Obtaining $\frac{\sin x}{x}$ via integral of Bessel function of the first kind

Question

How can I prove that,

$$\frac{\sin x}{x} = \int_0^{\pi/2} J_0(x\cos\theta)\cos\theta \hspace{0.11cm} \mathrm{d}\theta$$

Answer 1

Since $$J_{m}\left(x\right)=\sum_{k\geq0}\frac{\left(-1\right)^{k}}{4^{k}k!\left(k+m\right)!}x^{2k+m} $$ we have, exchanging series and integral (which is permitted due to the absolute convergence), $$ \int_{0}^{\pi/2}J_{0}\left(x\cos\left(\theta\right)\right)\cos\left(\theta\right)d\theta=\sum_{k\geq0}\frac{\left(-1\right)^{k}}{4^{k}k!^{2}}x^{2k}\int_{0}^{\pi/2}\cos^{2k+1}\left(\theta\right)d\theta $$ and the integral in the RHS is the Wallis integral which can be calculated (using Euler's Beta function) as $$\int_{0}^{\pi/2}\cos^{2k+1}\left(\theta\right)d\theta=\frac{\left(2k\right)!!}{\left(2k+1\right)!!}=\frac{4k!^{2}}{\left(2k+1\right)!} $$ so finally $$ \int_{0}^{\pi/2}J_{0}\left(x\cos\left(\theta\right)\right)\cos\left(\theta\right)d\theta=\sum_{k\geq0}\frac{\left(-1\right)^{k}}{4^{k}k!^{2}}x^{2k}\frac{4^{n}k!^{2}}{\left(2k+1\right)!} $$ $$=\sum_{k\geq0}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}x^{2k}=\frac{\sin\left(x\right)}{x}. $$

Answer 2

Note that $$J_0(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(n!)^2}\left(\frac{x}{2}\right)^{2n}$$ So $$\eqalign{ \int_{0}^{\pi/2}J_0(x\cos\theta)\cos\theta\,d\theta&= \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(n!)^24^n}\int_{0}^{\pi/2}\cos^{2n+1}\theta\,d\theta\cr &=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(n!)^24^n}\cdot \frac{(2^nn!)^2}{(2n+1)!}\cr &=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}=\frac{\sin x}{x}. } $$ Edit: Note that $$\eqalign{I_n-I_{n-1}&= -\int_{0}^{\pi/2}\cos^{2n-1}\theta\sin^2\theta\,d\theta\cr &=\left[\frac{\cos^{2n}\theta}{2n}\sin\theta \,d\theta\right]_0^{\pi/2}-\frac{1}{2n}\int_{0}^{\pi/2}\cos^{2n+1}\theta\,d\theta }$$ This proves that $$I_n=\frac{2n}{2n+1} I_{n-1}=\frac{(2n)(2n-2)\cdots(2)}{(2n+1)(2n-1)\cdots (3)(1)}I_0=\frac{2^{2n}(n!)^2}{(2n+1)!}$$

Answer 3

We may exploit the fact that $J_0$ is a very well-behaving analytic function and the integrals $\int_{0}^{\pi/2}\cos^n\theta\,d\theta$ are known. We get:

$$ \begin{eqnarray*}I(x)=\int_{0}^{\pi/2}J_0(x\cos\theta)\cos(\theta)\,d\theta&=&\sum_{k\geq 0}\frac{(-1)^k}{4^k k!^2}\int_{0}^{\pi/2}x^{2k}\cos^{2k+1}(\theta)\,d\theta\\&=&\sum_{k\geq 0}\frac{(-1)^k}{4^k k!^2}\cdot\frac{2^k k!}{(2k+1)!!}\,x^{2k}\\&=&\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)!}\,x^{2k}\\&=&\color{red}{\frac{\sin x}{x}}.\end{eqnarray*}$$