Difference between revisions of "Bessel function"

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'''Bessel function''' referes to a solution of the [[Bessel equation]]
+
'''Bessel function''' referes to a solution \(f\) of the [[Bessel equation]]
   
: $ \!\!\!\!\!\!\!\!\! (1) ~ ~ ~ f''(z)+f'(z)/z+(1-\nu/z^2)f(x) =0$
+
\( \!\!\!\!\!\!\!\!\! (1) ~ ~ ~ f''(z)+f'(z)/z+(1-\nu/z^2)f(x) =0\)
   
  +
with the following asymptotic behaviour:
  +
  +
\(\displaystyle
  +
f(x) \approx x^\nu \left( \frac{2^{-\nu}}{\mathrm{Factorial}(\nu)}+ O(x^2) \right)\)
  +
  +
In particular,
  +
  +
For \(\nu=0\), the initial condition is assumed \(~f(0)=1~\), \(~f'(0)=0\) , and
  +
  +
For \(\nu=1\), the initial condition is assumed \(~f(0)=0~\), \(~f'(0)=1\);
  +
  +
Such a solution \(f\) denoted with \(\mathrm{BesselJ}[\nu,x]\) or with \(J_\nu(x)\).
   
 
==BesselJ==
 
==BesselJ==
   
Due to singularity of the equation at $z=0$, the regular solution should have specific behavior. This solution is called
+
Due to singularity of the equation at \(z=0\), the regular solution should have specific behavior. This solution is called
$J_\nu$. For $\nu=0$ and $\nu=1$, there are specific implementations [[BesselJ0]] and [[BesselJ1]].
+
\(J_\nu\). For \(\nu=0\) and \(\nu=1\), there are specific implementations [[BesselJ0]] and [[BesselJ1]].
 
Many formulas about the Bessel functions below are borrowed from the handbook by [[Abramowirtz,Stegun]]
 
Many formulas about the Bessel functions below are borrowed from the handbook by [[Abramowirtz,Stegun]]
 
<ref name="a">
 
<ref name="a">
Line 16: Line 28:
 
==Integral representations==
 
==Integral representations==
   
: $ \!\!\!\!\!\!\!\!\!\! (9.1.20) ~ ~ ~ \displaystyle
+
: \( \!\!\!\!\!\!\!\!\!\! (9.1.20) ~ ~ ~ \displaystyle
 
J_\nu(z) = \frac{(z/2)^{\nu}}{\pi^{1/2} ~(\nu-1/2)!}
 
J_\nu(z) = \frac{(z/2)^{\nu}}{\pi^{1/2} ~(\nu-1/2)!}
 
~
 
~
Line 22: Line 34:
 
~
 
~
 
\cos(z \cos(t)) \sin(t)^{2 \nu} ~t~ \mathrm d t
 
\cos(z \cos(t)) \sin(t)^{2 \nu} ~t~ \mathrm d t
  +
\)
$
 
   
 
===Sonin representation===
 
===Sonin representation===
Line 30: Line 42:
 
</ref> suggest that
 
</ref> suggest that
   
: $\displaystyle J_\nu(z)=\frac{2}{\pi} \int_0 ^\infty \sin(x \cos(t) - \pi \nu/2) \cos(\nu t) \mathrm d t$
+
: \(\displaystyle J_\nu(z)=\frac{2}{\pi} \int_0 ^\infty \sin(x \cos(t) - \pi \nu/2) \cos(\nu t) \mathrm d t\)
   
: $\displaystyle Y_\nu(z)=\frac{-2}{\pi} \int_0 ^\infty \cos(x \cos(t) - \pi \nu/2) \cos(\nu t) \mathrm d t$
+
: \(\displaystyle Y_\nu(z)=\frac{-2}{\pi} \int_0 ^\infty \cos(x \cos(t) - \pi \nu/2) \cos(\nu t) \mathrm d t\)
   
 
and, in particular,
 
and, in particular,
   
: $\displaystyle J_0(x)=\frac{2}{\pi} \int_0^\infty \sin(x \cosh(t)) \mathrm d t$
+
: \(\displaystyle J_0(x)=\frac{2}{\pi} \int_0^\infty \sin(x \cosh(t)) \mathrm d t\)
   
: $\displaystyle Y_0(x)=\frac{-2}{\pi} \int_0^\infty \cos(x \cosh(t)) \mathrm d t$
+
: \(\displaystyle Y_0(x)=\frac{-2}{\pi} \int_0^\infty \cos(x \cosh(t)) \mathrm d t\)
   
 
Also,
 
Also,
: $\displaystyle J_\nu(x)=\frac{2 (x/2)^{-\nu}}{\pi^{1/2} \Gamma(1/2-\nu)}
+
: \(\displaystyle J_\nu(x)=\frac{2 (x/2)^{-\nu}}{\pi^{1/2} \Gamma(1/2-\nu)}
\int_1^\infty \frac{\sin(xt)~ \mathrm d t} { (t^2-1)^{\nu+1/2}}$
+
\int_1^\infty \frac{\sin(xt)~ \mathrm d t} { (t^2-1)^{\nu+1/2}}\)
   
: $\displaystyle Y_\nu(x)=-\frac{2 (x/2)^{-\nu}}{\pi^{1/2} \Gamma(1/2-\nu)}
+
: \(\displaystyle Y_\nu(x)=-\frac{2 (x/2)^{-\nu}}{\pi^{1/2} \Gamma(1/2-\nu)}
\int_1^\infty \frac{\cos(xt)~ \mathrm d t} { (t^2-1)^{\nu+1/2}}$
+
\int_1^\infty \frac{\cos(xt)~ \mathrm d t} { (t^2-1)^{\nu+1/2}}\)
   
   
However the reason of the suggested restriction $x>0$ is not clear.
+
However the reason of the suggested restriction \(x>0\) is not clear.
 
