A probabilistic integral $\int_{-\infty}^{\infty}e^{-x^2/2\sigma^2}\arcsin\left(1-2\left|\lfloor x\rceil-x\right|\right)\,dx$ In my probabilistic studies, a tough integral appeared. Note that $\lfloor x\rceil$ is not the floor function; it is the nearest integer function. Up to some constants, it appears in a Buffon-like situation.
$$
\int_{-\infty}^{\infty}{\rm e}^{-x^2/2\sigma^{2}}
\arcsin\left(1 - 2\left\vert\vphantom{\Large A}\,\left\lfloor\,x\,\right\rceil-x\,\right\vert\right)
\,{\rm d}x
$$
It is possible to evaluate this integral? I will be already very glad if the case $\sigma=1$ is settled. What makes this hard to me is the
$\left\vert\vphantom{\Large A}\left\lfloor\,x\,\right\rceil-x\,\right\vert$
part. A natural candidate would be a Fourier expansion, but even this seems hard.
A picture of the integrand.

After this analysis, I don't think we can do much better for this integral. Thank you for your time and detailed answers. 
 A: In essence, we're interested in the expectation value of $f(x)=\arcsin\left(1-2\big|\lfloor x\rceil-x\big|\right)$ with respect to a centered Gaussian distribution:
$$\text{E}[f(x)]=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty e^{-x^2/2\sigma^2} f(x)\,dx $$
Note that $f(x)$ is a $1$-periodic function that behaves as $g(x)=\arcsin(1-2|x|)$ for $|x|\leq \frac{1}{2}$. Taking $g(x)$ to have support only on this interval, we can write $f(x)$ in a form amenable to expansion via the Poisson summation formula:
\begin{align}
f(x)=\sum_{n=-\infty}^\infty g(x+n)=\sum_{k=-\infty}^\infty e^{2\pi i k x}\int_{-1/2}^{1/2} e^{-2\pi i k y}g(y)\,dy \tag{(1)}
\end{align}
Note that the integral in this expression is just the $k$-th Fourier coefficient $c_k$ of $f(x)$. So we have traded the periodicity of $f(x)$ for a weighted sum of double integrals. But these all factorize into single integrals, and the integrals over $x$ are all Gaussians: 
\begin{align}
\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty \exp\left(-\frac{x^2}{2\sigma^2}+2\pi i k x\right)\,dx
&=\frac{e^{-2\pi^2 k^2\sigma^2}}{\sqrt{2\pi \sigma^2}}
\int_{-\infty}^\infty \exp\left(-\frac{1}{2\sigma^2}\left(x-2 \pi i k\sigma^2  \right)^2\right)\,dx\\
&=e^{-2\pi^2 k^2\sigma^2}
\end{align} Note the convenient disappearance of the normalization.
Thus we may write the expectation value as
$$\text{E}[f(x)] = \sum_{k=-\infty}^\infty c_k \, e^{-2\pi^2 k^2\sigma^2} $$ Noting the quadratic dependence on $k$, this series is analogous to a Jacobi theta function except weighted by the Fourier coefficients $c_k$. To proceed further we need these coefficients: 
\begin{align}
c_k
&=2\,\text{Re}\left[\int_{0}^{1/2} e^{2\pi i k y} \arcsin(1-2y)\,dy\right] &(\text{symmetry of integrand})\\
&=\text{Re}\left[e^{2\pi i k}\int_{0}^{\pi/2} e^{-2\pi i k\sin\phi} \phi \cos\phi\,d\phi\right] &(\sin\phi=1-2y)
\end{align}
To proceed further, we use a Jacobi-Anger expansion to write the inner integral as
$$\int_{0}^{\pi/2} e^{-2\pi i k\sin\phi} \phi \cos\phi\,d\phi
=\sum\limits_{n=-\infty}^{\infty} J_n(-2\pi k) \int_{0}^{\pi/2} e^{i n \phi} \phi \cos\phi\,d\phi $$
This integral is certainly doable, but right now I don't know a nice formula for it and so halt for now. Even if is relatively nice, though, one still appears destined to end up with a doubly-infinite summation over Bessel functions and quadratic exponents. That does not seem encouraging.

UPDATE) Following the discussion in comments with @rajb245, I realized that his insight regarding the presence of the zeroth Struve H-function could be revealed directly. First, using symmetry along with the fact that $\cos(\pi k -\pi k u)=(-1)^k \cos{\pi k u}$, we rewrite $c_k$ using the substitution $y=1-2u$ to obtain 
$$c_k
=2\int_{0}^{1/2} \cos{2\pi k y} \arcsin(1-2y)\,dy
=(-1)^k \int_{0}^{1} \cos(\pi k u) \arcsin u\,du.$$ If $k=0$ then we may compute $c_0=\pi/2-1$. Otherwise, we can use the approach of this Ron Gordon answer, and integrate by parts:
\begin{align}
(-1)^k \int_{0}^{1} \cos(\pi k u) \arcsin u\,du
&=
(-1)^k \left[\frac{1}{\pi k}\sin(\pi k u) \arcsin u-\frac{1}{\pi k}\int \frac{\sin(\pi k u)}{\sqrt{1-u^2}}\,du\right]_0^1
\end{align}
The boundary term vanishes, and for the second term we recall the following integral representation of the zeroth Struve H-function:
$$\mathbf{H}_0(t)=\frac{2}{\pi}\int_0^1 \frac{\sin{t u}}{\sqrt{1-u^2}}\,du$$
Therefore $c_k=\dfrac{(-1)^{k-1}}{2k}\mathbf{H}_0(2\pi k)$ for $k\neq 0$. Note that $c)k=c_{-k}$ (consistent with $g(x)$ being even) since $\mathbf{H}_0(t)$ is an odd function.
We may now return to E$[f(x)]$ and write 

$$\text{E}[f(x)]
=\left(\frac{\pi}{2}-1\right)-\sum_{k=1}^\infty \dfrac{(-1)^{k}}{k}\mathbf{H}_0(2\pi k)e^{-2\pi^2 k^2\sigma^2}$$

