Hey Alireza, great question—this kind of mix of integrals, infinite series, and exponential-sinusoidal terms is exactly where Fourier analysis and special functions like Bessel functions intersect, which makes it super relevant for signal processing applications. Let’s break this down to get you closer to a Fourier series representation.

Step 1: Simplify the Original Expression Using Bessel Functions #

First, let’s recall the integral representation of the 0th-order Bessel function of the first kind:
$$J_0(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e{-jx\sin(\tau)}d\tau$$
Plugging this into your first term immediately simplifies it:
$$\frac{\pi}{6} \int_{-\pi}^{\pi}e{-jx\sin(\tau)}d\tau = \frac{\pi}{6} \cdot 2\pi J_0(x) = \frac{\pi^2}{3}J_0(x)$$

For the second infinite series term, we use the generating function for Bessel functions:
$$e^{-x\sin(\tau)} = \sum_{n=-\infty}^{\infty}(-1)n J_n(x)e^{jn\tau}$$
(This comes from the standard generating function $e^{z\sin\tau} = \sum_{n=-\infty}^\infty J_n(z)e^{jn\tau}$, substituting $z=-x$ and using $J_n(-z)=(-1)^nJ_n(z)$.)

Substitute this into the integral inside the series:
$$\int_{-\pi}^{\pi}e{2m\tau -x\sin(\tau)}d\tau = \int_{-\pi}^{\pi}e{2m\tau}\sum_{n=-\infty}^{\infty}(-1)n J_n(x)e^{jn\tau}d\tau$$
We can swap the sum and integral (thanks to uniform convergence of the Bessel generating series):
$$= \sum_{n=-\infty}^{\infty}(-1)n J_n(x)\int_{-\pi}^{\pi}e{j(2m+n)\tau}d\tau$$

The integral here is a standard orthogonality integral: it equals $2\pi$ if $2m + n = 0$ (i.e., $n=-2m$), and 0 otherwise. Substituting $n=-2m$, we get:
$$\int_{-\pi}^{\pi}e{2m\tau -x\sin(\tau)}d\tau = 2\pi (-1)^{-2m}J_{-2m}(x) = 2\pi J_{2m}(x)$$
(We used $(-1)^{-2m}=1$ and $J_{-2m}(x)=J_{2m}(x)$ since Bessel functions of even order are even.)

Now the second term simplifies to:
$$\sum_{m \neq 0, m=-\infty}^{+\infty}\frac{1}{2\pi m^2} \cdot 2\pi J_{2m}(x) = \sum_{m \neq 0}\frac{J_{2m}(x)}{m^2} = 2\sum_{m=1}^{{\infty}\frac{J_{2m}(x)}{m}2}$$
(The last step comes from pairing $m$ and $-m$, since $\frac{J_{2m}(x)}{m^2} = \frac{J_{-2m}(x)}{(-m)^2}$.)

So your original expression reduces to:
$$f(x) = \frac{\pi^2}{3}J_0(x) + 2\sum_{m=1}^{{\infty}\frac{J_{2m}(x)}{m}2}$$

Step 2: Convert to a Fourier Series #

Since $f(x)$ is an even function (both $J_0(x)$ and $J_{2m}(x)$ are even), its Fourier series will only contain cosine terms:
$$f(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty}a_k \cos(kx)$$
We just need to compute the coefficients $a_0$ and $a_k$.

Calculating $a_0$

$$a_0 = \frac{2}{\pi}\int_{0}^{\pi}f(x)dx = \frac{2}{\pi}\left( \frac{\pi^{2}{3}\int_{0}}{\pi}J_0(x)dx + 2\sum_{m=1}^{{\infty}\frac{1}{m}2}\int_{0}^{\pi}J_{2m}(x)dx \right)$$
The integrals of Bessel functions over $[0,\pi]$ can be expressed using infinite series (via the power series expansion of $J_n(x)$), but for practical use, you can look up tabulated values or use numerical integration.

Calculating $a_k$ for $k \geq 1$

$$a_k = \frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(kx)dx$$
Substitute the simplified $f(x)$ and swap sum and integral (justified by convergence):
$$a_k = \frac{2}{\pi}\left( \frac{\pi^{2}{3}\int_{0}}{\pi}J_0(x)\cos(kx)dx + 2\sum_{m=1}^{{\infty}\frac{1}{m}2}\int_{0}^{\pi}J_{2m}(x)\cos(kx)dx \right)$$

The integral $\int_{0}^{\pi}J_n(x)\cos(kx)dx$ can be evaluated using known identities. For example, using the Lommel integral formula or expressing the cosine as a combination of exponentials and using the Bessel integral representation. For integer $k$ and $n$, this integral often results in a combination of Bessel functions evaluated at $\pi$ or trigonometric terms.

Key Strategies for Simplification #

• Leverage Bessel Function Identities: As we did above, Bessel functions are the natural bridge between your original integral/series and more manageable forms. Their generating functions and orthogonality properties are incredibly useful here.
• Exploit Even Symmetry: Since $f(x)$ is even, you only need to compute cosine coefficients, cutting your work in half.
• Numerical Approximation: If closed-form coefficients are hard to find, numerical integration of the coefficient integrals (combined with truncating the Bessel series to a finite number of terms) will give you a practical Fourier series approximation for signal processing applications.

This approach ties directly to the mathematical beauty you mentioned, linking Fourier analysis, Bessel functions, and practical signal processing tools.

备注：内容来源于stack exchange，提问作者Alireza