贝塞尔方程解的积分恒等式证明求助：分部积分陷入循环

Hey there, let's work through this proof step by step—you're right that integration by parts can feel like it's looping if you don't lean into the Bessel equation property at the right moment. Let's break it down clearly:

First, let's rewrite the integral using the suggested substitution. If $z = \alpha x$, then $x = \frac{z}{\alpha}$ and $dx = \frac{dz}{\alpha}$. The limits of integration shift from $0 \to 1$ to $0 \to \alpha$, so the original integral becomes:
$$\int_0^1 x[J_n(\alpha x)]^2dx = \int_0^\alpha \frac{z}{\alpha} [J_n(z)]^2 \cdot \frac{dz}{\alpha} = \frac{1}{\alpha^2} \int_0^\alpha z [J_n(z)]^2 dz$$
Our goal now reduces to proving $\int_0^\alpha z [J_n(z)]^2 dz = \frac{\alpha^{2}{2}[J_n'(\alpha)]}2$, since substituting this back will give us the desired identity.

The Bessel equation for $J_n(z)$ is:
$$z^2 J_n''(z) + z J_n'(z) + (z^2 - n^2) J_n(z) = 0$$
We can rearrange this into a form that's more useful for our proof:
$$z \frac{d}{dz} \left( z J_n'(z) \right) = (n^2 - z^2) J_n(z)$$
This rearrangement lets us replace $z^2 J_n(z)$ with an expression involving derivatives of $J_n(z)$, which is key to breaking out of the integration loop.

Let's tackle $\int z [J_n(z)]^2 dz$. Choose:

• $u = [J_n(z)]^2$ (so $du = 2 J_n(z) J_n'(z) dz$)
• $dv = z dz$ (so $v = \frac{z^2}{2}$)

Applying integration by parts ($\int u dv = uv - \int v du$):
$$\int z [J_n(z)]^2 dz = \frac{z^{2}{2}[J_n(z)]}2 - \int \frac{z^2}{2} \cdot 2 J_n(z) J_n'(z) dz$$
Simplify the second term:
$$= \frac{z^{2}{2}[J_n(z)]}2 - \int z^2 J_n(z) J_n'(z) dz$$

Now take the integral $\int z^2 J_n(z) J_n'(z) dz$. From our rearranged Bessel equation, we know $z^2 J_n(z) = n^2 J_n(z) - z \frac{d}{dz}\left(z J_n'(z)\right)$. Substitute this in:
$$\int z^2 J_n(z) J_n'(z) dz = \int \left[ n^2 J_n(z) - z \frac{d}{dz}\left(z J_n'(z)\right) \right] J_n'(z) dz$$
Split this into two separate integrals:
$$= n^2 \int J_n(z) J_n'(z) dz - \int z J_n'(z) \frac{d}{dz}\left(z J_n'(z)\right) dz$$

Let's compute each part:

1. The first integral is straightforward: $\int J_n(z) J_n'(z) dz = \frac{1}{2}[J_n(z)]^2 + C$ (it's the integral of a derivative of a square).
2. For the second integral, let $w = z J_n'(z)$, so $dw = \frac{d}{dz}\left(z J_n'(z)\right) dz$. The integral becomes $\int w dw = \frac{1}{2}w^2 + C = \frac{1}{2}[z J_n'(z)]^2 + C$.

Substitute these back:
$$\int z^2 J_n(z) J_n'(z) dz = \frac{n^{2}{2}[J_n(z)]}2 - \frac{1}{2}[z J_n'(z)]^2 + C$$

Now substitute this back into our earlier expression:
$$\int z [J_n(z)]^2 dz = \frac{z^{2}{2}[J_n(z)]}2 - \left( \frac{n^{2}{2}[J_n(z)]}2 - \frac{1}{2}[z J_n'(z)]^2 \right) + C$$
Expand and simplify:
$$= \frac{z^{2}{2}[J_n(z)]}2 - \frac{n^{2}{2}[J_n(z)]}2 + \frac{1}{2}[z J_n'(z)]^2 + C$$
$$= \frac{1}{2}(z^2 - n^2)[J_n(z)]2 + \frac{1}{2}[z J_n'(z)]^2 + C$$

Remember that $\alpha$ is a zero of $J_n(z)$, so $J_n(\alpha) = 0$. Let's evaluate at the bounds:

• Upper limit ($z = \alpha$): The first term becomes $\frac{1}{2}(\alpha^2 - n^{2)[J_n(\alpha)]}2 = 0$, leaving only $\frac{1}{2}[\alpha J_n'(\alpha)]^2$.
• Lower limit ($z = 0$): For all $n \geq 0$, the asymptotic behavior of $J_n(z)$ as $z \to 0$ makes both terms vanish:
  • For $n = 0$, $J_0(z) \approx 1 - \frac{z^2}{4}$, so $(z^2 - 0)[J_0(z)]^2 \approx z^2 \to 0$, and $z J_0'(z) \approx -\frac{z^2}{2} \to 0$.
  • For $n \geq 1$, $J_n(z) \approx \frac{(z/2)^n}{n!}$, so $(z^2 - n^2)[J_n(z)]2$ decays to 0, and $z J_n'(z) \approx \frac{n(z/2)^n}{2n(n-1)!} \to 0$.

So the definite integral simplifies to:
$$\int_0^\alpha z [J_n(z)]^2 dz = \frac{1}{2}[\alpha J_n'(\alpha)]^2 - 0 = \frac{\alpha^{2}{2}[J_n'(\alpha)]}2$$

Recall from Step 1 that:
$$\int_0^1 x[J_n(\alpha x)]^2dx = \frac{1}{\alpha^2} \int_0^\alpha z [J_n(z)]^2 dz$$
Substitute our result:
$$= \frac{1}{\alpha^2} \cdot \frac{\alpha^{2}{2}[J_n'(\alpha)]}2 = \frac{1}{2}[J_n'(\alpha)]^2$$

That's it! The key was using the Bessel equation to rewrite the tricky integral instead of trying to do another round of integration by parts (which would cause the loop you encountered).

内容的提问来源于stack exchange，提问作者Deke