I found the following statement from wikipedia about Bessel functions $J_{\alpha}(x)$ but have no idea how to prove it.
If $x$ is held fixed at a non-zero value, then the Bessel functions are entire functions of $\alpha$.
Would you please help or give me some references. Thanks a lot!
The power series representation of the Bessel functions should do it. $$J_{\alpha}(x) = \sum_{m=0}^{\infty}\frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m+\alpha}=\sum_{m=0}^{\infty}\frac{(-x^2)^m}{2^{2m}m!}\frac{\exp(\alpha\ln\frac x2)}{\Gamma(m+\alpha+1)}$$ The exponential and reciprocal gamma are both entire, so after multiplying by something that doesn't depend on $\alpha$, each term is an entire function of $\alpha$. If a sequence of analytic functions converges uniformly on compact sets, the limit is also analytic (Use the Cauchy Integral formula to show the derivative also converges uniformly on compact sets), so all we need to do is show that the series converges well enough.
For that - well, just apply the ratio test. The ratio of the $m$th term to the one before is $\frac{-x^2}{4m(m+\alpha)}\to 0$ as $m\to\infty$ for any fixed $\alpha$ and $x$. Choose an arbitrary $R$, and then choose some $m>\max(2|R|,|x|)$. Then $\left|\frac{-x^2}{4m(m+\alpha)}\right|<\frac{|x|^2}{8R\cdot R} < \frac18$, and every term after the $m$th is multiplied by at most $\frac18$. If $A$ is the maximum of the $m$th term for $|\alpha|\le R$, then the $n$th term for $n>m$ is bounded by $8^{-(n-m)}A$, and the sum converges uniformly on this set. That's what we needed.