# Preface

This section shows that many properties of the Bessel functions of the first kind can be obtained with the aid of the Laplace transform.

# Bessel functions

The Bessel equation of order n

$t^2 y'' (t) + t\,y' (t) + \left( t^2 - n^2 \right) y (t) =0$
has a solution Jn(t) that is regular at t = 0. We denote by
$J_n^L (\lambda ) = {\cal L}\left[ J_n (t) \right] (\lambda ) = \int_0^{\infty} e^{-\lambda \,t} J_n (t) \,{\text d}t$
the Laplace transformation of the Bessel function. For n = 0, we have $$t\, y'' (t) + y' (t) + t\, y (t) =0 .$$ Application of the Laplace transformation to the latter gives
${\cal L} \left[ t\, y'' (t) \right] + {\cal L} \left[ y' (t) \right] + {\cal L} \left[ t\,y (t) \right] = 0 .$
Mathematica knows that
LaplaceTransform[y'[t], t, s]
s LaplaceTransform[y[t], t, s] - y[0]
For other terms, we use integration by parts.
\begin{align*} {\cal L} \left[ t\, y'' (t) \right] &= \int_0^{\infty} t\, y'' (t) \, e^{-\lambda \,t} {\text d} t = \left( t\, y' (t) \, e^{-\lambda \,t} \right)_{t=0}^{\infty} - \left[ y (t) \, \frac{\text d}{{\text d}t} \left( t\, e^{-\lambda \,t} \right) \right]_{t=0}^{\infty} + \int_0^{\infty} y(t) \, \frac{{\text d}^2}{{\text d}t} \left( t\, e^{-\lambda \,t} \right) ; \\ {\cal L} \left[ t\,y (t) \right] &= \int_0^{\infty} t\,y (t) \, e^{-\lambda \,t} {\text d} t = - \frac{\text d}{{\text d}\lambda}\, \int_0^{\infty} y (t) \,e^{-\lambda \,t} {\text d} t = - \frac{\text d}{{\text d}\lambda}\,{\cal L} \left[ y (t) \right] = - \frac{\text d}{{\text d}\lambda}\,y^L (\lambda ). \end{align*} The former requires more manipulations:
D[t*Exp[-s*t], t]
E^(-s t) - E^(-s t) s t
D[t*Exp[-s*t], t, t]
-2 E^(-s t) s + E^(-s t) s^2 t
${\cal L} \left[ t\, y'' (t) \right] = y(0) + \int_0^{\infty} t\,y (t) \left[ \lambda^2 t - 2 \lambda\right] e^{-\lambda \,t} {\text d} t .$
Therefore, we get
${\cal L} \left[ t\, y'' (t) \right] = y(0) - 2\lambda \, y^L (s) - \lambda^2 \frac{\text d}{{\text d}\lambda}\,y^L (s) .$
This leads to the differential equation of the first order for the Laplace transform of y(t)
$\left( 1 + \lambda^2 \right) \frac{\text d}{{\text d}\lambda}\,y^L (s) + \lambda \, y^L (s) = 0$
because we can set y(0) = 0. The general solution of the latter is known to be
${\cal L} \left[ y (t) \right] = y^L (\lambda ) = \frac{C}{\left( 1 + \lambda^2 \right)^{1/2}} ,$
where C is a n arbitrary constant that we set C = 1. Expanding the Laplace transform of the Bessel function into infinite series using binomial theorem $$\left( 1 + z \right)^{\alpha} = \sum_{k\ge 0} \binom{\alpha}{k} z^k ,$$ we get
${\cal L} \left[ J_0 (t) \right] = \frac{1}{\left( 1 + \lambda^2 \right)^{1/2}} = \frac{1}{\lambda} \left[ 1 - \frac{1}{2\lambda^2} + \frac{1 \cdot 3}{2!\, 2^2} \, \frac{1}{\lambda^4} - \frac{1 \cdot 3 \cdot 5}{3! \,2^3} \,\frac{1}{\lambda^6} + \cdots \right] .$
Then term-by-term integration yields
$J_0 (t) = 1 - \frac{(t/2)^2}{(1!)^2} + \frac{(t/2)^4}{(2!)^2} - \frac{(t/2)^6}{(3!)^2} + \cdots .$
This power series converges for all t because the signs of successive terms alternate and the general term turns to zero. Its Laplace transform is represented by uniformly convergent integral
$J_0^L (\lambda ) = \int_0^{\infty} J_0 (t)\, e^{-\lambda \,t} {\text d} t = \frac{1}{\left( 1 + \lambda^2 \right)^{1/2}} \qquad \Longrightarrow \qquad \int_0^{\infty} J_0 (t)\, {\text d} t = 1 .$
Taking the derivative, we have
$\frac{\text d}{{\text d}\lambda}\,J_0^L (\lambda ) = \int_0^{\infty} t\, J_0 (t) \, e^{-\lambda \,t} {\text d} t = - \frac{\lambda}{\left( 1 + \lambda^2 \right)^{3/2}}\qquad \Longrightarrow \qquad \int_0^{\infty} t\, J_0 (t) \, {\text d} t = 0 .$
From the Bessel equation $$t\, y'' (t) + y' (t) + t\, y (t) =0 ,$$ we derive
$\int_0^{\infty} t^2 J_0 (t) \, {\text d} t = -1 .$
Upon defining $$J_1 (t) = -J'_0 (t) ,$$ ,/span> we can show that this function is a solution of the Bessel equation of order 1: span class="math">$$t^2\, y'' (t) + t\,y' (t) + \left( t^2 -1 \right) y (t) =0 .$$ Its Laplace transform becomes
${\cal L} \left[ J_1 (t) \right] = {\cal L} \left[ - \frac{\text d}{{\text d}t} \,J_0 (t) \right] = - \left( \lambda\, J_0^L - J_0 (0) \right) = 1 - \frac{1}{\left( 1 + \lambda^2 \right)^{1/2}} .$