This represents my point of conclusion, since I haven't any clue whether this series can be simplified further. But this present form is quite useful: We have a constant leading term (i.e. the mean value of $\arcsin(1-2|x|)$ over $[-1/2,1/2]$) followed by a series of corrections which are each exponentially smaller than the last. Hence even taking the first correction will provide an excellent approximation. (It would probably be wise to validate this expression, either numerically or by computing the $\sigma\to0$ and $\sigma\to\infty$ limits.)
A: Hmm let's see if I'm correct, I did not check it entirely but it should work this way. Maybe there are some issues at $(\star)$ like missing constants.
We know about $\lfloor x \rceil$:
$$ \forall n\in\mathbb{N}\forall x\in \left[n-\frac{1}{2},n+\frac{1}{2}\right] \quad\quad\quad \left\lfloor x\right\rceil= n  \tag{$\alpha$}$$
So let's split up that integral:
$$ I:=\int\limits_{-\infty}^\infty \exp\left(-\frac{x^2}{2\sigma^2}\right)\arcsin\left(1-2\big|\lfloor x\rceil-x\big|\right) \text{d}x 
= {\sum_{n=-\infty}^\infty\int\limits_{n-\frac{1}{2}}^{n+\frac{1}{2}} \exp\left(-\frac{x^2}{2\sigma^2}\right)\arcsin\underbrace{\left(1-2\big|\lfloor x\rceil-x\big|\right)}_{=:A(x)} \text{d}x } $$
Now let's find a nice way of expressing $A$ throught $(\alpha)$:
$$ \forall n\in\mathbb{N}\forall x\in \left[n-\frac{1}{2},n+\frac{1}{2}\right]\quad\quad\quad A(x) = 1-2\big|n-x|\le 1 $$
In fact we know even more:
$$ A(x) = 1-2\big|n-x|\le 1 =1-2\big|x-n\big|=\left\{\begin{matrix} x-n+\frac{1}{2} & x\le n \\ n+\frac{1}{2}-x & x\ge n\end{matrix}\right. \tag{$\beta$}$$
So lets split the integral even further:
$$ I = {\sum_{n=-\infty}^\infty\int\limits_{n-\frac{1}{2}}^{n} \exp\left(-\frac{x^2}{2\sigma^2}\right)\arcsin\left(x-n+\frac{1}{2}\right) \text{d}x }+
{\int\limits_{n}^{n+\frac{1}{2}} \exp\left(-\frac{x^2}{2\sigma^2}\right)\arcsin\left(n+\frac{1}{2}-x\right) \text{d}x }$$
Now this is amazing because we can substitue $y=x-n+\frac{1}{2}$ and $z=n+\frac{1}{2}-x$ with $\text{d}x=\text{d}y = -\text{d}z$:
$$ I= {\sum_{n=-\infty}^\infty\underbrace{\int\limits_{0}^{\frac{1}{2}} \exp\left(-\frac{\left(-y+n-\frac{1}{2}\right)^2}{2\sigma^2}\right)\arcsin\left(y\right) \text{d}y }_{B(n)}}-
{\underbrace{\int\limits_{\frac{1}{2}}^{0} \exp\left(-\frac{\left(-n-\frac{1}{2}+z\right)^2}{2\sigma^2}\right)\arcsin\left(z\right) \text{d}z }_{C(n)}}$$
Now we can use the linearity of the integral that:
$$ \sum_{n=-\infty}^\infty B(n) = \int\limits_{0}^{\frac{1}{2}} \arcsin\left(y\right) \sum_{n=-\infty}^\infty\exp\left(-\frac{\left(y-n+\frac{1}{2}\right)^2}{2\sigma^2}\right)\text{d}y $$
Edit: this is Wrong
It's just an inequality 
Now we know for a monotonous function that:
$$ \sum_{n=-\infty}^\infty B(n) =\int\limits_{0}^\frac{1}{2}\arcsin(y)\sum_{n=-\infty}^\infty\exp\left(-\frac{\left(y-n+\frac{1}{2}\right)^2}{2\sigma^2}\right)  \text{d}y
= {\int\limits_{0}^\frac{1}{2}\arcsin(y)\int\limits_{-\infty}^\infty \exp\left(-\frac{\left(y-n+\frac{1}{2}\right)^2}{2\sigma^2}\right) \text{d}n \text{d}y}
=\sqrt{2\pi\sigma^2}\int\limits_{0}^\frac{1}{2}\arcsin(y)\text{d}y \tag{$\star$}$$
So the same is true for $C(n)$. So we get:
$$ I = 2\sqrt{2\pi\sigma^2}\int\limits_{0}^\frac{1}{2} \arcsin(x)\text{d}x $$
