Example 5. Expand the function $y=\left(1-x^{2}\right) x^{\nu}(\nu>-1)$ into a Fourier series on the interval $(0,1)$ with respect to the system of functions $J_{\nu}\left(\mu_{n} x\right), n=1,2, \ldots$, where $J_{\nu}$ is the Bessel function and $\mu_{n}$ is the zero of the function $J_{\nu}$.
Solution. Bessel functions $J_{\nu}(x)$ are defined by the equality

$$
J_{\nu}(x)=\sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!\Gamma(m+\nu+1)}\left(\frac{x}{2}\right)^{2 m+\nu}
$$

The functions $J_{\nu}\left(z_{1} x\right), J_{\nu}\left(z_{2} x\right), \ldots$ form an orthogonal basis in the space of functions defined on the interval $(0,1)$, with the scalar product

$$
(u, v)=\int_{0}^{1} u(x) v(x) x d x
$$

The desired expansion of the function $y=\left(1-x^{2}\right) x^{\nu}$ in a Fourier series on the interval $(0,1)$ with respect to the system of functions $J_{\nu}\left(z_{1} x\right), J_{\nu}\left(z_{2} x\right), \ldots$ is given by

$$
\left(1-x^{2}\right) x^{\nu}=\sum_{n=1}^{\infty} a_{n} J_{\nu}\left(\mu_{n} x\right), \quad x \in(0,1)
$$

where the coefficients $a_{1}, a_{2}, \ldots$ are determined by the Euler-Fourier formulas:

$$
a_{n}=\frac{\left(\left(1-x^{2}\right) x^{\nu}, J_{\nu}\left(\mu_{n} x\right)\right)}{\left(J_{\nu}\left(\mu_{n} x\right), J_{\nu}\left(\mu_{n} x\right)\right)}=\frac{\int_{0}^{1}\left(1-x^{2}\right) x^{\nu} J_{\nu}\left(\mu_{n} x\right) x d x}{\int_{0}^{1} J_{\nu}\left(\mu_{n} x\right)^{2} x d x}
$$

1. We check that

$$
\int_{0}^{1}\left(1-x^{2}\right)^{2} x^{2 \nu} x d x<\infty
$$

This inequality holds because the function $y=\left(1-x^{2}\right) x^{\nu}$ is bounded on $(0,1)$.

2. We compute the coefficients $a_{1}, a_{2}, \ldots$ using the Euler-Fourier formulas (19).

The integrals

$$
\int_{0}^{1}\left(1-x^{2}\right) x^{\nu} J_{\nu}\left(\mu_{n} x\right) x d x
$$

are computed using the formula

$$
z^{\nu+1} J_{\nu}(z)=\frac{d}{d z}\left(z^{\nu} J_{\nu+1}(z)\right)
$$

and integration by parts. We have

$$
\begin{gathered}
\int_{0}^{1}\left(1-x^{2}\right) x^{\nu} J_{\nu}\left(\mu_{n} x\right) x d x=\frac{1}{\mu_{n}^{\nu+1}} \int_{0}^{\mu_{n}}\left[1-\left(\frac{z}{\mu_{n}}\right)^{2}\right] z^{\nu+1} J_{\nu}(z) d z= \\
=\frac{1}{\mu_{n}^{\nu+1}} \int_{0}^{\mu_{n}}\left[1-\left(\frac{z}{\mu_{n}}\right)^{2}\right] \frac{d}{d z}\left(z^{\nu+1} J_{\nu+1}(z)\right) d z= \\
=\left.\frac{1}{\mu_{n}^{\nu+1}}\left[1-\left(\frac{z}{\mu_{n}}\right)^{2}\right] z^{\nu+1} J_{\nu+1}(z)\right|_{0} ^{\mu_{n}}+\frac{2}{\mu_{n}^{\nu+3}} \int_{0}^{\mu_{n}} z^{\nu+2} J_{\nu+1}(z) d z= \\
=\frac{2}{\mu_{n}^{2}} J_{\nu+2}\left(\mu_{n}\right)
\end{gathered}
$$

The integrals

$$
\int_{0}^{1} J_{\nu}\left(\mu_{n} x\right)^{2} x d x
$$

are computed using the formula

$$
\int J_{\nu}(\alpha x)^{2} x d x=\frac{x^{2}}{2}\left[J_{\nu}(\alpha x)^{2}-J_{\nu-1}(\alpha x) J_{\nu+1}(\alpha x)\right]+C
$$

Therefore,

$$
\int_{0}^{1} J_{\nu}\left(\mu_{n} x\right)^{2} x d x=-\frac{1}{2} J_{\nu-1}\left(\mu_{n}\right) J_{\nu+1}\left(\mu_{n}\right)
$$

For $\nu=0$, this equality simplifies to:

$$
\int_{0}^{1} J_{0}\left(\mu_{n} x\right)^{2} x d x=\frac{1}{2} J_{1}\left(\mu_{n}\right)^{2}
$$

since $J_{-k}(z)=(-1)^{k} J_{k}(z) \quad(k=0, \pm 1, \pm 2, \ldots)$.

Substituting the computed integrals into (19), we get

$$
a_{n}=-\frac{4}{\mu_{n}} \frac{J_{\nu+2}\left(\mu_{n}\right)}{J_{\nu-1}\left(\mu_{n}\right) J_{\nu+1}\left(\mu_{n}\right)}, \quad n=1,2, \ldots
$$

3. Substituting the found values of the coefficients $a_{1}, a_{2}, \ldots$ into (18). We obtain

$$
\left(1-x^{2}\right) x^{\nu}=-4 \sum_{n=1}^{\infty} \frac{J_{\nu+2}\left(\mu_{n}\right)}{\mu_{n} J_{\nu-1}\left(\mu_{n}\right) J_{\nu+1}\left(\mu_{n}\right)} J_{\nu}\left(\mu_{n} x\right)
$$

Answer. $\left(1-x^{2}\right) x^{\nu}=-4 \sum_{n=1}^{\infty} \frac{J_{\nu+2}\left(\mu_{n}^{2}\right)}{\mu_{n} J_{\nu-1}\left(\mu_{n}\right) J_{\nu+1}\left(\mu_{n}\right)} J_{\nu}\left(\mu_{n} x\right), x \in(0,1)$