Testing uniform convergence of improper integeal I have some trouble with showing uniform convergence of an improper integral, for which I normally use the comparison test.
I guess it won't be working in this case, so now I ask for help.
If we define for $t\in[0,1]$:
$$
I(t):=\int_0^\infty\frac{x\sin(2tx)}{x^2+1}dx.
$$
I want to show, that this integral converges uniformly. What I would do normally is:
$$
\left|\frac{x\sin(2tx)}{x^2+1}\right|\leq\frac{x}{x^2+1}=:g(x),
$$
but obviously $\int_0^\infty g(x)dx$ does not converge. Is there any chance to prove the uniform convergence?
I don't know if it helps, but I alredy solved the integral with complex analysis:
$$
I(t)=\frac{\pi}{2e^{2t}}.
$$
Thanks for your help!
 A: You are actually very close to the answer. Consider the function $I_R(t)$ defined by
$$ I_R(t) := \int_{0}^{R} \frac{x \sin(2tx)}{x^2 + 1} \, \mathrm{d}x $$
It is easy to check that $I_R$ is continuous on $[0, 1]$ for each $R \in (0, \infty)$. So, if $I_R$ converges uniformly on $[0, 1]$ as $R \to \infty$, then the limit function must be continuous as well. However,
$$ \lim_{R \to \infty} I_R(t) = I(t)
= \begin{cases}
0, & t = 0, \\[0.5em]
\frac{\pi}{2}e^{-2t}, & t > 0,
\end{cases} $$
which is not continuous on $[0, 1]$. Therefore the convergence is not uniform.

Here is another way of showing the above claim. As $R \to \infty$, we have
\begin{align*}
I(R^{-1}) - I_{\pi R}(R^{-1})
&= \int_{\pi R}^{\infty} \frac{x \sin(2x/R)}{x^2 + 1} \, \mathrm{d}x \\
&= \int_{\pi}^{\infty} \frac{u \sin(2u)}{u^2 + R^{-2}} \, \mathrm{d}u \tag{$x = Ru$} \\
&= \int_{\pi}^{\infty} \frac{\sin(2u)}{u} \, \mathrm{d}u
- \frac{1}{R^2} \int_{\pi}^{\infty} \frac{\sin(2u)}{u(u^2 + R^{-2})} \, \mathrm{d}u
\end{align*}
The second integral in the last line tends to $0$ as $R \to \infty$, in light of a simple estimate:
$$ \left| \frac{1}{R^2} \int_{\pi}^{\infty} \frac{\sin(2u)}{u(u^2 + R^{-2})} \, \mathrm{d}u \right|
\leq \frac{1}{R^2} \int_{\pi}^{\infty} \frac{\mathrm{d}u}{u^3} $$
Moreover, by substituting $s = 2u$ and decomposing the integral into subintervals of length $\pi$, we get
$$ \int_{\pi}^{\infty} \frac{\sin(2u)}{u} \, \mathrm{d}u
= \sum_{k=2}^{\infty} (-1)^k \int_{0}^{\pi} \frac{\sin s}{s + k\pi} \, \mathrm{d}s
> 0, $$
where the last step is a simple consequence of the alternating series estimation theorem. These together imply that
$$ \lim_{R \to \infty} \sup_{t \in [0, 1]} \left| I(t) - I_R(t) \right|
\geq \int_{\pi}^{\infty} \frac{\sin(2u)}{u} \, \mathrm{d}u > 0. $$
Therefore $I_R$ does not converge uniformly to $I$ on $[0, 1]$.
