# Simpler proof of an integral representation of Bessel function of the first kind $J_n(x)$

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...

• Since you used Gradshteyn as a "look up" reference, why not simply look up the representation of interest? I would suggest showing that the integral of interestsatisfies the ODE and associated boundary conditions for the Bessel function. Alternatively, substitute $\tau \to \tau -\pi/2$ and exploit the periodicity of the integrand to move from $(1)$ to $(2)$. Dec 18, 2015 at 21:49
• Unfortunately, I am not very strong at solving differential equations (having never taken a course on them), but I'll look into this one. I tried the substitution that you mentioned, but couldn't get it to work, since I couldn't find a way to get the right limits of the integral... Dec 18, 2015 at 22:20
• As for the limits, exploit periodicity. Dec 18, 2015 at 22:25
• Ahhh -- I think I got it! Of course, $e^{in\tau}$ has a period of $2\pi$ as long as $n$ is an integer. Since cosine also has a periodic of $2\pi$, everything works! Thank you, I'll update my post either tonight or tomorrow. Dec 19, 2015 at 0:01
• Pleased to hear! It's sometimes easy to miss a symmetry or other property such as periodicity. But those properties can often be exploited in very powerful ways. Dec 19, 2015 at 0:12

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

$$J_n(x)=\frac{e^{in\pi/2}}{2\pi}\int_0^{2\pi}e^{in\tau-x\cos \tau}\,d\tau$$

• Yup, I had the same substitutions in mind, except into (2) to show (1). It was really simple once recalled that $e^{in\tau}$ has a period of $2\pi$ over angle $\tau$ for integer $n$. Didn't get around to updating my post, but you should still get the credit for reminding me of that. :) Dec 19, 2015 at 12:30
• Much appreciative! Dec 19, 2015 at 14:42