Peerhaps, these expressions can be used to deduce the expansion suitable for the numerical implementation.
 
Peerhaps, these expressions can be used to deduce the expansion suitable for the numerical implementation.
   
==Expansion of $J_\nu$ at zero==
+
==Expansion of \(J_\nu\) at zero==
: $\!\!\!\!\!\!\!\!\!\!\!\!\! (9.1.10) ~ ~ ~ \displaystyle
+
: \(\!\!\!\!\!\!\!\!\!\!\!\!\! (9.1.10) ~ ~ ~ \displaystyle
 
J_\nu(z)=\left(\frac{z}{2}\right)^{\!\nu}~
 
J_\nu(z)=\left(\frac{z}{2}\right)^{\!\nu}~
 
\sum_{k=0}^{\infty} ~
 
\sum_{k=0}^{\infty} ~
 
\frac{(-z^2/4)^k}
 
\frac{(-z^2/4)^k}
{k!~ (\nu\!+\!k)!}$
+
{k!~ (\nu\!+\!k)!}\)
   
==Expansion of $Y_n$ at zero==
+
==Expansion of \(Y_n\) at zero==
   
The similar expansion for $Y_n$ at natural $n\!+\!1$ looks ugly:
+
The similar expansion for \(Y_n\) at natural \(n\!+\!1\) looks ugly:
   
: $\!\!\!\!\!\!\!\!\!\!\!\!\! (\mathrm{GR} 8.403) ~ ~ ~ \displaystyle
+
: \(\!\!\!\!\!\!\!\!\!\!\!\!\! (\mathrm{GR} 8.403) ~ ~ ~ \displaystyle
\pi Y_n(z)= 2 J_n(z) ( \ln(z/2) + C ) - \sum_{k=0}^{n-1} \frac{(n\!-\!k\!-\!1)!}{k!} (z/2)^{2k-n} -$
+
\pi Y_n(z)= 2 J_n(z) ( \ln(z/2) + C ) - \sum_{k=0}^{n-1} \frac{(n\!-\!k\!-\!1)!}{k!} (z/2)^{2k-n} -\)
: $ \displaystyle
+
: \( \displaystyle
 
- (z/2)^n \frac{1}{n!} \sum_{k=1}^n \frac{1}{k} - \sum_{k=0}^{\infty} \frac{(-1)^k (z/2)^{n+2k}}{k! ~(n\!+\!k)!}
 
- (z/2)^n \frac{1}{n!} \sum_{k=1}^n \frac{1}{k} - \sum_{k=0}^{\infty} \frac{(-1)^k (z/2)^{n+2k}}{k! ~(n\!+\!k)!}
 
\left(
 
\left(
 
\sum_{m=1}^{n+k} \frac{1}{m} + \sum_{m=1}^k \frac{1}{m} \right)
 
\sum_{m=1}^{n+k} \frac{1}{m} + \sum_{m=1}^k \frac{1}{m} \right)
  +
\)
$
 
where $C$ is [[Euler constant]], called also [[EulerGamma]]
+
where \(C\) is [[Euler constant]], called also [[EulerGamma]]
: $\displaystyle
+
: \(\displaystyle
C=- \int_0^\infty \exp(-t)~ \ln(t) ~\mathrm d t \approx 0.57721566490$
+
C=- \int_0^\infty \exp(-t)~ \ln(t) ~\mathrm d t \approx 0.57721566490\)
   
 
Up to year 2012, no beautiful representation for the expansion coefficients is available.
 
Up to year 2012, no beautiful representation for the expansion coefficients is available.
Line 86: Line 98:
 
!-->
 
!-->
   
: $ \displaystyle J_{\pm \nu}(z)=$
+
: \( \displaystyle J_{\pm \nu}(z)=\)
: $ \displaystyle = \sqrt{\frac{2}{\pi z}} \cos\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~
+
: \( \displaystyle = \sqrt{\frac{2}{\pi z}} \cos\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
\frac{ \Gamma(\nu+2k+1/2)}{(2k)!~ \Gamma(\nu-2k+1/2)} + R_1 \right) -$
+
\frac{ \Gamma(\nu+2k+1/2)}{(2k)!~ \Gamma(\nu-2k+1/2)} + R_1 \right) -\)
:$ \displaystyle - \sqrt{\frac{2}{\pi z}} \sin\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big)~ \left( \frac{1}{2z} ~ \sum_{k=0}^{n-1} ~
+
:\( \displaystyle - \sqrt{\frac{2}{\pi z}} \sin\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big)~ \left( \frac{1}{2z} ~ \sum_{k=0}^{n-1} ~
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
\frac{ \Gamma(\nu+2k+3/2)}{(2k\!+\!1)! ~\Gamma(\nu-2k-1/2)} + R_2 \right)$
+
\frac{ \Gamma(\nu+2k+3/2)}{(2k\!+\!1)! ~\Gamma(\nu-2k-1/2)} + R_2 \right)\)
   
: $ \displaystyle Y_{\pm \nu}(z)=$
+
: \( \displaystyle Y_{\pm \nu}(z)=\)
: $ \displaystyle = \sqrt{\frac{2}{\pi z}} \sin\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~
+
: \( \displaystyle = \sqrt{\frac{2}{\pi z}} \sin\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
\frac{ \Gamma(\nu+2k+1/2)}{(2k)!~ \Gamma(\nu-2k+1/2)} + R_1 \right) +$
+
\frac{ \Gamma(\nu+2k+1/2)}{(2k)!~ \Gamma(\nu-2k+1/2)} + R_1 \right) +\)
:$ \displaystyle + \sqrt{\frac{2}{\pi z}} \cos\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big)~ \left( \frac{1}{2z} ~ \sum_{k=0}^{n-1} ~
+
:\( \displaystyle + \sqrt{\frac{2}{\pi z}} \cos\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big)~ \left( \frac{1}{2z} ~ \sum_{k=0}^{n-1} ~
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
 
