirms that this is the correct answer.