Function $f(x)=\int_0^\infty\left|\sin(t)\cdot\sin(x\,t)\cdot e^{-t}\right|\,dt$ Let
$$f(x)=\int_0^\infty\Big|\sin(t)\cdot\sin(x\,t)\cdot e^{-t}\Big|\,dt,$$
where $|\dots|$ denotes the absolute value. We are concerned only with positive values of $x$ (i.e. let the domain of the function be $\mathbb{R}^+$).
The graph of this function looks roughly as follows (modulo possible errors in numeric algorithms used to plot it):

The function values at rational points can be evaluated in a closed form, e.g.
$$f\left(\frac32\right)=\frac2{145}\left(24+\frac{27\,\sqrt3\,\left(e^{\frac\pi3}+e^{-\frac\pi3}\right)-24\,\left(e^{\frac\pi3}-e^{-\frac\pi3}\right)-4}{e^\pi-e^{-\pi}}\right).$$

Questions: 


*

*Is the function $f(x)$ continuous? Is it smooth? Is it analytic?

*How many local extrema does it have?

*What is the $\lim\limits_{x\,\to\,\infty}f(x)$, if it exists?

*Can we find a closed-form value of $f(x)$ at an explicit irrational point (given by a closed-form expression)?

*Is there a general formula for values of $f(x)$ at rational points?

 A: The general formula for rational $x$ can be derived as
follows. Consider, first, a convergent sum of the form $\sum_k
\frac{p(k)}{q(k)}$, where $p$ and $q$ are polynomials, $\deg q\geq
\deg p+2$, and $q$ only has simple roots. If $\rho$ is a variable that
runs over the roots of $q$, then
$$ \frac1{q(x)} = \sum_{q(\rho)=0}\frac{1}{q'(\rho)}\frac1{x-\rho}, $$
so the sum can be written as
$$ \sum_k \frac{p(k)}{q(k)} = 
\sum_{q(\rho)=0} \frac1{q'(\rho)} \sum_k \frac{p(k)\bmod k-\rho}{k-\rho} + \lfloor p(k)/(k-\rho)\rfloor. $$
Now, $p(k)\bmod k-\rho = p(\rho)$ is a constant, 
$$ \sum_{k\geq0} \frac{z^k}{k-\rho} = \Phi(z,1,-\rho), $$
by definition of Lerch's transcendent function, and the fact that the original sum is convergent assures us we can ignore the terms $\sum_\rho\frac{\lfloor p(k)/(k-\rho)\rfloor}{q'(\rho)}$, as they must necessarily sum to zero.
Therefore, since $\Phi$ has an asymptotic expansion
$$ \Phi(z,1,-\rho) = -\log(1-z) - \gamma - \psi(-\rho), $$
($\gamma$ is Euler's gamma, and $\psi$ is the digamma function), and
since the fact that the original sum converges assures us that we can
ignore all terms of $\Phi(z,1,-\rho)$ independent of $\rho$ as we take
the limit $z\to1-$ to recover the sum, we therefore have
$$ \sum_{k\geq0} \frac{p(k)}{q(k)} = -\sum_{q(\rho)=0} \frac{p(\rho)}{q'(\rho)}\psi(-\rho). $$
The function $f$ has the form
$$ f(x) = -\frac{\coth\frac\pi2}{\pi} + \frac2\pi\sum_{n\geq0} S(n, \sin(2\pi nx), \cos(2\pi nx)), $$
where $S(n,\gamma,\delta) = \frac{p(n,\gamma)/q(n,\delta)}{\cosh\pi-\delta}$ is given as in sos440's answer by
$$ p(n,\gamma) = (2n^2x^2-1)\sinh\pi+2nx\gamma, \qquad q(n)=(4n^2-1)(4n^4x^4+1). $$
When $x=r/s$ is rational, this can be simplified. Rewrite the sum as 
$$ \sum_{n\geq0}S(n,\sin2\pi nx, \cos2\pi nx) = \sum_{0\leq\alpha<s} \frac{1}{\cosh\pi-\cos2\pi\alpha x} \sum_{m\geq0}
\frac{p(ms+\alpha, \sin2\pi\alpha x)}{q(ms+\alpha)}, $$
and now the inner sum over $m$ is a convergent sum of terms rational in $m$, so can be written as
$$ \sum_{q(\rho)=0}\frac{-1}{q'(\rho)} \frac1{s} \sum_{0\leq\alpha<s} \frac{p(\rho, \sin2\pi\alpha
   x)}{\cosh\pi-\cos2\pi\alpha x} \psi\left(\frac{\alpha-\rho}{s}\right),
$$
where $\rho$ runs over the roots of $q(n)$: $\{\pm\frac12,
\frac{e^{\pi i j/4}}{x\sqrt{2}}\}$, $j$ odd.
This is pretty much a closed form, it is a finite sum of digamma terms, but
you might not be able to simplify it much, especially by hand. For
example, Mathematica now gives:
$$
f({\textstyle\frac52}) = 
\left(
68-68 \sqrt{5}-500 \sqrt{10-2 \sqrt{5}} \cosh\frac{\pi }{5}+272 \cosh\frac{2 \pi }{5}-\left(125 \sqrt{10-2 \sqrt{5}}+125 \sqrt{5 \left(10-2 \sqrt{5}\right)}\right) \cosh\pi-\left(160+160 \sqrt{5}\right) \sinh\frac{\pi }{5}+160 \sinh\frac{3 \pi }{5}-160 \sinh\frac{7 \pi }{5}
\right)/\left(
689(-1+\sqrt{5}-4 \cosh\frac{2 \pi }{5})\sinh\pi\right)
$$
A: My calculation shows that
$$f(x) = \frac{1}{\pi}\coth\left(\frac{\pi}{2}\right)
+ \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{(2n^{2}x^{2} - 1) \sinh\pi + 2nx\sin (2\pi n x)}{(4n^{2} - 1)(4n^{4}x^{4} + 1) (\cosh \pi -\cos (2\pi n x))}. \tag{1} $$
This allows (at least theoretically) calculate the value of $f$. This shows that


