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Is there a way to compute/estimate the following integral?
$\int_0^\infty e^{-(x/c)^2}\left|\cos{x}\right|dx$
where $c$ is a real constant. I would like to know if it is of order $e^{-c^2/4}$ like the integral without absolute value. Or is there a sharp upperbound?
Thank you very much in advance.

anonymous
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  • i think that since $0\le|\cos x|\le 1$, then $0\le e^{-\left(\frac{x}{c}\right)^2}|\cos x|\le e^{-\left(\frac{x}{c}\right)^2}$ and $\displaystyle0\le\int_{0}^{+\infty}e^{-\left(\frac{x}{c}\right)^2}|\cos x|dx\le\int_{0}^{+\infty}e^{-\left(\frac{x}{c}\right)^2}dx$ – cand Sep 08 '14 at 18:25
  • Thanks for your answer. But i'm looking for a sharper bound, without killing the $\cos$ if possible. – anonymous Sep 08 '14 at 18:28
  • Well, of course it's of that order... Trying to think of a proof though. – ShakesBeer Sep 08 '14 at 18:38
  • Only thing I can think of at the moment is that $\cos^2(x) \leq \left| \cos(x) \right| $ which could give you a decent lower bound on the integral. – Spencer Sep 08 '14 at 18:45
  • @Spencer how exactly? (sorry to be annoying) – ShakesBeer Sep 08 '14 at 18:51
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    @coolydudey60, Do you mean how do we evaluate the integral with a $\cos^2(x)$? This can be done by replacing $\cos(x)=(\exp(ix)+\exp(-ix))/2$ and then completing the square in the exponents. Then its no harder than evaluating a normal gaussian integral. – Spencer Sep 08 '14 at 20:06
  • Do you have any idea for a sharp upperbound? – anonymous Sep 09 '14 at 06:51
  • Cross-post with http://mathoverflow.net/questions/180372/integration-of-gaussian-times-absolute-value-of-cosine/180377#comment452391_180377 – Robert Israel Sep 09 '14 at 17:48

2 Answers2

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Just so we can actually see what we're dealing with:

Let $\displaystyle I_c=\int_0^\infty e^{-(x/c)^2}\left|\cos{x}\right|dx$.

This is a plot of $I_c$ versus $c$.

Now, this is a plot of $\displaystyle \frac{I_c}{c}$ for $c \in (0,5]$.

enter image description here

Credits to MATLAB, as always.

Community
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ShakesBeer
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small things: we have that $-e^{-\left(\frac{x}{c}\right)^2}|\cos x|\le e^{-\left(\frac{x}{c}\right)^2}\cos x\le e^{-\left(\frac{x}{c}\right)^2}|\cos x|$, then we can also bound the intregral by: $$\int_{0}^{+\infty}e^{-\left(\frac{x}{c}\right)^2}\cos xdx\le\int_{0}^{+\infty}e^{-\left(\frac{x}{c}\right)^2}|\cos x|dx\le\int_{0}^{+\infty}e^{-\left(\frac{x}{c}\right)^2}dx$$

cand
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