1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook