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.



Integration of gaussian times absolute value of cosine
I presume you're interested in $c \to \infty$. Without absolute value you get
$\dfrac{\sqrt{\pi}}{2} c \exp(-c^2/4)$, which is not **quite** of order $\exp(-c^2/4)$. With absolute value, a lower bound is 
$$ \int\_0^\infty \exp(-(x/c)^2) \cos^2 x\; dx = \sqrt{\pi} c (1 + \exp(-c^2))/4$$
so it is not even $O(1)$. An upper bound is
$$ \int\_0^\infty \exp(-(x/c)^2)\; dx = \sqrt{\pi} c/2 $$


EDIT: For sharp bounds, consider that for $(n-1/2)\pi \le x \le (n+1/2)\pi$,


$$ e^{-((n-\frac12)\pi/c)^2} |\cos(x)| \ge e^{-(x/c)^2} |\cos(x)| \ge
e^{-((n+\frac12)\pi/c)^2} |\cos(x)|$$
so an upper bound is


$$1 + 2 \sum\_{n=1}^\infty e^{-(n-1/2)^2 \pi^2/c^2} = 1 + \theta\_2(0,e^{-\pi^2/c^2}) $$
where $\theta\_2$ is the second Jacobi theta function,
and a lower bound is
$$ e^{-\pi^2/(4c^2)} + 2 \sum\_{n=1}^\infty e^{-(n+1/2)^2 \pi^2/c^2}
= - e^{-\pi^2/(4 c^2)} + \theta\_2(0, e^{-\pi^2/c^2})$$
Using the Jacobi identities (or the Poisson summation formula)
$$\theta\_2(0,e^{-\pi^2/c^2}) = \dfrac{c}{\sqrt{\pi}} \left(1 + 2 \sum\_{n=1}^\infty (-1)^n e^{-n^2 c^2}\right)$$
In particular, the integral is $ c/\sqrt{\pi} + O(1)$. I expect that
further analysis could change this $O(1)$ to something much better. 


EDIT: OK, here's something much better. Write the integral as


$$\eqalign{J &= \dfrac{1}{2} \int\_{-\infty}^\infty e^{-(x/c)^2} |\cos(x)|\; dx\cr
 &= \dfrac{1}{2} \sum\_{n=-\infty}^\infty \int\_{n\pi}^{(n+1)\pi} e^{-(x/c)^2} |\cos(x)|\; dx\cr
&= \dfrac{1}{2} \sum\_{n=-\infty}^\infty \int\_0^{\pi/2} \left( e^{-(t+n\pi)^2/c^2} + e^{-((n+1)\pi - t)^2/c^2}\right) \cos(t)\; dt\cr
&= \sum\_{n=-\infty}^\infty a(n)}$$


Poisson summation formula says this is $\sum\_{k=-\infty}^\infty \widehat{a}(k)$ where
$$ \eqalign{\widehat{a}(k) &= \int\_{-\infty}^\infty e^{-2\pi i k s} a(s)\ ds\cr
&= \dfrac{1}{2} \int\_0^{\pi/2} \int\_{-\infty}^\infty \left( e^{-(t+s\pi)^2/c^2} + e^{-((s+1)\pi - t)^2/c^2}\right) \cos(t) e^{-2\pi i ks}\; ds\; dt\cr
&= (-1)^k \dfrac{ c}{\sqrt{\pi} (1 - 4 k^2)} e^{-c^2 k^2}}$$


Thus $$J = \dfrac{c}{\sqrt{\pi}} \left(1 + \dfrac{1}{3} e^{-c^2} + O\left(e^{-4 c^2}\right)\right)$$ 


