Inequality with absolute value of a complex integral

  • Thread starter bernardbb
  • Start date
  • #1
4
0
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.
 

Answers and Replies

  • #2
316
0
Use

[tex]\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx[/tex]
 
  • #3
4
0
As far as I know, that only holds if f(x) is real, which it is not.
 
  • #4
316
0
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.
 

Related Threads on Inequality with absolute value of a complex integral

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
11
Views
878
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
13K
  • Last Post
Replies
10
Views
2K
Replies
2
Views
9K
Replies
1
Views
4K
Top