Prove:
$$\int_{0}^{\infty} \frac{\sin(x)}{x} dx=\frac{\pi}{2}$$
starting with the facts that:
$$\int_{-\pi}^{\pi}D_N(\theta)=2\pi, \ \ \ \ \ \text{and }\ f(\theta)=\frac{1}{\sin(\frac{\theta}{2})}-\frac{2}{\theta}\ \text{ is continuous on} \ [-\pi, \pi] $$
(apply the Riemann-Lebesgue lemma)
I realize there are many different proofs of this fact already but I haven't seen one using the given facts.
What I have so far is that as is given:
$$\int_{- \pi}^{\pi} \frac{\sin((N+\frac{1}{2})\theta)}{\sin(\frac{\theta}{2})} =2\pi$$
and in order to use the Riemann-Lebesgue lemma I need to find a useful function $g(x)$ and use $\lim_{n \to \infty} \hat{g}(n)=0$. Given what I know, it seems like if I were able to have:
$$\hat{g}(n) =\int_{0}^{n} \frac{\sin(x)}{x}dx -\frac{\pi}{2} $$
then I would be done, so I attempted to find the Fourier coefficients of the given $f(\theta)$ but this did not seem particularly fruitful.
Best Answer
You can start by integrating the trigonometric form of the Dirichlet kernel.
$$\int_{-\pi}^{\pi}D_N(x)dx=2\pi=\int_{-\pi}^{\pi}\frac{\sin((N+1/2)x)}{\sin(x/2)}dx$$
Rewrite the integrand to make use of the second fact from the prompt.
$$\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{1}{\sin(x/2)}-\frac{2}{x}+\frac{2}{x})dx=2\pi$$
Split up the integral and consider the limit as $N\to\infty$.
$$\lim_{N\to\infty}\left[\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{1}{\sin(x/2)}-\frac{2}{x})dx+\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{2}{x})dx\right]=2\pi$$
The first integral vanishes in the limit per the Riemann-Lebesgue lemma. The second has an even integrand, so take the part from zero to $\pi$ and simplify.
$$\lim_{N\to\infty}\int_{0}^{\pi}\sin((N+1/2)x)(\frac{1}{x})dx=\frac{\pi}{2}$$
Finally, change variables to $y=(N+1/2)x$.
$$\lim_{N\to\infty}\int_{0}^{(N+1/2)\pi}\frac{\sin(y)}{y}dy=\frac{\pi}{2}$$