For arbitrary positive integer n, we obtain the Laplace transform of Jn from the corresponding transformation of Bessel's differential equation:

${\cal L} \left[ t\,\frac{\text d}{{\text d}t} \left( t\,\frac{{\text d} J_n (t)}{{\text d}t} \right) \right] + {\cal L} \left[ \left( t^2 - n^2 \right) J_n (t) \right] = 0 .$
We have
\begin{align*} {\cal L} \left[ t\,\frac{\text d}{{\text d}t} \left( t\,\frac{{\text d} J_n (t)}{{\text d}t} \right) \right] &= \frac{\text d}{{\text d}\lambda} \,\lambda {\cal L} \left[ t\,\frac{{\text d} J_n (t)}{{\text d}t} \right] = \frac{\text d}{{\text d}\lambda} \, \frac{\text d}{{\text d}\lambda} \,J_n^L (\lambda ) , \\ {\cal L} \left[ t^2 J_n (t) \right] &= \frac{{\text d}^2}{{\text d} \lambda^2} \, J_n^L (\lambda ) . \end{align*}
Therefore, application of the Laplace transform to the Bessel equation gives the differential equation for the Laplace transform of the Bessel function:
$\left( 1 + \lambda^2 \right) \frac{{\text d}^2}{{\text d} \lambda^2} \, J_n^L (\lambda ) + 3\lambda \, \frac{{\text d}}{{\text d} \lambda} \,J_n^L + J_n^L (\lambda ) = n^2 J_n^L (\lambda ) .$
Making substitution $$\phi (\lambda ) = \left( 1 + \lambda^2 \right)^{1/2} J_n^L (\lambda ) ,$$ we obtain
\begin{align*} \frac{{\text d}}{{\text d} \lambda} \,J_n^L &= \left( 1 + \lambda^2 \right)^{-1/2} \phi' - \lambda \left( 1 + \lambda^2 \right)^{-3/2} \phi (\lambda ) , \\ \frac{{\text d}^2}{{\text d} \lambda^2} \, J_n^L (\lambda ) &= \left( 1 + \lambda^2 \right)^{-1/2} \phi'' - 2\lambda \left( 1 + \lambda^2 \right)^{-3/2} \phi' (\lambda ) - \left( 1 + \lambda^2 \right)^{-3/2} \phi + 3\lambda^2 \left( 1 + \lambda^2 \right)^{-1/2} \phi . \end{align*}
Then for new dependent variable, we get the differential equation
$\left( 1 + \lambda^2 \right)^{1/2} \phi'' + \lambda \left( 1 + \lambda^2 \right)^{-1/2} \phi' = n^2 \left( 1 + \lambda^2 \right)^{1/2} \phi (\lambda ) ,$
which upon multiplication by an integrating factor $$\phi' \left( 1 + \lambda^2 \right)^{1/2} ,$$ results in the exact equation
$\left( 1 + \lambda^2 \right)^{1/2} \phi' \, \frac{\text d}{{\text d}\lambda} \left[ \phi' \left( 1 + \lambda^2 \right)^{1/2} \right] = n^2 \phi (\lambda ) \, \phi' (\lambda ) .$
Integration yields
$\left[ \phi' \left( 1 + \lambda^2 \right)^{1/2} \right]^2 = n^2 \left[ \phi (\lambda ) \right]^2 + C ,$
where C is a constant of integration. Setting C = 0, we have
$\phi' \left( 1 + \lambda^2 \right)^{1/2} = \pm n\, \phi (\lambda ) .$
Checking with knowing formula for J1, we have to choose minus sign and get
$\left( 1 + \lambda^2 \right)^{1/2} \frac{{\text d} \phi}{{\text d} \lambda} = -n\,\phi (\lambda ) \qquad\Longrightarrow \qquad \frac{{\text d} \phi}{\phi} = -n \left( 1 + \lambda^2 \right)^{-1/2} {\text d} \lambda ,$
with integral
$\phi (\lambda ) = C \left[ \left( 1 + \lambda^2 \right)^{1/2} - \lambda \right]^n .$
Now we obtain explicit expression for the Laplace transform of Bessel's function of first kind (upon setting C = 1):
${\cal L} \left[ J_n (t) \right] (\lambda ) = \frac{\left[ \left( 1 + \lambda^2 \right)^{1/2} - \lambda \right]^n}{\left( 1 + \lambda^2 \right)^{1/2}} , \qquad n=1,2,\ldots .$
For large λ, we have $$\displaystyle {\cal L} \left[ J_n (t) \right] (\lambda ) \sim \frac{1}{2^n \lambda^{n+1}} + O\left( \lambda^{-n-3} \right) , \] which leads to $J_n (t) \sim \frac{t^n}{2^n n!} + O \left( t^{n+2} \right) .$ Thus, Jn(0) = 0 (n = 1, 2, 3, ...), J'n(0) = 0 (n = 2, 3, ...), but J'1(0) = -1. Setting λ = 0 in the Laplace transform, we derive $\int_0^{\infty} J_n (t) \,{\text d}t = 1 \qquad \mbox{and} \qquad \int_0^{\infty} t^{-1} J_n (t) \,{\text d}t = \frac{1}{n} \qquad (n=0,1,2,\ldots ).$ From the identity $\frac{1}{\left( 1 + \lambda^2 \right)^{1/2} - \lambda} + \left( 1 + \lambda^2 \right)^{1/2} - \lambda = 2 \left( 1 + \lambda^2 \right)^{1/2} ,$ we obtain, upon on multiplication by \( \left[ \left( 1 + \lambda^2 \right)^{1/2} - \lambda \right]^n ,$$
$J_{n-1}^L (\lambda ) + J_{n+1}^L (\lambda ) = 2n {\cal L} \left[ J_n (t)/t \right]$
which, on inversion, gives the recurrence relation
$t \left[ J_{n+1} (t) + J_{n-1} (t) \right] = 2n\, J_n (t) .$
Similarly, the identity
$Similarly, the identity \[ \frac{1}{\left( 1 + \lambda^2 \right)^{1/2} - \lambda} - \left( 1 + \lambda^2 \right)^{1/2} - \lambda = 2 \lambda ,$
yields
$J_{n-1}^L - J_{n+1}^L = 2\lambda\,J_n^L \qquad\Longrightarrow \qquad J_{n-1} (t) - J_{n+1} (t) = 2\, J'_n (t) \qquad (n=1,2,3,\ldots ),$
because $$J_n (0) =0 \ (n=1,2,\ldots ).$$

