I am preparing for an exam, and in finding an old exam for my course, I have run into following problem:
We are to show that the function $$u(x)=\int_0^x \frac{\sin(x-y)}{1+y^2} dy$$
is either not uniformly continuous or uniformly continuous on the interval $[0,\infty)$. My current approach is to first expand this integral as $$u(x)=\int_0^x \frac{\sin(x)\cos(y)-\cos(x)\sin(y)}{1+y^2} dy$$ as a hint implies we are to do. Next, my thought is to assume the function is Riemann integrable on some bounded interval $[0,a<\infty]\subset [0,\infty)$, which implies $f$ is locally bounded. Since on $[0,a]$, $f$ is locally bounded and we suppose Riemann integrability, then $\forall \epsilon>0 \hspace{0.1 cm} \exists \delta>0: mesh(P)<\delta \Rightarrow U(f,P)-L(f,P)<\epsilon$ for all partitions $P$ of $[0,a]$.
This is where I am stuck. I do not know how to proceed from here, or if this is even a correct approach.
