While doing research in electrical engineering, I derived the following integral representation of the Bessel function of the first kind:
$$J_n(x)=\frac{e^{in\pi/2}}{2\pi}\int_0^{2\pi}e^{i(n\tau-x\cos\tau)}\mathrm{d}\tau\tag{1}$$
My derivation, which I include below, is long and ugly. I am wondering if there is a more elegant proof of (1) using basic facts about other integral representations of the Bessel function, trig identities, and, perhaps, clever integration techniques. The integral representation for Bessel function (found on wikipedia page) that looks similar to mine is:
$$J_n(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(n\tau-x\sin\tau)}\mathrm{d}\tau. \tag{2}$$
Gradshteyn and Ryzhik (G&R) give the same expression in a slightly different form as formula 8.411.1 in the 7th edition. However, I failed to convert (2) into (1) using simple substitution. Does anyone have any other ideas?
My LONG proof of (1)
First, I use Euler's formula to break the integrand in (1) into in-phase and quadrature components, and apply the angle sum identities:
$$\begin{align}e^{i(n\tau-x\cos\tau)}&=\cos(n\tau-x\cos\tau)+i\sin(n\tau-x\cos\tau)\\
&=\cos(n\tau)\cos(x\cos\tau)\tag{a}\\
&\phantom{=}+\sin(n\tau)\sin(x\cos\tau)\tag{b}\\
&\phantom{=}+i\sin(n\tau)\cos(x\cos\tau)\tag{c}\\
&\phantom{=}-i\cos(n\tau)\sin(x\cos\tau)\tag{d}
\end{align}$$
Now let's integrate (a)-(d) in turn. First, for (a), note that:
$$\begin{align}\int_{\pi}^{2\pi}\cos(n\tau)\cos(x\cos\tau)\mathrm{d}\tau&=\int_0^{\pi}\cos(n(\tau+\pi))\cos(x\cos(\tau+\pi))\mathrm{d}\tau\\
&=(-1)^n\int_0^{\pi}\cos(n\tau)\cos(x\cos\tau)\mathrm{d}\tau\tag{a1},
\end{align}$$
where (a1) is due to the negation of the cosine (and sine) from the shift by odd multiples of $\pi$, or, formally, $\cos(\theta+n\pi)=(-1)^n\cos\theta$. By formula 3.715.18 in G&R 7th ed:
$$\int_0^{\pi}\cos(n\tau)\cos(x\cos\tau)\mathrm{d}\tau=\pi\cos\left(\frac{n\pi}{2}\right)J_n(x).$$
Thus,
$$\begin{align}\int_0^{2\pi}\cos(n\tau)\cos(x\cos\tau)\mathrm{d}\tau&=(1+(-1)^n)\pi\cos\left(\frac{n\pi}{2}\right)J_n(x)\\
&=2\pi\cos\left(\frac{n\pi}{2}\right)J_n(x),\tag{a2}
\end{align}$$
where (a2) is because when $n$ is odd, $\cos\left(\frac{n\pi}{2}\right)=0$, making the double-multiplication by zero in this case unnecessary.
Now let's integrate (b). First consider odd $n$:
$$\int_0^{2\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau=2\int_0^{\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau,$$
because $\sin(n(\tau+\pi))\sin(x\cos(\tau+\pi))=\sin(n\tau)\sin(x\cos\tau)$ due to the negation of the sine (and cosine) from the shift by odd multiples of $\pi$, or, formally, $\sin(\theta+n\pi)=(-1)^n\sin\theta$ and the fact that $\sin(-\theta)=-\sin\theta$. Furthermore,
$$\begin{align}\int_0^{\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau&=\int_0^{\pi/2}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau+\int_{\pi/2}^{\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau\\
&=\int_0^{\pi/2}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau\\
&\phantom{=}+\int_0^{\pi/2}\sin(n(\pi-\tau))\sin(x\cos(\pi-\tau))\mathrm{d}\tau\tag{b1}\\
&=\int_0^{\pi/2}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau-\int_0^{\pi/2}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau=0\tag{b2},
\end{align}$$
where (b1) is due to the substitution of $\tau=\pi-\tau'$ (the prime is dropped after substitution is made) and (b2) is since $\sin(n\pi-\theta)=\sin(\theta)$ for odd $n$ and $\cos(\pi-\theta)=-\cos(\theta)$.
Now let's integrate (b) with even $n$:
$$\begin{align}\int_0^{2\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau&=\int_0^{\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau+\int_{\pi}^{2\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau\\
&=\int_0^{\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau\\
&\phantom{=}+\int_0^{\pi}\sin(n(2\pi-\tau))\sin(x\cos(2\pi-\tau))\mathrm{d}\tau\tag{b3}\\
&=\int_0^{\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau-\int_0^{\pi}\sin(n\tau)\sin(x\cos\tau)\mathrm{d}\tau=0\tag{b4},
\end{align}$$
where (b3) is due to the substitution of $\tau=2\pi-\tau'$ (again, the prime is dropped after substitution is made) and (b4) is since $\sin(2n\pi-\theta)=\sin(-\theta)=-\sin(\theta)$ for an integer $n$ and $\cos(2\pi-\theta)=\cos(\theta)$.