Eliminating Jn+1(t) from the above recurrence relations, we obtain

$t\, J_{n-1} (t) = n\,J_n (t) + t\, J'_n (t) \qquad\Longrightarrow \qquad \frac{\text d}{{\text d}t} \left[ t^n J_n (t) \right] = t^n J_{n-1} (t) .$
On the other hand, elimination of Jn-1(t) yields
$\frac{\text d}{{\text d}t} \left[ \frac{J_n (t)}{t^n} \right] = - \frac{J_{n+1} (t)}{t^n} \qquad\mbox{or} \qquad \frac{1}{t}\, \frac{\text d}{{\text d}t} \left[ \frac{J_n (t)}{t^n} \right] = - \frac{J_{n+1} (t)}{t^{n+1}} .$
Application of the above relation m times gives
$\left[ \frac{1}{t}\,\frac{\text d}{{\text d}t} \right]^m \left[ \frac{J_n (t)}{t^n} \right] = (-1)^m \frac{J_{n+m} (t)}{t^{n+m}} .$
Putting n = 0 in the latter shows
$J_m (t) = (-1)^m \left[ \frac{1}{t}\,\frac{\text d}{{\text d}t} \right]^m J_0 (t) , \qquad m=1,2,\ldots .$
1. Chorlton, F., Studies in Bessel functions via Laplace transforms, International Journal of Mathematical Education in Science and Technology, 1998, Vol. 29, No. 3, pp. 437--473; doi: https://doi.org/10.1080/0020739980290312
2. Titchmarsh, E.C., The Theory of Functions, 1939, second edition, Oxford: Oxford University Press.