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Improper Integral (Calc II)

  1. Jun 28, 2011 #1
    Been doing some calculus review to knock the rust off for this coming fall semester and I got stuck...

    1. The problem statement, all variables and given/known data

    From Stewart's book (Early Transcendentals: 6E): (7.8 pg517 #69)

    Determine how large the number "a" has to be so that:

    [itex]\int[/itex][itex]\stackrel{\infty}{a}[/itex][itex]\frac{1}{x^{2}+1}[/itex]dx <.001

    2. Relevant equations


    3. The attempt at a solution

    Ok, I can easily picture the graph and the area under it. I figured I'd integrate, use "a" for my lower bound and "t" for the upper bound, then by using the potential equation it's just a simple matter of solving for "a" while taking the limit of said equation as t goes to infinity.

    I managed to get:

    arctan (t) - arctan (a) < 1/1000 (I suck at "latex" but this should technically be the limit of those arctans as t -> infinity < .001)

    Here is where I think I'm screwing up... I take the tangent of both sides:

    tan [arctan t -arctan a] < tan (1/1000) - and I'm stuck, I know I can't just apply the tangent function independently to both parameters giving me t - a < tan (.001) is there some trig identity I'm not thinking of..?
  2. jcsd
  3. Jun 28, 2011 #2


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    Take the limit as t → ∞ .

    [tex]\lim_{t\to\infty}\,\arctan(t) = ?[/tex]
  4. Jun 29, 2011 #3
    Ugh... Sorry to waste your time. The limit is pi/2, all I had to do was take the limit and I was already done lol. For some reason I thought there would be no limit for the arctan and that was bothering me.
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