\left(\frac{-1}{4z^2}\right)^{\!\! k}
\frac{ \Gamma(\nu+2k+3/2)}{(2k\!+\!1)! ~\Gamma(\nu-2k-1/2)} + R_2 \right)$
+
\frac{ \Gamma(\nu+2k+3/2)}{(2k\!+\!1)! ~\Gamma(\nu-2k-1/2)} + R_2 \right)\)
   
: $ \displaystyle H_{\nu}(z)=$
+
: \( \displaystyle H_{\nu}(z)=\)
: $ \displaystyle = \sqrt{\frac{2}{\pi z}} \exp\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~
+
: \( \displaystyle = \sqrt{\frac{2}{\pi z}} \exp\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~
 
\left(\frac{\mathrm i}{2z}\right)^{\!\! k}
 
\left(\frac{\mathrm i}{2z}\right)^{\!\! k}
 
\frac{ \Gamma(\nu+n+1/2)}{(2n)!~ \Gamma(\nu-n+1/2)} +
 
\frac{ \Gamma(\nu+n+1/2)}{(2n)!~ \Gamma(\nu-n+1/2)} +
 
\theta_1 \left(\frac{\mathrm i}{2z}\right)^{\!\! k}
 
\theta_1 \left(\frac{\mathrm i}{2z}\right)^{\!\! k}
\frac{ \Gamma(\nu+n+1/2)}{(2n)!~ \Gamma(\nu-n+1/2)} \right) $
+
\frac{ \Gamma(\nu+n+1/2)}{(2n)!~ \Gamma(\nu-n+1/2)} \right) \)
   
: $ \displaystyle
+
: \( \displaystyle
 
|R_1|< \left|
 
|R_1|< \left|
 
\frac{\Gamma(\nu+2n+1/2)}{(2z)^{2n} ~ (2n)! ~ \Gamma(\nu-2n+1/2)}
 
\frac{\Gamma(\nu+2n+1/2)}{(2z)^{2n} ~ (2n)! ~ \Gamma(\nu-2n+1/2)}
\right|$
+
\right|\)
   
: $ \displaystyle
+
: \( \displaystyle
 
|R_2|< \left|
 
|R_2|< \left|
 
\frac{\Gamma(\nu+2n+3/2)}{(2z)^{2n+1} ~ (2n\!+\!1)! ~ \Gamma(\nu-2n-1/2)}
 
\frac{\Gamma(\nu+2n+3/2)}{(2z)^{2n+1} ~ (2n\!+\!1)! ~ \Gamma(\nu-2n-1/2)}
\right|$
+
\right|\)
   
: while $\Im(z)\le 0$, the esitmate $|\theta_1| < 1 $
+
: while \(\Im(z)\le 0\), the esitmate \(|\theta_1| < 1 \)
   
For half–natural $\nu$, the singularity of $\Gamma$ terminates the series and they become finite sums.
+
For half–natural \(\nu\), the singularity of \(\Gamma\) terminates the series and they become finite sums.
   
 
==Expansion at infinity from [[Abramowitz,Stegun]]==
 
==Expansion at infinity from [[Abramowitz,Stegun]]==
Let $\mu=4 \nu^2$. Define two series $P_\nu(z)$ and $Q_\nu(z)$ with
+
Let \(\mu=4 \nu^2\). Define two series \(P_\nu(z)\) and \(Q_\nu(z)\) with
   
: $\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.09) ~ ~ ~ \displaystyle
+
: \(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.09) ~ ~ ~ \displaystyle
 
P_\nu(z)=1
 
P_\nu(z)=1
 
-\frac{(\mu\!-\!1)(\mu\!-\!9)}{2! ~ (8z)^2}
 
-\frac{(\mu\!-\!1)(\mu\!-\!9)}{2! ~ (8z)^2}
 
+\frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)(\mu\!-\!49)}{4! ~ (8z)^4}
 
+\frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)(\mu\!-\!49)}{4! ~ (8z)^4}
- ..$
+
- ..\)
: $\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.10) ~ ~ ~ \displaystyle
+
: \(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.10) ~ ~ ~ \displaystyle
 
Q_\nu(z)=
 
Q_\nu(z)=
 
\frac{(\mu\!-\!1)}{1! ~ (8z)}
 
\frac{(\mu\!-\!1)}{1! ~ (8z)}
 
+\frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)}{3! ~ (8z)^3}
 
+\frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)}{3! ~ (8z)^3}
 
- \frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)(\mu\!-\!49)(\mu\!-\!81)}{5! ~ (8z)^5}
 
- \frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)(\mu\!-\!49)(\mu\!-\!81)}{5! ~ (8z)^5}
+..$
+
+..\)
Let $x=z-\Big(\nu/2+\pi/4\Big)\pi~$;
+
Let \(x=z-\Big(\nu/2+\pi/4\Big)\pi~\);
 
then
 
then
: $\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.05) ~ ~ ~ \displaystyle
+
: \(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.05) ~ ~ ~ \displaystyle
J_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) \cos(z)-Q_\nu(z) \sin(z) \Big)$
+
J_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) \cos(z)-Q_\nu(z) \sin(z) \Big)\)
   
: $\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.06) ~ ~ ~ \displaystyle
+
: \(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.06) ~ ~ ~ \displaystyle
Y_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) \sin(z)+Q_\nu(z) \cos(z) \Big)$
+
Y_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) \sin(z)+Q_\nu(z) \cos(z) \Big)\)
   
: $\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.07) ~ ~ ~ \displaystyle
+
: \(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.07) ~ ~ ~ \displaystyle
H_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) + \mathrm i ~ Q_\nu(z) \Big) \mathrm e^{\mathrm i z}$
+
H_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) + \mathrm i ~ Q_\nu(z) \Big) \mathrm e^{\mathrm i z}\)
   
while $|\mathrm{Arg}(z)|<\pi$.
+
while \(|\mathrm{Arg}(z)|<\pi\).
   
