Special Functions – Bessel functions – Integral representations

First we will prove

Theorem:

\[J_{\nu}(z)=\frac{1}{\Gamma(\nu+1/2)\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \int_{-1}^1e^{izt}(1-t^2)^{\nu-1/2}\,dt,\quad\text{Re}(\nu)>-1/2.\tag1\]

Proof: We start with

\[\int_{-1}^1e^{izt}(1-t^2)^{\nu-1/2}\,dt=\sum_{n=0}^{\infty}\frac{(iz)^n}{n!}\int_{-1}^1t^n(1-t^2)^{\nu-1/2}\,dt.\]

Note that the latter integral vanishes when \(n\) is odd. For \(n=2k\) we obtain using \(t^2=u\)

\[\int_{-1}^1t^{2k}(1-t^2)^{\nu-1/2}\,dt=2\int_0^1u^k(1-u)^{\nu-1/2}\,\frac{du}{2\sqrt{u}} =\int_0^1u^{k-1/2}(1-u)^{\nu-1/2}\,du=B(k+1/2,\nu+1/2)=\frac{\Gamma(k+1/2)\Gamma(\nu+1/2)}{\Gamma(\nu+k+1)}.\]

Now we use Legendre's duplication formula to find that

\[\Gamma(k+1/2)=\frac{\sqrt{\pi}}{2^{2k-1}}\cdot\frac{\Gamma(2k)}{\Gamma(k)} =\frac{\sqrt{\pi}}{2^{2k}}\cdot\frac{\Gamma(2k+1)}{\Gamma(k+1)}=\frac{\sqrt{\pi}}{2^{2k}}\cdot\frac{(2k)!}{k!}.\]

Hence we have

\[\int_{-1}^1e^{izt}(1-t^2)^{\nu-1/2}\,dt=\sum_{k=0}^{\infty}\frac{(iz)^{2k}}{(2k)!}\cdot \frac{\Gamma(k+1/2)\Gamma(\nu+1/2)}{\Gamma(\nu+k+1)} =\Gamma(\nu+1/2)\sqrt{\pi}\,\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)\,k!}\left(\frac{z}{2}\right)^{2k}.\]

This proves the theorem.

We also have Poisson's integral representations:

Theorem:

\[J_{\nu}(z)=\frac{1}{\Gamma(\nu+1/2)\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu}\int_0^{\pi}e^{iz\cos\theta}\,(\sin\theta)^{2\nu}\,d\theta =\frac{1}{\Gamma(\nu+1/2)\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \int_0^{\pi}\cos(z \cos\theta)\,(\sin\theta)^{2\nu}\,d\theta,\quad\text{Re}(\nu)>-1/2.\]

Proof: Use the substitution \(t=\cos\theta\) to obtain

\[\int_{-1}^1e^{izt}(1-t^2)^{\nu-1/2}\,dt=\int_{\pi}^0e^{iz\cos\theta}(1-\cos^2\theta)^{\nu-1/2}\cdot(-\sin\theta)\,d\theta =\int_0^{\pi}e^{iz\cos\theta}(\sin\theta)^{2\nu}\,d\theta.\]

Further we have

\[e^{iz\cos\theta}=\cos(z \cos\theta)+i\sin(z \cos\theta)\]

and

\[\int_0^{\pi}\sin(z \cos\theta)(\sin\theta)^{2\nu}\,d\theta=0.\]

This shows that Poisson's integral representations follow from the integral representation (1).

Another (Barnes-type) integral represenation is

Theorem:

\[J_{\nu}(z)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\frac{\Gamma(-s)}{\Gamma(\nu+s+1)}\left(\frac{z}{2}\right)^{\nu+2s}\,ds, \quad\text{Re}(\nu)>-1,\quad\text{Re}(z)>0.\]

Proof: Here we use Cauchy's residue theorem to conclude that the right-hand side equlas

\[\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)\,k!}\left(\frac{z}{2}\right)^{\nu+2k}=J_{\nu}(z).\]

Remarks:

1) The Fourier transform is defined by

\[F(z):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-izt}f(t)\,dt\]

with inversion formula

\[f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{izt}F(z)\,dz.\]

This implies that the Fourier transform of the function

\[f(t)=\left\{\begin{array}{ll}(1-t^2)^{\nu-1/2}, & |t|\le 1\\[2.5mm]0, & |t|>1\end{array}\right.\]

is

\[F(z)=\frac{\Gamma(\nu+1/2)}{\sqrt{2}}\left(\frac{z}{2}\right)^{-\nu}J_{\nu}(z).\]

2) Instead of the substitution \(t=\cos\theta\) in (1) one can also use the substitution \(t=\sin\theta\), which leads to slightly different forms of Poisson's integral representations. In fact we have

\begin{align*} J_{\nu}(z)&=\frac{1}{\Gamma(\nu+1/2)\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \int_{-\pi/2}^{\pi/2}e^{iz\sin\theta}\,(\cos\theta)^{2\nu}\,d\theta\\[2.5mm] &=\frac{1}{\Gamma(\nu+1/2)\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \int_{-\pi/2}^{\pi/2}\cos(z \sin\theta)\,(\cos\theta)^{2\nu}\,d\theta\\[2.5mm] &=\frac{2}{\Gamma(\nu+1/2)\sqrt{\pi}}\left(\frac{z}{2}\right)^{\nu} \int_0^{\pi/2}\cos(z \sin\theta)\,(\cos\theta)^{2\nu}\,d\theta,\quad\text{Re}(\nu)>-1/2. \end{align*}

---

Last modified on September 30, 2021