I am trying to evaluate the integral$$\int_{0}^{\infty}\frac{\sin\left(k\ln\left(t\right)\right)}{\sqrt{t}}\,\left\{\frac{1}{t}\right\} \, dt,$$ where $\left\{\cdots\right\}$ denotes fractional part function and $k$ is a real number.
Let $t = e^u$, then $dt = e^u \, du$ and the integral becomes:$$ \int_{-\infty}^{\infty} \frac{\sin(ku)}{\sqrt{e^u}}\{e^{-u}\} e^u \, du=\int_{-\infty}^{\infty} \sin(ku) \{e^{-u}\} e^{u/2} \, du.$$
The fractional part function $\{x\} = x - \lfloor x \rfloor$ may be written as:
$$ \{x\} = \frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\sin(2 \pi n x)}{ n}.$$So the above integral becomes:
\begin{align}& \int_{-\infty}^{\infty} \sin(ku) \{e^{-u}\} e^{u/2} \, du\\[3mm] = & \ \int_{-\infty}^{\infty} \sin(ku) e^{u/2}\left[\frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\sin\left(2 \pi n e^{-u}\right)}{ n}\right] \, du.\end{align}
Is this correct? What after this?