Prove $f(x)=\int_{-1}^{x}\sin{(xt)e^{-t^2}}dt$ is uniformly continuous on $I=[0,\infty)$
This should be relatively straight forward, but I just wanted to make sure I got it right.
$\textbf{Proof}:$ We need to show that $\forall \varepsilon>0, \exists \delta>0;\forall x,y\in I$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon$
Let $x,y\in I$ and $\varepsilon>0$ be given. Then, \begin{align} |f(x)-f(y)|&=\left|\int_{-1}^{x}\sin{(xt)e^{-t^2}}dt-\int_{-1}^{y}\sin{(yt)e^{-t^2}}dt\right| \\ &= \left| \int_{y}^{x}\left[\sin{(xt)}+\sin{(yt)}\right]e^{-t^2}dt \right| \\ &\leq \int_{y}^{x}\left|\left[\sin{(xt)}+\sin{(yt)}\right]e^{-t^2}\right| dt \\ &\leq 2\int_{y}^{x}\left|e^{-t^2}\right| dt \\ &\leq 2\int_{y}^{x} dt \\ &\leq 2|x-y| \end{align} Now, if $|x-y|<\delta$, choosing $\delta=\varepsilon/2$ completes our proof, so the function is uniformly continous on $I$.
Does this proof seem correct?
I think it’s almost all completely right, especially the use of $\varepsilon,\delta$ argument but there is a mistake in the second equal sign (as Conan mentioned). Now you can use the following argument: $$|f(x)-f(y)|=\left|\int_{-1}^x\sin(xt)e^{-t^2}dt-\int_{-1}^y\sin(yt)e^{-t^2}dt\pm \int_{-1}^x\sin(yt)e^{-t^2}dt\right|\le$$ $$\le\left|\int_{-1}^x(\sin(xt)-\sin(yt))e^{-t^2}dt\right|+\left|\int_y^x \sin(yt)e^{-t^2} dt\right|\le$$ $$\le\int_{-1}^x\left|\sin(xt)-\sin(yt)\right|e^{-t^2}dt+ \int_y^x \left|\sin(yt)e^{-t^2} \right|dt \le $$ , by the Lagrange theorem together with the fact that $|\sin'(x)|=|\cos(x)|\le 1$, we get: $$\le\int_{-1}^\infty\left|x-y\right|te^{-t^2}dt+ \int_y^x 1\;dt \le\;\; ... $$ p.s. note that now $\int_{-1}^\infty te^{-t^2}dt$ it's simply a constant.