Finding interval of uniform convergence for $\sin^2(tx)/x^2$

Question

I need to find the interval of uniform convergence of $$\int_0^\infty \frac{\sin^2(tx)}{x^2} \, dx .$$ Splitting the integral in $$\int_0^1\frac{\sin^2(tx)}{x^2} \, dx + \int_1^\infty \frac{\sin^2(tx)}{x^2} \, dx,$$ it's clear we can apply the M-Weirstrass test to the second integral, comparing it to $\frac{1}{x^2}$. But how do I compute the first one? I've tried several change of variables and nothing works...

Answer 1

$$ \int_0^\infty \frac{\sin^2(tx)}{x^2} \, dx $$ In a comment you say you want to figure out for which values of $t$ the integral converges. But you begin by speaking of "uniform convergence". The term "uniform convergence" has a standard definition, and that is not it at all.

You split it into two parts: $$ \int_0^1 \frac{\sin^2(tx)}{x^2} \,dx + \int_1^\infty \frac{\sin^2(tx)}{x^2} \,dx $$ You have $$ 0 \le \frac{\sin^2(tx)}{x^2} \le \frac 1 {x^2} \text{ and } \int_1^\infty \frac 1 {x^2}\,dx <+\infty, $$ so that "converges".

With $\displaystyle\int_0^1 \frac{\sin^2(tx)}{x^2} \, dx,$ you have $\displaystyle \frac{\sin^2(tx)}{x^2} = t^2\left( \frac{\sin(tx)}{tx} \right)^2 \to t^2 \text{ as } x\downarrow 0,$ so this can be extended to a continuous function on the closed interval $[0,1]$ by making it equal to $t^2$ at $x=0.$ Thus you have the integral of a continuous function over a closed bounded interval. That converges regardless of the value of $t.$

I put scare quotes around "converges" above because the ways in which that could be defined will bear some examination. One sense is that an integral $\displaystyle\int_A g(x)\,dx$ "converges" if $\displaystyle\int_A|g(x)|\,dx<+\infty.$ That holds in this instance. Another sense is that "converges" may mean $\displaystyle\lim_{a\,\to\,+\infty} \int_0^a$ exists. In this case, since the function being integrated is everywhere nonnegative and so $\displaystyle\int_0^a$ is an increasing function of $a,$ this becomes merely a matter of that function of $a$ being bounded, and that also holds in this case.

So the bottom line is that this integral is well behaved in any of the senses above no matter what number $t$ is.