Let $f$ be a monotonic and bounded function defined on [0,1) and $\lim_{x\rightarrow 0^+}f(x)$ exists. Prove \begin{align} \lim_{\lambda\rightarrow \infty}\int_0^1 f(x)\frac{\sin\lambda x}{x}dx = \frac{\pi}{2}f(0^+) \end{align}
I am thinking of dividing this integral into two parts: \begin{align} \int_0^1 f(x)\frac{\sin\lambda x}{x}dx = \int_0^\delta+\int_\delta^1 f(x)\frac{\sin\lambda x}{x}dx \end{align}
It is easy to show the second part goes to zero, however I don't really know how to deal with the first part. In other words, I need to estimate \begin{align} \lim_{\lambda\rightarrow\infty}\int_0^\delta (f(x)-f(0^+))\frac{\sin\lambda x}{x}dx = 0 \end{align} However $\frac{\sin\lambda x}{x}$ is oscillating and is not absolutely integrable.
Note that we can write the kernel $\frac{\sin(\lambda x)}{x}$ as the integral
$$\frac{\sin(\lambda x)}{x}=\frac12\int_{-\lambda }^\lambda e^{i x t}\,dt\tag 1$$
Using $(1)$ we can write
$$\begin{align} \int_0^1 f(x)\frac{\sin(\lambda x)}{x}\,dx&=\frac12\int_0^1 f(x)\int_{-\lambda }^\lambda e^{i x t}\,dt\,dx\\\\ &=\frac12\int_{-\lambda }^\lambda \int_0^1 f(x)e^{i x t}\,dx\,dt\\\\ &=\frac12\int_{-\lambda }^\lambda \int_{-\infty}^\infty f(x)\xi_{[0,1]}(x)e^{i x t}\,dx\,dt\\\\ \end{align}$$
Letting $\lambda \to\infty$ yields
$$\lim_{\lambda \to \infty}\int_0^1 f(x)\frac{\sin(\lambda x)}{x}\,dx=\frac12\int_{-\infty }^\infty \int_{-\infty}^\infty f(x)\xi_{[0,1]}(x)e^{i x t}\,dx\,dt\tag2$$
Since $f(x)\xi_{[0,1]}(x)\in L^1$, the inner integral on the right-hand side of $(2)$ is the Fourier Transform of $f(x)\xi_{[0,1]}(x)$. And the outer integral is $2\pi$ times the inverse Fourier Transform of the Fourier Transform, evaluated at $0$.
Inasmuch as $f(x)\xi_{[0,1]}(x)$ is discontinuous at $0$, we conclude See this on the Fourier Inversion Theorem that
$$\begin{align} \lim_{\lambda \to \infty}\int_0^1 f(x)\frac{\sin(\lambda x)}{x}\,dx&=\pi \left(\frac12 (f(0^+)\xi_{[0,1]}(0^+)+\frac12f(0^-)\xi_{[0,1]}(0^-)\right)\\\\ &=\frac\pi2 f(0^+) \end{align}$$
as was to be shown!