 
==Asymptotic expansion for modulus and phase==
 
==Asymptotic expansion for modulus and phase==
   
: $ \!\!\!\!\!\!\!\!\! (9.2.19) ~ ~ ~ J_\nu(z)= M \cos(\theta) ~~, ~~$ $~ Y_\nu(z)= M \sin(\theta) ~~, ~~$
+
: \( \!\!\!\!\!\!\!\!\! (9.2.19) ~ ~ ~ J_\nu(z)= M \cos(\theta) ~~, ~~\) \(~ Y_\nu(z)= M \sin(\theta) ~~, ~~\)
   
: $ \!\!\!\!\!\!\!\!\! (9.2.17) ~ ~ ~ M = \sqrt{J_\nu(z)^2 + Y_\nu(z)^2}$
+
: \( \!\!\!\!\!\!\!\!\! (9.2.17) ~ ~ ~ M = \sqrt{J_\nu(z)^2 + Y_\nu(z)^2}\)
   
In certain range, while $ |\Re(\theta)|<\pi$, also
+
In certain range, while \( |\Re(\theta)|<\pi\), also
: $ \!\!\!\!\!\!\!\!\! (9.2.0) ~ ~ ~ \theta = \mathrm{atan2}( Y_\nu(z)/ J_\nu(z))$
+
: \( \!\!\!\!\!\!\!\!\! (9.2.0) ~ ~ ~ \theta = \mathrm{atan2}( Y_\nu(z)/ J_\nu(z))\)
   
$M$ is called "modulus" and $\theta$ is called "argument"
+
\(M\) is called "modulus" and \(\theta\) is called "argument"
 
<ref name="a">
 
<ref name="a">
 
http://people.math.sfu.ca/~cbm/aands/page_365.htm
 
http://people.math.sfu.ca/~cbm/aands/page_365.htm
 
Abramovitz, Stegun. Handbook on mathematical functions.
 
Abramovitz, Stegun. Handbook on mathematical functions.
</ref>. Let $\mu=4\nu^2$.
+
</ref>. Let \(\mu=4\nu^2\).
   
At large values of the argument, it worth to expand $M$ and $\theta$ instead of $J$:
+
At large values of the argument, it worth to expand \(M\) and \(\theta\) instead of \(J\):
   
: $ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.28) \displaystyle ~ ~ ~ M^2 = \frac{2}{\pi z} \left( 1 +
+
: \( \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.28) \displaystyle ~ ~ ~ M^2 = \frac{2}{\pi z} \left( 1 +
 
\frac{1}{2} \frac{\mu \!-\!1}{(2z)^2} +
 
\frac{1}{2} \frac{\mu \!-\!1}{(2z)^2} +
 
\frac{1\cdot 3}{2 \cdot 4} \frac{(\mu \!-\!1)(\mu\!-\!9)}{(2z)^4} +
 
\frac{1\cdot 3}{2 \cdot 4} \frac{(\mu \!-\!1)(\mu\!-\!9)}{(2z)^4} +
 
\frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)}{(2z)^6} +..
 
\frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)}{(2z)^6} +..
 
\right)
 
\right)
  +
\)
$
 
   
: $ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.29) \displaystyle ~ ~ ~ \theta = z - \left(\frac{\nu}{2}+\frac{1}{4}\right) \pi
+
: \( \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.29) \displaystyle ~ ~ ~ \theta = z - \left(\frac{\nu}{2}+\frac{1}{4}\right) \pi
 
+ \frac{\mu\!-\!1}{2(2z)}
 
+ \frac{\mu\!-\!1}{2(2z)}
 
+ \frac{ (\mu\!-\!1) (\mu-25)}{6(4z)^3}
 
+ \frac{ (\mu\!-\!1) (\mu-25)}{6(4z)^3}
+ \frac{ (\mu\!-\!1) (\mu^2-114\mu+1073)}{5(4z)^5}
+
+ \frac{ (\mu\!-\!1) (\mu^2-114\mu+1073)}{5(4z)^5} \) \(+\) \(~\, \displaystyle \frac{ (\mu\!-\!1) (5\mu^3-1535\mu^2+54703\mu-375733) }{14(4z)^7} + ..\)
  +
+ \frac{ (\mu\!-\!1) (5\mu^3-1535\mu^2+54703\mu-375733) }{14(4z)^7} + ..$
 
