Integration of gaussian times absolute value of cosine 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.
 A: 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 $$J = \dfrac{c}{\sqrt{\pi}} \left(1 + \dfrac{1}{3} e^{-c^2} + O\left(e^{-4 c^2}\right)\right)$$ 
