Alternate proof of Dirichlet integral $\frac{\sin(x)}{x}$.

Question

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.

Answer 1

Here's a simplified version of a proof with which Richard Feynman popularised differentiation under the integral sign among physicists, who to this day call it Feynman's trick. (He integrated $\int_0^\infty\sin x\,e^{-xy}dx$ with respect to $y$, but we can make it simpler.) Using $\dfrac{1}{x}=\int_0^\infty e^{-xy}dy$ we have $$\int_0^\infty\frac{\sin x}{x}dx=\int_0^\infty dy\int_0^\infty dx \,e^{-xy}\sin x=\int_0^\infty \frac{dy}{1+y^2}=\frac{\pi}{2}.$$

Answer 2

Rewrite the original integral $$\int^{\infty}_0\frac{\sin x}{x}\,dx=\sum_{k=1}^{\infty}\int^{k\pi}_{(k-1)\pi}\frac{\sin x}{x}\,dx$$ You can show that the above alternating series is convergent by some appropriate convergence test. Denote by $$S_n:=\sum_{k=1}^{n}\int^{k\pi}_{(k-1)\pi}\frac{\sin x}{x}\,dx$$ Let $b\geqslant\pi$ then there is an integer $n$ such that $n\pi\leqslant b<(n+1)\pi$. Then we have the following $$S_n\leqslant \int^b_0\frac{\sin x}{x}\,dx\leqslant S_{n+1}$$ Since the alternating series above is convergent i.e. $\lim_n S_n$ exists then it follows that $$\lim_{b\to\infty}\int^b_0\frac{\sin x}{x}\,dx=\lim_nS_n=\int^{\infty}_0\frac{\sin x}{x}\,dx$$ From the given condition $$\int^{\pi}_{-\pi}\frac{\sin(n+1/2)x}{\sin x/2}\,dx=2\pi$$ and the deinition of $f(x)$ we get $$2\int^{\pi}_{-\pi}\frac{\sin(n+1/2)x}{x}\,dx=\int^{\pi}_{-\pi}\frac{\sin(n+1/2)x}{\sin x/2}\,dx-\int^{\pi}_{-\pi}f(x)\sin(n+1/2)x\,dx$$ Since $f(x)$ is continous on $[-\pi,\pi]$ then it is integrable on this interval which is compact (by Weierstrass maximum is attained so it is bounded etc.). By Lebesgue-Riemann lemma then it follows that $$\lim_{n\to\infty}\int^{\pi}_{-\pi}f(x)\sin(n+1/2)x\,dx=0$$ Therefore we obtain $$2\lim_n\int^{\pi}_{-\pi}\frac{\sin(n+1/2)x}{x}\,dx=\lim_n\int^{\pi}_{-\pi}\frac{\sin(n+1/2)x}{\sin x/2}\,dx=\lim_n 2\pi=2\pi$$ This finally yields $$\pi=\lim_n\int^{\pi}_{-\pi}\frac{\sin(n+1/2)x}{x}\,dx=\lim_n\int^{(n+1/2)\pi}_{-(n+1/2)\pi}\frac{\sin x}{x}\,dx=\int^{\infty}_{-\infty}\frac{\sin x}{x}\,dx$$ But $\sin x/x$ is an even function (since both $\sin x$ and $x$ are odd) therefore $$\int^{\infty}_{-\infty}\frac{\sin x}{x}\,dx=2\int^{\infty}_{0}\frac{\sin x}{x}\,dx\Rightarrow \int^{\infty}_{0}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$$

Answer 3

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 $y=(N+1/2)x$.

$$\lim_{N\to\infty}\int_{0}^{(N+1/2)\pi}\frac{\sin(y)}{y}dy=\frac{\pi}{2}$$