Let \begin{equation}g(t)=\begin{cases}\frac{\sin{\frac{1}{2}}t}{t}, & t \not =0 \\ \frac{1}{2}, &t=0 \end{cases} \end{equation} Calculate $\text{lim}_{m\to \infty}\frac{1}{\pi}\int_{-\pi}^{\pi}D_{m}(t)g(t)dt$ where $D_{m}(t)$ is Dirchlet's kernel.
What I have done is to use the identity $D_{m}(t)=\frac{\sin{(m+\frac{1}{2})}t}{2\sin{\frac{1}{2}t}}$. This substitution gave me \begin{equation} \frac{\sin{(m+\frac{1}{2})t}}{2t} \tag{*}\end{equation} Now, the integral of $(*)$ is an non-elementary function $\frac{1}{2}\text{Si}(\frac{1}{2}(2m+1)t)$ (this can be found using Taylor expansion),which evaluated becomes $\text{Si}((m+\frac{1}{2})\pi)$. As $m \to \infty$ we have $\text{Si}((m+\frac{1}{2})\pi)=\frac{\pi}{2}$ So my answer is that $\text{lim}_{m\to \infty}\frac{1}{\pi}\int_{-\pi}^{\pi}D_{m}(t)g(t)dt=\frac{1}{2}$.
My question to you is whether or not there are other methods to solve this problem? In particular, I would like to know if there are methods that don't involve $\text{Si}$-function.
If you know about Dirichlet's kernel, you probably know about Fourier series and their convergence. If not, please disregard the following.
Consider the $2\,\pi$ periodic extension of $g$. It is continuous and piecewise $C^1$. Dirichlet's convergence theorem for Fourier series implies that the Fourier series of $g$ converges point-wise to $g$. The $m$-th partial sum of the Fourier series of $g$ is given by the convolution of $D_m$ and $g$: $$ \frac{1}{\pi}\int_{-\pi}^\pi D_m(t-x)g(t)\,dt. $$ Thus $$ \lim_{m\to\infty}\frac{1}{\pi}\int_{-\pi}^\pi D_m(t)g(t)\,dt=g(0)=\frac12. $$