Let's define the sequence $$ a_n=\int_0^1\sin (2 \pi nx)\, \mathrm{e}^{-x}dx. $$ This sequence converges to $0$, as $n\to\infty$, and this could be shown using integration by parts.
Now I was thinking if the exponential is changed by $\mathrm{e}^{-\sqrt x}$ if it still converges to 0. Which is true by Wolfram. But I couldn't come up with a proof.
In order to show that $$ \lim_{n\to\infty}\int_0^1 \sin(2\pi n x)\,\mathrm{e}^{-\sqrt{x}}\,dx=0, $$ let $\varepsilon>0$. Then integration by parts provides $$ \int_{\varepsilon/2}^1 \sin(2\pi n x)\,\mathrm{e}^{-\sqrt{x}}\,dx=-\frac{\cos(2\pi n)-\cos(2\pi n \varepsilon)}{2\pi n}+\frac{1}{4\pi n}\int_{\varepsilon/2}^1 x^{-1/2}\cos(2\pi n x)\,\mathrm{e}^{-\sqrt{x}}\,dx, $$ and hence $$ \Big|\int_{\varepsilon/2}^1 \sin(2\pi n x)\Big|\le \frac{1}{\pi n}+\frac{1}{2\pi n}(1-\sqrt{\varepsilon}). $$ Clearly, there exists an $n_0\in\mathbb N$, such that the left hand side of the above is less than $\varepsilon/2$, whenever $n\ge n_0$.
Thus, for $n\ge n_0$,
$$ \Big|\int_0^1 \sin(2\pi n x)\,\mathrm{e}^{-\sqrt{x}}\,dx\Big|\le \Big|\int_0^\varepsilon \sin(2\pi n x)\,\mathrm{e}^{-\sqrt{x}}\,dx\Big|+ \Big|\int_\varepsilon^1 \sin(2\pi n x)\,\mathrm{e}^{-\sqrt{x}}\,dx\Big|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} $$