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Prove F(-x) = -F(x) for all x

  1. Oct 21, 2011 #1
    =[itex]\frac{arctan x}{x}[/itex] for x different from 0
    = 1 for x equal to 0

    F(x) is a definite integral from 0 to x, but I couldn't find the code for it, so just assume it is from 0 to x in the equation below.

    F(X) = [itex]\int f(t) dt[/itex]

    Now, the task is to prove that F(-x) = -F(x).

    This means we need to prove:

    [itex]\int f(t) dt[/itex] : definite from 0 to -x


    -[itex]\int f(t) dt[/itex] : definite from 0 to x

    We have tried some basic manipulation of integrals, but came nowhere.

    If anyone can give a hint of how to prove this, or tell us if we have understood the problem wrong, we would be most grateful.

  2. jcsd
  3. Oct 21, 2011 #2


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    The way to include limits of integration in LaTeX code is to use the subscript & superscript with the integral symbol: \int_{0}^{x} in your case. Using the \displaystyle code will make Large integral symbols and fractions when using [​itex​] & [​/itex​] tags.

    The code:
    Code (Text):
    [itex]\displaystyle -\int_{0}^{x} f(t) dt[/itex]
    gives you [itex]\displaystyle -\int_{0}^{x} f(t) dt[/itex] .
  4. Oct 21, 2011 #3


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    A quick way to look at this: what is the symmetry (about the y-axis) of the function [itex]\frac{arctan x}{x}[/itex]? (It is the ratio of two functions -- what are their symmetries and so what is the symmetry of the ratio?). Then, what does it mean to integrate [itex]\int_{0}^{x} f(t) dt[/itex] and [itex]\int_{0}^{-x} f(t) dt[/itex] for a function f(x) with such a symmetry?
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