Compute $ \int_{\mathbb R} |\sin x| e^{-(x-a)^2 / b^2} dx $ For $a, b > 0$, I want to compute $$\int_{\mathbb R} |\sin x| e^{-(x-a)^2 / b^2} dx.$$
What I have tried:
Since $\int_{\mathbb R} |\sin x| e^{-(x-a)^2 / b^2} dx= \sum_{k \in \mathbb Z} \int_{0}^{\pi} (\sin x) e^{-(x-a-k\pi)^2 / b^2} dx$, I tried calculating $\int_{0}^{\pi}(\sin x) e^{-(x-a)^2 / b^2} dx$, but I cannot proceed further. Any help is appreciated.
Ps. Even a good bound is welcomed.
 A: I will compute $\int_{\mathbb R} |\sin x| g(x) dx$ instead where $g$ is the Gaussian density with mean $\mu$ and variance $\sigma^2$.
By the Poisson summation formula, it is equal to $$\frac{4}{\pi} \Big(1-2 \sum_{k \geq 1} \frac{e^{-2\sigma^2 k^2}}{4k^2 -1 }\cos (2k\mu)\Big).$$
This follows from the following consideration.
Notice $\int_{0}^{\pi} (\sin x) e^{-(x-\mu-k\pi)^2 / 2\sigma^2}/\sqrt{2\pi\sigma^2} dx= (f \ast g)(\mu+k\pi)$ where $f(x)= 1_{[0,\pi]}(x) \sin x$ and $\ast$ is the convolution. We want to calculate $\sum_{k \in \mathbb Z} (f\ast g) (\mu+k\pi)$. By the Poisson summation formula, it is equal to $$\sqrt{\frac{2}{\pi}}\sum_{k \in \mathbb Z} \hat f(2k)\hat g(2k)e^{i2k \mu}.$$ It is easy to check that $\hat f(2k)= \frac{2}{1-4k^2}$ and $\hat g(2k)= \sqrt{\frac{2}{\pi}}e^{2\sigma^2 k^2 }$, so we are done by noticing that our result must be a real number.
A: I don't know if this is enough.
$$ \begin{split}
\int_{\mathbb R} |\sin x| e^{-(x-a)^2 / b^2} dx & \le \int_{\mathbb R} e^{-(x-a)^2 / b^2} dx \\ [c := \frac{b}{\sqrt{2}} ] \qquad & = \int_{\mathbb R} e^{\frac{-(x-a)^2}{2c^2}} dx\\ & =  \sqrt{2\pi c^2} \int_{\mathbb R} \frac{1}{\sqrt{2\pi c^2}} e^{\frac{-(x-a)^2}{2c^2}} dx  \\ & = \sqrt{2\pi c^2}
\end{split}$$
Where the last equalty is valid since  $\int_{\mathbb R} \frac{1}{\sqrt{2\pi c^2}} e^{\frac{-(x-a)^2}{2c^2}} dx = 1$ since the inner function is the density function of a normal distribution with mean $a$ and variance $c^2$
A: Not sure how helpful this is but perhaps you can use the Fourier Series of $|\sin(x)|$?
$$|\sin(x)|=\frac{2}{\pi}\left(1+\sum_{n=2}^{\infty}\frac{\cos(\pi n)+1}{1-n^2} \cos(nx) \right)$$