  +
==Integrals==
  +
Some examples with Mathematica are shown below:
  +
  +
<nomathjax>
  +
Integrate[BesselJ[0, x y] y, {y, 0, 1}]
  +
</nomathjax> \(~\)
  +
does formula
  +
  +
\(\displaystyle \int_0^1 J_0(x y) \, y\, \mathrm d y = \frac{ J_1(x)}{x}\)
  +
  +
<nomathjax>
  +
Simplify[Integrate[1/Sqrt[1-x^2] BesselJ[0, k x], {x, 0, 1}], k > 0]
  +
</nomathjax> \(~\)
  +
does formula
  +
  +
\(\displaystyle \int_0^1 \frac{1}{\sqrt{1-x^2}}\, J_0(kx) \, \mathrm d x = \frac{\pi}{2} J_0(k/2)^2\)
  +
  +
<nomathjax>
  +
Simplify[Integrate[1/Sqrt[1-x^2] BesselJ[0, k x] x, {x, 0, 1}], k > 0]
  +
</nomathjax> \(~\)
  +
does formula
  +
  +
\(\displaystyle \int_0^1 \frac{1}{\sqrt{1-x^2}} \, J_0(kx) \, x \, \mathrm d x = \frac{\sin(k)}{k}\)
  +
  +
These expressions can be used for testing of the numerical implementations of the [[Bessel transform]].
  +
Wolfram suggests more functions such that their Bessel transforms are expressed in terms of special functions at
  +
http://mathworld.wolfram.com/HankelTransform.html
  +
  +
==Derivatives of the Bessel==
  +
D[BesselJ[v, x], x] \(~ ~\) gives
  +
  +
1/2 (BesselJ[-1 + v, x] - BesselJ[1 + v, x])
  +
  +
In "short" notations, this I can be written with
  +
  +
\({J_v}^{\prime}(x)=\frac{1}{2}\big(
  +
J_{v-1}(x)-J_{v+1}(x)
  +
\big)\)
  +
  +
==Zeros of the Bessel==
  +
  +
Zeros of the Bessel function are demote with identifier BesselJZero; first argument indicates the order of the Bessel function; second one base sense of the radial coordinate.
  +
For the numerical evaluation, the asymptotic expression can be used,
  +
  +
zo[k_] = Series[BesselJZero[v, k], {k, Infinity, 1}]
  +
  +
\(\pi \left(k+\frac{v}{2}-\frac{1}{4}\right)-\frac{(2
  +
v-1) (2 v+1)}{8 \pi
  +
\left(k+\frac{v}{2}-\frac{1}{4}\right)}\)
  +
  +
For the precise evaluation at integer \(v\), three steps of the Newton iteration to adjust the value seem to be sufficient, at least for the "double" precision.
  +
The [[C++]] implementation can be denoted [[jnz]] or, as in Mathematica, [[BesselJZero]]. The example and the test:
  +
<poem><nomathjax><nowiki>
  +
#include<math.h>
  +
#include<stdio.h>
  +
#define DB double
  +
#define DO(x,y) for(x=0;x<y;x++)
  +
  +
DB jnp(int n,DB x){ return .5*( jn(n-1,x)-jn(n+1,x) ) ; } // Derivative of n th Bessel
  +
  +
DB jnz(int v, int k){ DB x,t; t=M_PI*(k+.5*v-.25); x= t - (v*v-.25)*.5/t;
  +
x-= jn(v,x)/jnp(v,x); // Newton adjustment of the root
  +
x-= jn(v,x)/jnp(v,x);
  +
x-= jn(v,x)/jnp(v,x);
  +
return x; } // the k th zero of v th Bessel
  +
  +
int main(){ int n,v,k; DB x,y; // evaluation of zeros of Bessel of order 0,1,2,3 and the testing
  +
for(n=1;n<101;n++){ printf("%3d",n);
  +
DO(v,4){x=jnz(v,n); y=jn(v,x);
  +
printf("%21.16lf %19.16lf",x,y);
  +
}
  +
printf("\n");
  +
}
  +
return 0;
  +
}
  +
</nowiki></nomathjax></poem>
  +
The first 10 lines of the output are copupasted below:
  +
  +
1 2.4048255576957729 -0.0000000000000001 3.8317059702075125 -0.0000000000000000 5.1356223018406828 -0.0000000000000001 6.3801618959239841 -0.0000000000000002
  +
2 5.5200781102863106 -0.0000000000000000 7.0155866698156188 0.0000000000000000 8.4172441403998643 -0.0000000000000001 9.7610231299816697 -0.0000000000000000
  +
3 8.6537279129110125 -0.0000000000000001 10.1734681350627216 0.0000000000000001 11.6198411721490587 0.0000000000000002 13.0152007216984344 -0.0000000000000000
  +
4 11.7915344390142813 -0.0000000000000001 13.3236919363142228 -0.0000000000000001 14.7959517823512599 -0.0000000000000002 16.2234661603187682 0.0000000000000000
  +
5 14.9309177084877867 -0.0000000000000001 16.4706300508776344 -0.0000000000000003 17.9598194949878263 0.0000000000000000 19.4094152264350122 -0.0000000000000001
  +
6 18.0710639679109235 0.0000000000000002 19.6158585104682430 0.0000000000000002 21.1169970530218443 -0.0000000000000002 22.5827295931044425 0.0000000000000001
  +
7 21.2116366298792585 0.0000000000000001 22.7600843805927724 -0.0000000000000001 24.2701123135731009 0.0000000000000003 25.7481666992949769 0.0000000000000001
  +
8 24.3524715307493018 -0.0000000000000001 25.9036720876183821 -0.0000000000000001 27.4205735499845566 -0.0000000000000001 28.9083507809217579 0.0000000000000000
  +
9 27.4934791320402532 0.0000000000000002 29.0468285349168553 -0.0000000000000000 30.5692044955163986 -0.0000000000000002 32.0648524070977103 -0.0000000000000001
  +
10 30.6346064684319757 0.0000000000000001 32.1896799109744052 0.0000000000000002 33.7165195092226995 -0.0000000000000001 35.2186707386101148 0.0000000000000000
   
 
==References==
 
==References==

Latest revision as of 18:25, 30 July 2019

Bessel function referes to a solution \(f\) of the Bessel equation

\( \!\!\!\!\!\!\!\!\! (1) ~ ~ ~ f''(z)+f'(z)/z+(1-\nu/z^2)f(x) =0\)

with the following asymptotic behaviour:

\(\displaystyle f(x) \approx x^\nu \left( \frac{2^{-\nu}}{\mathrm{Factorial}(\nu)}+ O(x^2) \right)\)

In particular,

For \(\nu=0\), the initial condition is assumed \(~f(0)=1~\), \(~f'(0)=0\) , and

For \(\nu=1\), the initial condition is assumed \(~f(0)=0~\), \(~f'(0)=1\);

Such a solution \(f\) denoted with \(\mathrm{BesselJ}[\nu,x]\) or with \(J_\nu(x)\).

