$$
\mbox{Let}\ \operatorname{S}\left(x\right) =
\sum_{n = 1}^{\infty}\sin\left(\,\sqrt{\,{n}\,}\,\right)
{\rm e}^{-nx},\quad \forall x > 0
$$
$\operatorname{S}\left(x\right)$ exists $\forall x > 0$
$\left(~\left\vert\,\sin\left(\sqrt{n}\right)
{\rm e}^{-nx}\,\right\vert \leq {\rm e}^{-nx}\ \mbox{and}\ \sum{\rm e}^{-nx}\ \mbox{geometric series}~\right)$
I wonder if the limit
$\displaystyle\lim_{x \to 0^{+}}\,\,\operatorname{S}\left(x\right)$ exists and how to calculate it.
I don't see , if it's useful, how to use
$\displaystyle\int_{0}^{\infty}\sin(\,\sqrt{\,{t}\,}\,){\rm e}^{-pt}
\,{\rm d}t =
\frac{{\rm e}^{-1/\left(4p\right)}}{2p}\,\sqrt{\frac{\pi}{p}}$
Best Answer
I will use the original notation $S(x)=\sum_{n=1}^\infty e^{-nx}\sin\sqrt{n}$. Then we have (see below) $$S(0^+):=\lim_{x\to 0^+}S(x)=-\frac{1}{2\pi\sqrt2}\sum_{n=1}^\infty\frac{1}{n^{3/2}}\cos\left(\frac\pi4+\frac1{8n\pi}\right).$$
While this converges pretty slowly, PARI/GP's
sumnumis still able to compute it: $$S(0^+)=-0.203568606528057117563426728913068844242517762840958\cdots$$Also, as suggested by @reuns in a comment, this can be rewritten as $$S(0^+)=-\frac1{4\pi}\sum_{n=0}^\infty\frac{(-1)^{\lceil n/2\rceil}}{n!(8\pi)^n}\zeta\left(n+\frac32\right)$$ (using the power series of $\sqrt2\cos(z+\pi/4)=\cos z-\sin z$ and exchanging the sums). With a high-precision computation of $\zeta$ (which PARI/GP is capable of), this is clearly superior.
To obtain the series above, we invert the $\int_0^\infty e^{-st}\sin\sqrt{t}\,dt=\ldots$ from the OP: $$\sin\sqrt{t}=\frac1{4i\sqrt\pi}\int_{c-i\infty}^{c+i\infty}s^{-3/2}\exp\left(st-\frac1{4s}\right)ds,$$ where $c>0$ is arbitrary; using this for $0<c<x$, we obtain $$S(x)=\frac1{4i\sqrt\pi}\int_{c-i\infty}^{c+i\infty}\frac{s^{-3/2}e^{-1/(4s)}}{e^{x-s}-1}\,ds.$$
Now, for an odd positive integer $N$, and $0<c<x<T$, let $C_{N,T}$ be the rectangular contour with vertices at $c\pm N\pi i$ and $T\pm N\pi i$ (and standard orientation). Then $\int_{c-i\infty}^{c+i\infty}=-\lim\limits_{N,T\to\infty}\int_{C_{N,T}}$ (the integrals along the "far" sides tend to $0$). Thus $S(x)$ equals $-\sqrt\pi/2$ times the sum (over $n\in\mathbb{Z}$) of the residues of the integrand at its poles $s=x+2n\pi i$: $$S(x)=\frac{\sqrt\pi}2\sum_{n\in\mathbb{Z}}(x+2n\pi i)^{-3/2}e^{-1/(4x+8n\pi i)}.$$ Here we can take $x\to0^+$ termwise (dominated convergence). This gives the expected result.