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Inequality with absolute value of a complex integral

  1. Mar 6, 2009 #1
    I'm stuck trying to prove a step inside a lemma from Serre; given is

    0<a<b
    0<x

    To prove:

    [tex]|\int_{a}^{b}e^{-tx}e^{-tiy}dt|\leq\int_{a}^{b}e^{-tx}dt[/tex]

    I've tried using Cauchy-Schwartz for integrals, but this step is too big (using Mathematica, it lead to a contradiction); something simpler must do the trick.
    Thanks in advance.
     
  2. jcsd
  3. Mar 7, 2009 #2
    Use

    [tex]\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx[/tex]
     
  4. Mar 7, 2009 #3
    As far as I know, that only holds if f(x) is real, which it is not.
     
  5. Mar 7, 2009 #4
    It also holds for complex-valued functions. For Riemann sums this is just the triangle inequality, and in the general case one can use an approximation argument.
     
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