I'm currently in a complex analysis class and doing a little project on asymptotics. We are using the Stein and Shakarchi Complex book for reference. In Appendix A: Asymptotics of the Stein book, the Bessel function is given in the form for $\nu >-1/2$: $$ J_{\nu}(s) = \frac{(s/2)^\nu}{\Gamma(\nu + 1/2)\Gamma(1/2)}\int_{-1}^1e^{isx}(1-x^2)^{\nu-1/2}\text{d}x $$ From differential equations class, I'm most used to the bessel function being presented as: $$J_{\nu}(s) = (\frac{s}{2})^\nu \sum_{j=0} ^{\infty} \frac{-1^j (\frac{s}{2})^{2j}}{j! \Gamma(j+\nu+1)}$$ I'm having a lot of trouble going from this second definition to the definition presented in Stein. I was playing around with using Fourier transforms to get the integral, but I still don't see a connection. Could anyone be of help to point out how to make this transition?
I appreciate any help.
Good question!
There are several ways of showing the equivalence.
The most straightforward is as follows. Let $I $ denote the integral. We have
$ \displaystyle I=2\int_{0}^{1}{\cos(sx)(1-x^2)^{\nu-1/2}dx}.\tag*{} $
Why? Take real parts of the integral and then use the evenness of the integrand.
Now let $ x=t^{1/2} $ and expand the cosine in its power series. You will need three things:
$ \displaystyle \int_{0}^{1}{t^{r-1/2}(1-t)^{\nu-1/2}dt}=B(r+1/2,\nu+1/2)=\Gamma(r+1/2)\Gamma(\nu+1/2)/\Gamma(r+\nu+1),\tag*{} $
where $B$ is the beta function, and
$ \displaystyle (2r)!=r!\Gamma(r+1/2)2^{2r-1}/\sqrt{\pi}.\tag*{}$
The result will be the desired series.
There is no way of avoiding dealing with gamma functions in the derivation since they occur in the series definition of the Bessel function.
Another more arcane (and slicker) way is to use Laplace transforms.
$ \displaystyle \mathfrak{L}\bigg[t^{\nu}J_{\nu}(t)\bigg]=(p^2+1)^{-\nu-1/2},\tag*{}$
which is easily shown by taking the Laplace transform term by term of the power series for $ J_{\nu}(t)$ .
One inverts this transform by integrating the function $ e^{pt}(p^2+1)^{-\nu-1/2}$ along a contour in the complex plane from $ -i{\infty}\;\text{to}\;{i\infty} $ making little semicircles around $ -i,i $ . The integrand is multivalent and one must be careful to assign the correct value to it along the various segments of the integration path.
When all the smoke clears, the result will be the desired integral.