What are BesselJ functions?

Solution 1:

$\qquad\qquad\qquad\qquad\qquad\qquad$ What are Bessel J functions ?

You are probably familiar with the fact that $e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}$, right ? Well then, Bessel functions

are basically what happens when we ask ourselves, “What is $~\displaystyle\sum_{n=0}^\infty\frac{x^n}{(n!)^2}~$ ?” But $(n!)^2=n!\cdot n!$,

so we then go a step further, by generalizing the question even more, and asking “What is

$\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!~(n+a)!}~$ ?” This is more or less how the Bessel I function is born. Then we ask

ourselves, “What would happen if the series were allowed to oscillate or alternate ?”, i.e.,

“What is $~\displaystyle\sum_{n=0}^\infty\frac{(-x)^n}{n!~(n+a)!}~$ ?” And this is how the Bessel J function comes into existence.

Very similar to how $e^{-x}=\displaystyle\sum_{n=0}^\infty\frac{(-x)^n}{n!}$, for example.

$$\begin{align} e^x&=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!} \\\\ \bigg(1-\frac{\Gamma(a,x)}{\Gamma(a)}\bigg)~e^x&=\displaystyle\sum_{n=0}^\infty\frac{x^{n+a}}{(n+a)!} \\\\ I_a(2x)&=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}~\frac{x^{n+a}}{(n+a)!} \\\\ J_a(2x)&=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}~\frac{x^{n+a}}{(n+a)!}~(-1)^n \\\\ L_a(2x)&=\displaystyle\sum_{n=0}^\infty\frac{x^{n+\frac12}}{\Big(n+\frac12\Big)!}~\frac{x^{n+\frac12+a}}{\Big(n+\frac12+a\Big)!} \\\\ H_a(2x)&=\displaystyle\sum_{n=0}^\infty\frac{x^{n+\frac12}}{\Big(n+\frac12\Big)!}~\frac{x^{n+\frac12+a}}{\Big(n+\frac12+a\Big)!}~(-1)^n \end{align}$$

See also Struve functions for more information. Speaking of which, notice that the last two

identities can be rewritten in the following $($non-standard, but rather intuitive$)$ manner :

$$\begin{align} L_a(2x)&=\displaystyle\sum_{n=\tfrac12}^\infty\frac{x^n}{n!}~\frac{x^{n+a}}{(n+a)!} \\\\ H_a(2x)&=\displaystyle\sum_{n=\tfrac12}^\infty\frac{x^n}{n!}~\frac{x^{n+a}}{(n+a)!}~(-1)^n \end{align}$$

Solution 2:

$~\quad~$ Bessel and Struve functions also appear in the following context: What happens when

we evaluate $($definite$)$ integrals of the form $\displaystyle\int_0^\lambda f\Big(g(x)\Big)~dx$, where $\big\{f,g\big\}\in\big\{\sin,~\sinh,$

$\cos,~\cosh\big\},~$ and $\lambda$ is either $\dfrac\pi2$ or $\infty$, depending on whether g is either a trigonometric or

a hyperbolic function. Thus, for $a>0$ we have the following identities:

$$ \int_0^\tfrac\pi2~\sin~\big(a~\sin x\big)~dx~=~\int_0^\tfrac\pi2~\sin~\big(a~\cos x\big)~dx~=~\frac\pi2~H_0(a) \\ \int_0^\tfrac\pi2 \sinh\big(a~\sin x\big)~dx~=~\int_0^\tfrac\pi2\sinh\big(a~\cos x\big)~dx~=~\frac\pi2~L_0(a) $$

$$ \int_0^\tfrac\pi2~\cos~\big(a~\sin x\big)~dx~=~\int_0^\tfrac\pi2~\cos~\big(a~\cos x\big)~dx~=~\frac\pi2~J_0(a) \\ \int_0^\tfrac\pi2 \cosh\big(a~\sin x\big)~dx~=~\int_0^\tfrac\pi2\cosh\big(a~\cos x\big)~dx~=~\frac\pi2~I_0(a) $$

$$\begin{align} \int_0^\infty\sin(a~\sinh x)~dx~&=~\quad\dfrac\pi2~\Big(I_0(a)-L_0(a)\Big) \\ \int_0^\infty\sin(a~\cosh x)~dx~&=~\quad\dfrac\pi2\cdot J_0(a) \\ \int_0^\infty\cos(a~\sinh x)~dx~&=~\quad~\quad~K_0(a) \\ \int_0^\infty\cos(a~\cosh x)~dx~&=~-\dfrac\pi2\cdot Y_0(a) \end{align}$$