Now let's integrate (c). Consider odd $n$ (let's omit the imaginary unit):
$$\begin{align}\int_0^{2\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau&=\int_0^{\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau+\int_{\pi}^{2\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau\\
&=\int_0^{\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau\\
&\phantom{=}+\int_0^{\pi}\sin(n(2\pi-\tau))\cos(x\cos(2\pi-\tau))\mathrm{d}\tau\tag{c1}\\
&=\int_0^{\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau-\int_0^{\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau=0,\tag{c2}\\
\end{align}$$
where (c1) is due to the substitution of $\tau=2\pi-\tau'$ (the prime is dropped after substitution is made) and (c2) is since $\sin(2n\pi-\theta)=\sin(-\theta)=-\sin(\theta)$ for integer $n$ and $\cos(2\pi-\theta)=\cos(\theta)$.
Now integrate (c) with even $n$ (again, let's omit the imaginary unit):
$$\int_0^{2\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau=2\int_0^{\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau,$$
because $\sin(n(\tau+\pi))\cos(x\cos(\tau+\pi))=\sin(n\tau)\sin(x\cos\tau)$ due to $\sin(\theta+n\pi)=\sin\theta$ for even $n$, and $\cos(-\theta)=\cos\theta$. Furthermore,
$$\begin{align}\int_0^{\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau&=\int_0^{\pi/2}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau+\int_{\pi/2}^{\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau\\
&=\int_0^{\pi/2}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau\\
&\phantom{=}+\int_0^{\pi/2}\sin(n(\pi-\tau))\cos(x\cos(\pi-\tau))\mathrm{d}\tau\tag{c3}\\
&=\int_0^{\pi/2}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau-\int_0^{\pi/2}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau=0\tag{c4},
\end{align}$$
where (c3) is due to the substitution of $\tau=\pi-\tau'$ (the prime is dropped after substitution is made) and (c4) is since $\sin(n\pi-\theta)=\sin(-\theta)=-\sin(\theta)$ for even $n$ and $\cos(-\theta)=\cos(\theta)$.
Finally, we integrate (d), omitting the negative imaginary unit for now. First, note that
$$\begin{align}\int_{\pi}^{2\pi}\cos(n\tau)\sin(x\cos\tau)\mathrm{d}\tau&=\int_0^{\pi}\cos(n(\tau+\pi))\sin(x\cos(\tau+\pi))\mathrm{d}\tau\\
&=(-1)^{n+1}\int_0^{\pi}\cos(n\tau)\sin(x\cos\tau)\mathrm{d}\tau\tag{d1},
\end{align}$$
where (d1) is due to the negation of the cosine (and sine) from the shift by odd multiples of $\pi$, or, formally, $\cos(\theta+n\pi)=(-1)^n\cos\theta$ and by the fact that $\sin(-\theta)=-\sin\theta$. By formula 3.715.13 in G&R 7th ed:
$$\int_0^{\pi}\cos(n\tau)\sin(x\cos\tau)\mathrm{d}\tau=\pi\sin\left(\frac{n\pi}{2}\right)J_n(x).$$
Thus,
$$\begin{align}\int_0^{2\pi}\sin(n\tau)\cos(x\cos\tau)\mathrm{d}\tau&=(1+(-1)^{n+1})\pi\sin\left(\frac{n\pi}{2}\right)J_n(x)\\
&=2\pi\sin\left(\frac{n\pi}{2}\right)J_n(x),\tag{d2}
\end{align}$$
where (d2) is because when $n$ is even, $\sin\left(\frac{n\pi}{2}\right)=0$, making the double-multiplication by zero in this case unnecessary.
Combining all the terms, using Euler's formula, and solving for $J_n(x)$, we arrive at (1). Surely there is a better way…
Best Answer
In Equation $(1)$ of the OP, enforce the substitution $\tau\to \tau -\pi/2$. Then, we obtain
$$\begin{align} \frac{e^{in\pi/2}}{2\pi}\int_0^{2\pi}e^{in\tau-x\cos \tau}\,d\tau&=\frac1{2\pi}\int_{-\pi/2}^{3\pi/2}e^{in\tau-x\sin \tau}\,d\tau\\\\ &=\frac1{2\pi}\int_{-\pi/2}^\pi e^{in\tau-x\sin \tau}\,d\tau+\int_\pi^{3\pi/2}e^{in\tau-x\sin \tau}\,d\tau \tag{A} \end{align}$$
Now, enforce the substitution $\tau\to \tau +2\pi$ in the second integral on the right-hand side of $(A)$. Then,
$$\begin{align} \frac{e^{in\pi/2}}{2\pi}\int_0^{2\pi}e^{in\tau-x\cos \tau}\,d\tau&=\frac1{2\pi}\int_{-\pi/2}^\pi e^{in\tau-x\sin \tau}\,d\tau+\int_{-\pi}^{-\pi/2}e^{in\tau-x\sin \tau}\,d\tau \\\\ &=\frac1{2\pi}\int_{-\pi}^{\pi}e^{in\tau-x\sin \tau}\,d\tau \tag{B} \end{align}$$
Comparing $(B)$ to Equation $(2)$ in the OP, we find that
$$J_n(x)=\frac{e^{in\pi/2}}{2\pi}\int_0^{2\pi}e^{in\tau-x\cos \tau}\,d\tau$$