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October 9th, 2019, 12:12 PM   #1
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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.
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October 9th, 2019, 03:14 PM   #2
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Cosine is on the order of 1. Wouldn't it be the same values for which integral of e^sx converges, i.e., $s<0$?
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October 10th, 2019, 12:04 AM   #3
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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.
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October 10th, 2019, 04:45 AM   #4
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Quote:
Originally Posted by idontknow View Post
$\displaystyle |e^{sx}\cos(sx)|\leq e^{sx}$
Find me a value of s or x where this inequality is false.
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October 11th, 2019, 05:01 AM   #5
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
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.
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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 View Post
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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