One way to define Bessel functions is

$$ e^{\frac{x}{2}(t-\frac{1}{t})}=\sum_{n=-\infty}^{+\infty}J_{n}(x)t^n. $$

How do I prove that?

I can't see a way of writing the L.H.S. as a geometrical (or Laurent) series on $t$ to compare the coefficients with the definition of $J_n(x)$. Searching for a more deep relation for the comparison, I arrived at

$$ \sum_{n,k=0}^{\infty}\frac{(-1)^{n-k}}{2^{n}k!(n-k)!}x^{n}t^{2k-n}=J_0(x)+\sum_{l=1}^{\infty}J_{l}(x)\left[t^{l}+\frac{(-1)^{l}}{t^{l}}\right] $$

Where the L.H.S. comes from the expansion of $e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}$ about $x=0$ and of $\left(t^2-1\right)^{n}$ about $t^2=0$.

Of course, when analysing the coefficients I would make use of the series form of the Bessel functions:

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

Let me know if there are mistakes or if what I did is just not useful here.



2 Answers 2


Here comes a big hint:

Start with the expansions $$ \begin{aligned} e^{xt/2}&=1+\frac{xt}{2}+\frac{1}{2!}\Bigl(\frac{xt}{2}\Bigr)^2+\frac{1}{3!}\Bigl(\frac{xt}{2}\Bigr)^3+\cdots\quad\text{and}\\ e^{-x/2t}&=1-\frac{x}{2t}+\frac{1}{2!}\Bigl(\frac{x}{2t}\Bigr)^2 -\frac{1}{3!}\Bigl(\frac{x}{2t}\Bigr)^3+\cdots. \end{aligned} $$ Multiply them, and look for the coefficient in front of $t^n$ in the product.

  • 1
    $\begingroup$ (+1) This is an even faster way :) $\endgroup$ Jun 21, 2016 at 17:58

By replacing $t$ with $e^{i\theta}$, we want to prove the identity: $$ e^{ix\sin\theta} = \sum_{n\in\mathbb{Z}} J_n(x)\,e^{in\theta} \tag{1}$$ that (by Fourier inversion) is equivalent to proving that: $$ J_n(x) = \frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(ix\sin\theta-n\theta\right)\,d\theta. \tag{2}$$ The fastest way is probably to notice that both your series and the RHS of $(2)$ are solutions of the differential equation: $$ x^2\, f''(x)+x\, f'(x)+(x^2-n^2)\,f(x)=0\tag{3} $$ through termwise differentiation and integration by parts.


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