Functions of class $C^{1}$?

Here I will only show that $u$ is of class $C^{1}$ since the same arguments apply for $v$ too.

Let $x,y\in\mathbb{R}$ and $t\geq0$ , then by the mean value theorem, we have $\sin(xt)-\sin(yt)=t\cos(tz_{t})(x-y)$ , where $z_{t}$ is between $x$ and $y$ , which yields $u(x)-u(y)=\int_{0}^{\infty}\mathrm{e}^{-t}\sqrt{t}\cos(tz_{t})(x-y)\mathrm{d}t.$ Thus, for all $x,y\in\mathbb{R}$ , we have $|u(x)-u(y)|\leq\sqrt{\pi}|x-y|/2$ since $\int_{0}^{\infty}\mathrm{e}^{-t}\sqrt{t}\mathrm{d}t=\sqrt{\pi}/2$ . Given $\varepsilon>0$ , we may let $\delta:=2\varepsilon/\sqrt{\pi}$ to have $|u(x)-u(y)|<\varepsilon$ for all $x,y\in\mathbb{R}$ with $|x-y|<\delta$ . This shows that $u$ is continuous.

For the next step, we shall show that for $x\in\mathbb{R}$ , the integral $\int_{0}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\sin(tx)\mathrm{d}t$ converges uniformly, i.e., for every $\varepsilon>0$ there exists $r>0$ such that $\bigg|\int_{s}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\sin(tx)\mathrm{d}t\bigg|$ $<\varepsilon$ for all $s\geq r$ and $x\in\mathbb{R}$ .

Since $\int_{0}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\mathrm{d}t=\sqrt{\pi}$ , for any given $\varepsilon>0$ , there exits $r>0$ such that $\int_{r}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\mathrm{d}t<\varepsilon$ , which yields $\bigg|\int_{s}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\sin(tx)\mathrm{d}t\bigg|$ $\leq\int_{s}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\mathrm{d}t<\varepsilon$ for all $s\geq r$ . Therefore, the integral converges uniformly, which allows us to change the order of the differential operator and the integral. Then, for all $x\in\R$ , we have $u^{\prime}(x)=\int_{0}^{\infty}\mathrm{e}^{-t}\frac{1}{\sqrt{t}}\big(\sin(tx)\big)^{\prime} \mathrm{d}t$ $=\int_{0}^{\infty}\mathrm{e}^{-t}\sqrt{t}\cos(tx)\mathrm{d}t$ showing that $u^{\prime}$ exists. We can show that $u^{\prime}$ is continuous as well, i.e., $u$ is in $C^{1}$ .