*

*The series $\text{(1)}$ defines a holomorphic function, say $\tilde{f}$, on $\Bbb{C} \setminus \Bbb{R}$ with poles
$$ \frac{\pm 1 \pm i}{2n} : n = 1, 2, \cdots
\quad\text{and}\quad
\frac{k}{n} \pm \frac{i}{2n} : k \in \Bbb{Z}, n = 1, 2, \cdots. $$
Because these poles accumulate to $\Bbb{R}$ as $n\to\infty$, I'm not sure if $\tilde{f}$ and $f$ can be amalgamated to yield a nice function.

*This sum converges uniformly on $\Bbb{R}$: putting $y = nx$, 
$$ \left| \frac{(2n^{2}x^{2} - 1) \sinh\pi + 2nx\sin (2\pi n x)}{(4n^{4}x^{4} + 1) (\cosh \pi -\cos (2\pi n x))} \right| \leq \frac{(2y^{2} + 1)\sinh \pi + 2|y|}{(4y^{4}+1)(\cosh \pi - 1)}, $$
which is uniformly bounded by some constant $C > 0$. Thus we can take limit $x \to \infty$ pointwise to $\text{(1)}$, proving that
$$ \lim_{x\to\infty} f(x) = \frac{1}{\pi}\coth\left(\frac{\pi}{2}\right). $$

*The graph of its derivative is shown below: I'm unable to find any clear pattern that the zeros of $f'(x)$ must satisfy. One exception is that some zeros seem to be very close to integers.


To derive $\text{(1)}$, I just exploited the following formula:
$$ \left| \sin x \right| = \frac{2}{\pi} \sum_{n \in \Bbb{Z}} \frac{1 - \cos(2nx)}{4n^{2} - 1}, \quad x \in \Bbb{R} $$
Since each summand is non-negative, Tonelli says that we can plug this to the integral and interchange the order of limiting operators to obtain
$$ f(x) = \frac{4}{\pi^{2}} \sum_{m \in \Bbb{Z}} \sum_{n \in \Bbb{Z}} \frac{1}{(4m^{2}-1)(4n^{2} - 1)} \int_{0}^{\infty} (1 - \cos 2mt)(1 - \cos 2nxt) e^{-t} \, dt. $$
Calculating the integral and utilizing the fact $\sum_{n\in\Bbb{Z}} (4n^{2}-1)^{-1} = 0$, this reduces to
$$ f(x) = \frac{4}{\pi^{2}} \sum_{m \in \Bbb{Z}} \sum_{n \in \Bbb{Z}} \frac{1}{(4m^{2}-1)(4n^{2} - 1)(4(nx+m)^{2} + 1)}. $$
Further simplification then gives $\text{(1)}$.

EDIT.


*

*Numerical observation suggest that $f$ is $C^{2}(\Bbb{R}^{+})$, and heuristic investigation suggests that $f^{(3)}$ has jump discontinuity at every point in $\frac{1}{2}\Bbb{Z} \cup \frac{1}{3}\Bbb{Z}$.



