Inequality with absolute value of a complex integral

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The discussion centers on proving an inequality involving a complex integral, specifically showing that the absolute value of the integral is less than or equal to the integral of the real part. The user initially attempted to apply the Cauchy-Schwartz inequality but encountered a contradiction. Another participant clarifies that the inequality can be applied to complex-valued functions, not just real functions, and suggests using the triangle inequality or approximation arguments for Riemann sums. This highlights the importance of understanding the properties of integrals involving complex functions in proving the desired inequality. The conversation emphasizes the need for simpler approaches in complex analysis.
bernardbb
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I'm stuck trying to prove a step inside a lemma from Serre; given is

0<a<b
0<x

To prove:

|\int_{a}^{b}e^{-tx}e^{-tiy}dt|\leq\int_{a}^{b}e^{-tx}dt

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.
 
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Use

\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx
 
As far as I know, that only holds if f(x) is real, which it is not.
 
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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