BesselJ

Due to singularity of the equation at \(z=0\), the regular solution should have specific behavior. This solution is called \(J_\nu\). For \(\nu=0\) and \(\nu=1\), there are specific implementations BesselJ0 and BesselJ1. Many formulas about the Bessel functions below are borrowed from the handbook by Abramowirtz,Stegun [1]; the numeration of formulas from there is used below.

Integral representations

\( \!\!\!\!\!\!\!\!\!\! (9.1.20) ~ ~ ~ \displaystyle J_\nu(z) = \frac{(z/2)^{\nu}}{\pi^{1/2} ~(\nu-1/2)!} ~ \int_0^\pi ~ \cos(z \cos(t)) \sin(t)^{2 \nu} ~t~ \mathrm d t \)

Sonin representation

The Mehler,Sonin formulas [2] suggest that

\(\displaystyle J_\nu(z)=\frac{2}{\pi} \int_0 ^\infty \sin(x \cos(t) - \pi \nu/2) \cos(\nu t) \mathrm d t\)
\(\displaystyle Y_\nu(z)=\frac{-2}{\pi} \int_0 ^\infty \cos(x \cos(t) - \pi \nu/2) \cos(\nu t) \mathrm d t\)

and, in particular,

\(\displaystyle J_0(x)=\frac{2}{\pi} \int_0^\infty \sin(x \cosh(t)) \mathrm d t\)
\(\displaystyle Y_0(x)=\frac{-2}{\pi} \int_0^\infty \cos(x \cosh(t)) \mathrm d t\)

Also,

\(\displaystyle J_\nu(x)=\frac{2 (x/2)^{-\nu}}{\pi^{1/2} \Gamma(1/2-\nu)} \int_1^\infty \frac{\sin(xt)~ \mathrm d t} { (t^2-1)^{\nu+1/2}}\)
\(\displaystyle Y_\nu(x)=-\frac{2 (x/2)^{-\nu}}{\pi^{1/2} \Gamma(1/2-\nu)} \int_1^\infty \frac{\cos(xt)~ \mathrm d t} { (t^2-1)^{\nu+1/2}}\)


However the reason of the suggested restriction \(x>0\) is not clear. Peerhaps, these expressions can be used to deduce the expansion suitable for the numerical implementation.

Expansion of \(J_\nu\) at zero

\(\!\!\!\!\!\!\!\!\!\!\!\!\! (9.1.10) ~ ~ ~ \displaystyle J_\nu(z)=\left(\frac{z}{2}\right)^{\!\nu}~ \sum_{k=0}^{\infty} ~ \frac{(-z^2/4)^k} {k!~ (\nu\!+\!k)!}\)

Expansion of \(Y_n\) at zero

The similar expansion for \(Y_n\) at natural \(n\!+\!1\) looks ugly:

\(\!\!\!\!\!\!\!\!\!\!\!\!\! (\mathrm{GR} 8.403) ~ ~ ~ \displaystyle \pi Y_n(z)= 2 J_n(z) ( \ln(z/2) + C ) - \sum_{k=0}^{n-1} \frac{(n\!-\!k\!-\!1)!}{k!} (z/2)^{2k-n} -\)
\( \displaystyle - (z/2)^n \frac{1}{n!} \sum_{k=1}^n \frac{1}{k} - \sum_{k=0}^{\infty} \frac{(-1)^k (z/2)^{n+2k}}{k! ~(n\!+\!k)!} \left( \sum_{m=1}^{n+k} \frac{1}{m} + \sum_{m=1}^k \frac{1}{m} \right) \)

where \(C\) is Euler constant, called also EulerGamma

\(\displaystyle C=- \int_0^\infty \exp(-t)~ \ln(t) ~\mathrm d t \approx 0.57721566490\)

Up to year 2012, no beautiful representation for the expansion coefficients is available.

Expansion at infinity by Gradshtein,Ryzhik

Gradshtein,Ryzhik [3] suggest the following expansions (See 8.4.5.5)

