Suppose $f\colon[0,a]\to \mathbb{R}$ with $a>0$, is monotonically increasing on $[0,a]$ and $f(x) \geq 0$ for all $x\in [0,a]$. Calculate
$$\lim_{p\to +\infty} \int_{0}^{a}f(x)\frac{\sin(px)}{x}\,\mathrm dx.$$
My idea is as follows. Let $t=px$, then we get $$\int_{0}^{a}f(x)\frac{\sin(px)}{x}\,\mathrm dx= \int_{0}^{ap}f\left(\frac{t}{p}\right)\frac{\sin t}{t}\,\mathrm dt=\int_{0}^{+\infty}\chi_{[0,ap]}(t)f\left(\frac{t}{p}\right)\frac{\sin t}{t}dt. $$
If we change the order of limit and integral, I obtain
$$\begin{align*}\lim_{p\to +\infty} \int_{0}^{a}f(x)\frac{\sin(px)}{x}\,\mathrm dx&=\lim_{p\to +\infty} \int_{0}^{+\infty}\chi_{[0,ap]}(t) f\left(\frac{t}{p}\right)\frac{\sin t}{t}\,\mathrm dt\\&=\int_{0}^{+\infty} \lim_{p\to +\infty} \chi_{[0,ap]}(t) f\left(\frac{t}{p}\right)\frac{\sin t}{t}\,\mathrm dt\\&=\int_{0}^{+\infty}f(0^+)\frac{\sin t}{t}dt=\frac{\pi}{2}f(0^+).\end{align*}$$
But I can't explain why we can change the order there. Can I use the Lebesgue Dominated Convergence theorem? Or perhaps use other methods to solve this question?
Best Answer
Since $f$ is monotone increasing and the integral is not affected by altering the values of $f$ at countably many points, we may replace $f$ by its right-continuous version
$$ g(x) = \lim_{y \to x^+} f(y) $$
with the convention that $g(a) = f(a)$.
Let $\mu$ denote the Stieltjes measure on $[0, a]$ corresponding to $g$, so that $\mu([0, x]) = g(x) - g(0)$ for any $x \in [0, a]$. Then, with the sine integral function $\operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, \mathrm{d}t$, we have
\begin{align*} \int_{0}^{a} f(x) \frac{\sin(px)}{x} \, \mathrm{d}x &= \int_{0}^{a} g(x) \frac{\sin(px)}{x} \, \mathrm{d}x \\ &= \int_{0}^{a} \left( g(0) + \int_{[0, a]} \mathbf{1}_{\{t \leq x\}} \, \mu(\mathrm{d}t) \right) \frac{\sin(px)}{x} \, \mathrm{d}x \\ &= \operatorname{Si}(pa)g(0) + \int_{[0, a]} \left( \int_{0}^{a} \mathbf{1}_{\{t \leq x\}}\frac{\sin(px)}{x} \, \mathrm{d}x \right) \, \mu(\mathrm{d}t) \\ &= \operatorname{Si}(pa)g(0) + \int_{[0, a]} \left( \operatorname{Si}(pa) - \operatorname{Si}(pt) \right) \, \mu(\mathrm{d}t). \end{align*}
Now we know that
$\mu$ is a finite measure with $\mu([0, a]) = f(a) - f(0) < \infty$, and
$\operatorname{Si}(x)$ is bounded and $\operatorname{Si}(x) \to \frac{\pi}{2}$ as $x \to \infty$.
Hence by the dominated convergence theorem,
\begin{align*} \int_{0}^{a} f(x) \frac{\sin(px)}{x} \, \mathrm{d}x &\to \frac{\pi}{2}g(0) + \int_{[0, a]} \left( \frac{\pi}{2} - \frac{\pi}{2}\mathbf{1}_{\{t > 0\}} \right) \, \mu(\mathrm{d}t) \\ &= \frac{\pi}{2}g(0), \end{align*}
where the last equality follows from $\mu(\{0\}) = 0$.