next up previous
Next: Bessel functions with order Up: No Title Previous: An integral for the

Integral representations of the Bessel function

In section 6 we found the power series (47) for the Bessel function $J_\nu (x)$. By means of the Hankel representation (143) in the last section we can now easily find an integral representation for the Bessel function,

$\displaystyle J_\nu (x)$ = $\displaystyle \sum_{r=0}^\infty (-1)^r\frac{\left(\frac{x}{2}\right)^{\nu+2r}}
  = $\displaystyle \frac{i}{2\pi}\left(\frac{x}{2}\right)^\nu\int_\infty^{(0+)}dt~ e...
{(-t)^{\nu+1}}~ e^{-t+\frac{x^2}{4t}}.$ (144)

Often the contour is turned $\pi$, so that it runs from $-\infty$ to a small circel enclosing zero, and then returning to $-\infty$. Then

 \begin{displaymath}J_\nu (x)=\frac{-i}{2\pi}\left(\frac{x}{2}\right)^\nu\int_{-\infty}^{(0+)}
{t^{\nu+1}}~ e^{t-\frac{x^2}{4t}}.
\end{displaymath} (145)

This can be brought to another form by taking x>0 and substituting t=ux/2. Then

\begin{displaymath}J_\nu (x)=\frac{1}{2\pi i}\int_{-\infty}^{(0+)}
{u^{\nu+1}}~ e^{\frac{1}{2}x(u-\frac{1}{u})}.
\end{displaymath} (146)

We now deform the contour so that it runs from $-\infty$ to -1, encircles 0 through the unit circle, and then returns from -1 to $-\infty$. Thus

\begin{displaymath}\int_{-\infty}^{(0+)}=\int_{-\infty}^{-1}+\int_{\rm unit~ circle}+\int_{-1}^
\end{displaymath} (147)

In the first integral on the right we have $u=ze^{-i\pi},~1<z<+\infty $, in the second $u=e^{i\theta},~-\pi <\theta <\pi ,$ and in the third $u=ze^{+i\pi}$. Thus

\begin{displaymath}J_\nu (x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-i\nu\theta}~e^{\f...
\end{displaymath} (148)

In the last integral we substitute $z=e^\theta$ and obtain the standard integral representation

 \begin{displaymath}J_\nu (x)=\frac{1}{\pi}\int_0^\pi\cos (\nu\theta-x\sin\theta)...
...\nu\pi)}{\pi}\int_0^\infty e^{-\nu\theta-x\sinh\theta}d\theta.
\end{displaymath} (149)

If $\nu$ is an integer equal to n, the last term disappears and we get

\begin{displaymath}J_n (x)=\frac{1}{\pi}\int_0^\pi\cos (n\theta-x\sin\theta)d\th...
...rac{1}{2\pi}\int_{-\pi}^\pi e^{-in\theta+ix\sin\theta}d\theta.
\end{displaymath} (150)

From the integral representation (149) we can obtain an asymptotic formula for $x\rightarrow \infty$ by means of the saddle point method. First, let us notice that the second integral on the right hand side does not have a saddle, because if we differentiate the exponent we get $\cosh \theta=-\nu/x$, which has no solution in the range of integration from 0 to infinity. The first integral on the right hand side of (149) can be written as

\begin{displaymath}\frac{1}{2\pi}\int_{-\pi}^\pi e^{-i\nu \theta+ix\sin\theta}d\theta
\end{displaymath} (151)

and has two saddle points. Differentiating the exponent we find them at $\cos\theta=\nu/x\rightarrow 0$, corresponding to $\theta=\pm \pi/2$. The values for sin$\theta$ are therefore $\pm 1$. Using the formulas (136) with the function g given by $e^{-i\nu\theta}$, we immediately get

\begin{displaymath}J_\nu (x)\approx \sqrt{\frac{2}{\pi x}}~\sum_{\pm}~e^{\mp i\nu\frac{\pi}{2}
\pm ix}e^{\mp i\frac{\pi}{4}}.
\end{displaymath} (152)

Here it was used that the second derivative of $\sin \theta$, namely $-\sin \theta$, becomes $\mp 1$ for $\sin\theta=\pm 1$. This was important in (136) for the sign of the term $i\pi/4$ in the exponent. Thus, summing over the saddle points we have the asymptotic expansion

 \begin{displaymath}J_\nu (x)\approx\sqrt{\frac{2}{\pi x}}~\cos\left(x-\nu\frac{\pi}{2}-
\end{displaymath} (153)

showing that the Bessel function have damped oscillations for large values of the argument x.

From the integral representation (145) it is easy to derive some important recursion relations by differentiation. The reader should have no difficulty in showing that

 \begin{displaymath}\frac{dx^{-\nu}J_\nu (x)}{dx}=-x^{-\nu}~J_{\nu +1}(x),
\end{displaymath} (154)

and hence

 \begin{displaymath}\frac{dJ_\nu (x)}{dx}=\frac{\nu}{x}~J_\nu (x)-J_{\nu+1} (x).
\end{displaymath} (155)

From (154) and
$\displaystyle J_{\nu -1}(x)+J_{\nu +1}(x)$ = $\displaystyle \frac{1}{2\pi i}\int_{-\infty}^{(0+)}
{\partial}{\partial u}~e^{\frac{1}{2}x(u-\frac{1}{u})}$  
  = $\displaystyle \frac{2\nu}{x}\frac{1}{2\pi i}\int_{-\infty}^{(0+)}\frac{du}{u^{\nu+1}}~
e^{\frac{1}{2}x(u-\frac{1}{u})},$ (156)


\begin{displaymath}J_{\nu -1}(x)+J_{\nu +1}(x)=\frac{2\nu}{x}~J_\nu (x),
\end{displaymath} (157)

we can easily derive

\begin{displaymath}J'_\nu (x)=\frac{1}{2}[J_{\nu-1}(x)-J_{\nu+1}(x)],~~{\rm and}~~J'_\nu (x)=
J_{\nu -1}(x)-\frac{\nu}{x}~J_\nu (x),
\end{displaymath} (158)

which are useful relations.

We can trivially rewrite (154) as

\begin{displaymath}\frac{d}{xdx}\left(x^{-\nu}J_\nu (x)\right)=-x^{-\nu-1}~J_{\nu +1}(x).
\end{displaymath} (159)

By induction it is then simple to show

 \begin{displaymath}x^{-\nu-r}~J_{\nu +r}(x)=(-1)^r\frac{d^r}{(xdx)^r}\left(x^{-\nu}
J_\nu (x)\right).
\end{displaymath} (160)

next up previous
Next: Bessel functions with order Up: No Title Previous: An integral for the
Mette Lund