\( \displaystyle J_{\pm \nu}(z)=\)
\( \displaystyle = \sqrt{\frac{2}{\pi z}} \cos\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~ \left(\frac{-1}{4z^2}\right)^{\!\! k} \frac{ \Gamma(\nu+2k+1/2)}{(2k)!~ \Gamma(\nu-2k+1/2)} + R_1 \right) -\)
\( \displaystyle - \sqrt{\frac{2}{\pi z}} \sin\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big)~ \left( \frac{1}{2z} ~ \sum_{k=0}^{n-1} ~ \left(\frac{-1}{4z^2}\right)^{\!\! k} \frac{ \Gamma(\nu+2k+3/2)}{(2k\!+\!1)! ~\Gamma(\nu-2k-1/2)} + R_2 \right)\)
\( \displaystyle Y_{\pm \nu}(z)=\)
\( \displaystyle = \sqrt{\frac{2}{\pi z}} \sin\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~ \left(\frac{-1}{4z^2}\right)^{\!\! k} \frac{ \Gamma(\nu+2k+1/2)}{(2k)!~ \Gamma(\nu-2k+1/2)} + R_1 \right) +\)
\( \displaystyle + \sqrt{\frac{2}{\pi z}} \cos\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big)~ \left( \frac{1}{2z} ~ \sum_{k=0}^{n-1} ~ \left(\frac{-1}{4z^2}\right)^{\!\! k} \frac{ \Gamma(\nu+2k+3/2)}{(2k\!+\!1)! ~\Gamma(\nu-2k-1/2)} + R_2 \right)\)
\( \displaystyle H_{\nu}(z)=\)
\( \displaystyle = \sqrt{\frac{2}{\pi z}} \exp\!\Big(z-(\pm 2 \nu\!+\!1)\pi/4 \Big) ~ \left( ~ \sum_{k=0}^{n-1} ~ \left(\frac{\mathrm i}{2z}\right)^{\!\! k} \frac{ \Gamma(\nu+n+1/2)}{(2n)!~ \Gamma(\nu-n+1/2)} + \theta_1 \left(\frac{\mathrm i}{2z}\right)^{\!\! k} \frac{ \Gamma(\nu+n+1/2)}{(2n)!~ \Gamma(\nu-n+1/2)} \right) \)
\( \displaystyle |R_1|< \left| \frac{\Gamma(\nu+2n+1/2)}{(2z)^{2n} ~ (2n)! ~ \Gamma(\nu-2n+1/2)} \right|\)
\( \displaystyle |R_2|< \left| \frac{\Gamma(\nu+2n+3/2)}{(2z)^{2n+1} ~ (2n\!+\!1)! ~ \Gamma(\nu-2n-1/2)} \right|\)
while \(\Im(z)\le 0\), the esitmate \(|\theta_1| < 1 \)

For half–natural \(\nu\), the singularity of \(\Gamma\) terminates the series and they become finite sums.

Expansion at infinity from Abramowitz,Stegun

Let \(\mu=4 \nu^2\). Define two series \(P_\nu(z)\) and \(Q_\nu(z)\) with

\(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.09) ~ ~ ~ \displaystyle P_\nu(z)=1 -\frac{(\mu\!-\!1)(\mu\!-\!9)}{2! ~ (8z)^2} +\frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)(\mu\!-\!49)}{4! ~ (8z)^4} - ..\)
\(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.10) ~ ~ ~ \displaystyle Q_\nu(z)= \frac{(\mu\!-\!1)}{1! ~ (8z)} +\frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)}{3! ~ (8z)^3} - \frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)(\mu\!-\!49)(\mu\!-\!81)}{5! ~ (8z)^5} +..\)

Let \(x=z-\Big(\nu/2+\pi/4\Big)\pi~\); then

\(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.05) ~ ~ ~ \displaystyle J_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) \cos(z)-Q_\nu(z) \sin(z) \Big)\)
\(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.06) ~ ~ ~ \displaystyle Y_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) \sin(z)+Q_\nu(z) \cos(z) \Big)\)
\(\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.07) ~ ~ ~ \displaystyle H_\nu(z_)=\sqrt{\frac{2}{\pi z}}\Big(P_\nu(z) + \mathrm i ~ Q_\nu(z) \Big) \mathrm e^{\mathrm i z}\)

while \(|\mathrm{Arg}(z)|<\pi\).

Asymptotic expansion for modulus and phase

\( \!\!\!\!\!\!\!\!\! (9.2.19) ~ ~ ~ J_\nu(z)= M \cos(\theta) ~~, ~~\) \(~ Y_\nu(z)= M \sin(\theta) ~~, ~~\)
\( \!\!\!\!\!\!\!\!\! (9.2.17) ~ ~ ~ M = \sqrt{J_\nu(z)^2 + Y_\nu(z)^2}\)

In certain range, while \( |\Re(\theta)|<\pi\), also

\( \!\!\!\!\!\!\!\!\! (9.2.0) ~ ~ ~ \theta = \mathrm{atan2}( Y_\nu(z)/ J_\nu(z))\)

\(M\) is called "modulus" and \(\theta\) is called "argument" [1]. Let \(\mu=4\nu^2\).

At large values of the argument, it worth to expand \(M\) and \(\theta\) instead of \(J\):

\( \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.28) \displaystyle ~ ~ ~ M^2 = \frac{2}{\pi z} \left( 1 + \frac{1}{2} \frac{\mu \!-\!1}{(2z)^2} + \frac{1\cdot 3}{2 \cdot 4} \frac{(\mu \!-\!1)(\mu\!-\!9)}{(2z)^4} + \frac{1\cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{(\mu\!-\!1)(\mu\!-\!9)(\mu\!-\!25)}{(2z)^6} +.. \right) \)
\( \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! (9.2.29) \displaystyle ~ ~ ~ \theta = z - \left(\frac{\nu}{2}+\frac{1}{4}\right) \pi + \frac{\mu\!-\!1}{2(2z)} + \frac{ (\mu\!-\!1) (\mu-25)}{6(4z)^3} + \frac{ (\mu\!-\!1) (\mu^2-114\mu+1073)}{5(4z)^5} \) \(+\) \(~\, \displaystyle \frac{ (\mu\!-\!1) (5\mu^3-1535\mu^2+54703\mu-375733) }{14(4z)^7} + ..\)

Integrals

Some examples with Mathematica are shown below:

Integrate[BesselJ[0, x y] y, {y, 0, 1}] \(~\) does formula

\(\displaystyle \int_0^1 J_0(x y) \, y\, \mathrm d y = \frac{ J_1(x)}{x}\)

Simplify[Integrate[1/Sqrt[1-x^2] BesselJ[0, k x], {x, 0, 1}], k > 0] \(~\) does formula

