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 October 9th, 2019, 12:12 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Integral convergence For which values of s the integral converges ? $\displaystyle \int_{0}^{\infty}e^{sx}\cos(sx)dx \; , s\in \mathbb{R}$. Last edited by idontknow; October 9th, 2019 at 12:18 PM.
 October 9th, 2019, 03:14 PM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 Cosine is on the order of 1. Wouldn't it be the same values for which integral of e^sx converges, i.e., $s<0$? Thanks from idontknow and SDK
 October 10th, 2019, 12:04 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Integral converges only if $\displaystyle |e^{sx}\cos(sx)|\leq e^{sx}$. Now the RHS must converge or $\displaystyle s<0$ . $\displaystyle e^{-sx}|_{0}^{\infty }$ converges then the integral converges . Last edited by idontknow; October 10th, 2019 at 12:14 AM.
October 10th, 2019, 04:45 AM   #4
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Quote:
 Originally Posted by idontknow $\displaystyle |e^{sx}\cos(sx)|\leq e^{sx}$
Find me a value of s or x where this inequality is false.

October 11th, 2019, 05:01 AM   #5
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Quote:
 Originally Posted by DarnItJimImAnEngineer Find me a value of s or x where this inequality is false.
$\displaystyle \int_{D} e^{sx}cos(sx) < \int_{D} e^{sx} =\frac{e^{sx}}{s} \displaystyle |_{0}^{\infty}$ diverges , so it's own reciprocal expression converges , which is $\displaystyle se^{-sx}=s(e^{x})^{-s}$.
So $\displaystyle s<0$.

Last edited by idontknow; October 11th, 2019 at 05:05 AM.

October 11th, 2019, 05:36 AM   #6
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That was essentially my argument from the beginning.
In your second post, you said,
Quote:
 Originally Posted by idontknow Integral converges only if $\displaystyle |e^{sx}\cos(sx)|\leq e^{sx}$.
and I asked you to give me an example where this would not be true. It's true for all combinations of real s and x, isn't it?

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