- The Bessel functions arise in the study of spherical symmetries and the Fourier transform. See Chapter 6 in Book I. Prove that the following power series identity holds for Bessel functions of real order \(\nu > - 1 / 2\)
\[ J _ {\nu} (x) = \frac {(x / 2) ^ {\nu}}{\Gamma (\nu + 1 / 2) \sqrt {\pi}} \int_ {- 1} ^ {1} e ^ {i x t} (1 - t ^ {2}) ^ {\nu - (1 / 2)} d t = \left(\frac {x}{2}\right) ^ {\nu} \sum_ {m = 0} ^ {\infty} \frac {(- 1) ^ {m} \left(\frac {x ^ {2}}{4}\right) ^ {m}}{m ! \Gamma (\nu + m + 1)} \]
whenever \(x > 0\) . In particular, the Bessel function \(J _ { \nu }\) satisfies the ordinary diferential equation
\[ \frac {d ^ {2} J _ {\nu}}{d x ^ {2}} + \frac {1}{x} \frac {d J _ {\nu}}{d x} + \left(1 - \frac {\nu^ {2}}{x ^ {2}}\right) J _ {\nu} = 0. \]
[Hint: Expand the exponential \(e ^ { i x t }\) in a power series, and express the remaining integrals in terms of the gamma function, using Exercise 7.]
Solution
1. Expand \(e^{ixt} = \sum_{m=0}^{\infty} \frac{(ixt)^m}{m!}\).
power series of the exponential.
2. Then\[\int_{-1}^{1} e^{ixt}(1 - t^2)^{\nu - 1/2}\,dt = \sum_{m=0}^{\infty} \frac{(ix)^m}{m!}\int_{-1}^{1} t^m (1 - t^2)^{\nu - 1/2}\,dt.\]
1, integrating term by term.
3. For \(m\) odd, \(\int_{-1}^{1} t^m (1 - t^2)^{\nu - 1/2}\,dt = 0\) (the integrand is odd).
symmetry.
4. For \(m = 2k\) even, \(\int_{-1}^{1} t^{2k}(1 - t^2)^{\nu - 1/2}\,dt = \frac{\Gamma(k + 1/2)\Gamma(\nu + 1/2)}{\Gamma(\nu + k + 1)}\).
beta function (substituting \(u = t^2\)).
5. Hence\[\int_{-1}^{1} e^{ixt}(1 - t^2)^{\nu - 1/2}\,dt = \sum_{k=0}^{\infty} \frac{(ix)^{2k}}{(2k)!} \cdot \frac{\Gamma(k + 1/2)\Gamma(\nu + 1/2)}{\Gamma(\nu + k + 1)}.\]
<1>2–<1>4.
6. Using \(\Gamma(k + 1/2) = \frac{(2k)!}{4^k k!}\sqrt{\pi}\) and \((ix)^{2k} = (-1)^k x^{2k}\):\[\int_{-1}^{1} e^{ixt}(1 - t^2)^{\nu - 1/2}\,dt = \Gamma(\nu + 1/2)\sqrt{\pi}\sum_{k=0}^{\infty} \frac{(-1)^k (x/2)^{2k}}{k!\,\Gamma(\nu + k + 1)}.\]
5 and the identity for \(\Gamma(k + 1/2)\).
7. Multiplying by \(\frac{(x/2)^\nu}{\Gamma(\nu + 1/2)\sqrt{\pi}}\):\[J_\nu(x) = \frac{(x/2)^\nu}{\Gamma(\nu + 1/2)\sqrt{\pi}}\int_{-1}^{1} e^{ixt}(1 - t^2)^{\nu - 1/2}\,dt = \left(\frac{x}{2}\right)^\nu \sum_{k=0}^{\infty} \frac{(-1)^k (x/2)^{2k}}{k!\,\Gamma(\nu + k + 1)}.\]
6.
8. This is the standard power series for \(J_\nu(x)\).
7.
9. Q.E.D.
8.