\(\displaystyle \int_0^1 \frac{1}{\sqrt{1-x^2}}\, J_0(kx) \, \mathrm d x = \frac{\pi}{2} J_0(k/2)^2\)

Simplify[Integrate[1/Sqrt[1-x^2] BesselJ[0, k x] x, {x, 0, 1}], k > 0] \(~\) does formula

\(\displaystyle \int_0^1 \frac{1}{\sqrt{1-x^2}} \, J_0(kx) \, x \, \mathrm d x = \frac{\sin(k)}{k}\)

These expressions can be used for testing of the numerical implementations of the Bessel transform. Wolfram suggests more functions such that their Bessel transforms are expressed in terms of special functions at http://mathworld.wolfram.com/HankelTransform.html

Derivatives of the Bessel

D[BesselJ[v, x], x] \(~ ~\) gives

1/2 (BesselJ[-1 + v, x] - BesselJ[1 + v, x])

In "short" notations, this I can be written with

\({J_v}^{\prime}(x)=\frac{1}{2}\big( J_{v-1}(x)-J_{v+1}(x) \big)\)

Zeros of the Bessel

Zeros of the Bessel function are demote with identifier BesselJZero; first argument indicates the order of the Bessel function; second one base sense of the radial coordinate. For the numerical evaluation, the asymptotic expression can be used,

zo[k_] = Series[BesselJZero[v, k], {k, Infinity, 1}]

\(\pi \left(k+\frac{v}{2}-\frac{1}{4}\right)-\frac{(2 v-1) (2 v+1)}{8 \pi \left(k+\frac{v}{2}-\frac{1}{4}\right)}\)

For the precise evaluation at integer \(v\), three steps of the Newton iteration to adjust the value seem to be sufficient, at least for the "double" precision. The C++ implementation can be denoted jnz or, as in Mathematica, BesselJZero. The example and the test:


#include<math.h>
#include<stdio.h>
#define DB double
#define DO(x,y) for(x=0;x<y;x++)

DB jnp(int n,DB x){ return .5*( jn(n-1,x)-jn(n+1,x) ) ; } // Derivative of n th Bessel

DB jnz(int v, int k){ DB x,t; t=M_PI*(k+.5*v-.25); x= t - (v*v-.25)*.5/t;
                x-= jn(v,x)/jnp(v,x); // Newton adjustment of the root
                x-= jn(v,x)/jnp(v,x);
                x-= jn(v,x)/jnp(v,x);
                return x; } // the k th zero of v th Bessel

int main(){ int n,v,k; DB x,y; // evaluation of zeros of Bessel of order 0,1,2,3 and the testing
for(n=1;n<101;n++){ printf("%3d",n);
                DO(v,4){x=jnz(v,n); y=jn(v,x);
                        printf("%21.16lf %19.16lf",x,y);
                        }
                printf("\n");
                }
return 0;
}

The first 10 lines of the output are copupasted below:

 1   2.4048255576957729 -0.0000000000000001   3.8317059702075125 -0.0000000000000000   5.1356223018406828 -0.0000000000000001   6.3801618959239841 -0.0000000000000002
 2   5.5200781102863106 -0.0000000000000000   7.0155866698156188  0.0000000000000000   8.4172441403998643 -0.0000000000000001   9.7610231299816697 -0.0000000000000000
 3   8.6537279129110125 -0.0000000000000001  10.1734681350627216  0.0000000000000001  11.6198411721490587  0.0000000000000002  13.0152007216984344 -0.0000000000000000
 4  11.7915344390142813 -0.0000000000000001  13.3236919363142228 -0.0000000000000001  14.7959517823512599 -0.0000000000000002  16.2234661603187682  0.0000000000000000
 5  14.9309177084877867 -0.0000000000000001  16.4706300508776344 -0.0000000000000003  17.9598194949878263  0.0000000000000000  19.4094152264350122 -0.0000000000000001
 6  18.0710639679109235  0.0000000000000002  19.6158585104682430  0.0000000000000002  21.1169970530218443 -0.0000000000000002  22.5827295931044425  0.0000000000000001
 7  21.2116366298792585  0.0000000000000001  22.7600843805927724 -0.0000000000000001  24.2701123135731009  0.0000000000000003  25.7481666992949769  0.0000000000000001
 8  24.3524715307493018 -0.0000000000000001  25.9036720876183821 -0.0000000000000001  27.4205735499845566 -0.0000000000000001  28.9083507809217579  0.0000000000000000
 9  27.4934791320402532  0.0000000000000002  29.0468285349168553 -0.0000000000000000  30.5692044955163986 -0.0000000000000002  32.0648524070977103 -0.0000000000000001
10  30.6346064684319757  0.0000000000000001  32.1896799109744052  0.0000000000000002  33.7165195092226995 -0.0000000000000001  35.2186707386101148  0.0000000000000000

References

  1. 1.0 1.1 http://people.math.sfu.ca/~cbm/aands/page_365.htm Abramovitz, Stegun. Handbook on mathematical functions.
  2. http://dlmf.nist.gov/10.9 Digital library of mathematical functions
  3. http://books.google.co.jp/books?id=aBgFYxKHUjsC&pg=PA859&hl=ja&source=gbs_toc_r&cad=4#v=onepage&q&f=false Izrail Solomonovich Gradshtein, Iosif Moiseevich Ryzhik, Alan Jeffrey, Daniel Zwillinger. Table of Integrals, Series, And Products.

http://en.wikipedia.org/wiki/Bessel_function

http://en.citizendium.org/wiki/Bessel_functions

Keywords

BesselJ0, BesselJ1, BesselY0, BesselY1,