Prove this identity about limit of integral with parameter Here is a question I met in an exam several days ago,I failed to give an answer that time and still feels confused now:
Suppose $f$ is monotonely increasing on $[0,1]$,prove that 
$$\lim_{y\to \infty}\int_{0}^{1}f(x)\frac{\sin(xy)}{x}dx={\pi\over 2}f(0+)$$
It's easy to see that we can assume $f(0+)=0$,but how to do the estimate next?
Should I try to find some appropriate $\delta\gt 0$ and estimate them separately?
 A: This is a special case of Jordan's theorem, which follows from the Riemann-Lebesgue lemma, but $f(x)$ must be bounded on $[0, 1]$. You can find the proof in Apostol's Mathematical Analysis (p. 315). By the way, you cannot just assume that $f(0+) = 0$.
A: A bit more details than @glebovg's answer follow. For whatever $\delta$ we pick, we can write
$$\int_0^1 f(x) \frac{\sin(xy)}{x} dx = \int_0^\delta f(x) \frac{\sin(xy)}{x} dx + \int_\delta^1 \frac{f(x)}{x} \sin(xy) dx.$$
The second term goes to zero as $y \to \infty$ by the Riemann-Lebesgue lemma, provided $f(x)/x$ is integrable on $[\delta,1]$.
For the first term, note that we can assume $f$ is right-continuous, since it is increasing and we can change its value at countably many points without changing the integrals. Thus given $\varepsilon > 0$ we can find $\delta > 0$ so that for $0 < x < \delta$ we have $|f(x) - f(0^+)| < \varepsilon$. So if we let $e(x)=|f(x)-f(0^+)|$, we have $e(x)<\varepsilon$ on this interval. So
$$\left | \int_0^\delta f(x) \frac{\sin(xy)}{x} dx - f(0^+) \int_0^\delta \frac{\sin(xy)}{x} dx \right | < \int_0^\delta e(x) \frac{\sin(xy)}{x} dx < \varepsilon \int_0^\delta \frac{\sin(xy)}{x} dx.$$
The last step is to show that
$$\lim_{y \to \infty} \int_0^\delta \frac{\sin(xy)}{x} dx = \frac{\pi}{2}.$$
Let $u=xy$, $du=ydx$, $dx=du/y$, $u(0)=0$, $u(\delta)=y\delta$, $x=u/y$. Now
$$\int_0^\delta \frac{\sin(xy)}{x} dx = \int_0^{y\delta} \frac{\sin(u)}{u/y} du/y = \int_0^{y\delta} \frac{\sin(u)}{u} du.$$
And we know that as $y \to \infty$ this goes to $\frac{\pi}{2}$ by standard methods. Compiling everything we get the desired result.
