2
$\begingroup$

Can someone tell me how the Bessel function be this form.

$$\frac{2}{π}\int_0^1 \frac{\cos (xt)} {\sqrt{1-t^2}} dt = J_0(x)$$

$\endgroup$
  • 2
    $\begingroup$ How do you want to define $J_0(x)$? $\endgroup$ – Robert Israel Feb 9 '17 at 2:51
1
$\begingroup$

$\newcommand{\dif}{\mathrm d}$ $$\begin{align*} \int_0^1 \frac{\cos (xt)}{\sqrt{1-t^2}} \dif t &= \int_0^1 \sum_{j=0}^\infty \frac{(-1)^j (xt)^{2j}}{(2j)!\sqrt{1-t^2}} \dif t \\ &= \sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}} \end{align*}$$ Observe that (this part needs some justification) by some identities of the $\Gamma$ function, $$ \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}} = \frac{\sqrt\pi \Gamma(j+1/2)}{2\Gamma(j+1)} = \frac{\sqrt\pi}{j!}\frac{(2j)!\sqrt\pi}{4^j j! } = \pi\frac{(2j)!}{2 \cdot 4^j (j!)^2} $$ Hence $$\sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}} = \frac{\pi}{2}\sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \frac{(2j)!}{2 \cdot 4^j (j!)^2} = \frac{\pi}{2} \sum_{j=0}^\infty \frac{(-1)^j}{(j!)^2} \left( \frac x 2 \right)^{2j} = \frac{\pi}{2} \mathrm J_0(x) $$ A good exercise would be to use the completeness of $\mathrm L^{\!1}$ to justify the switch of the order of summation.


Justification of the integral $ \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}}$:

Via the diffeomorphism $s \mapsto \sqrt s$, and the $\mathrm B$ identity $\mathrm B(x,y) \Gamma(x+y)=\Gamma(x) \Gamma(y)$, $$\begin{align*} \int_0^1 \frac{t^{2j}}{\sqrt{1-t^2}} \dif t &= \int_0^1 \frac{1}{2s^{1/2}}\frac{s^j}{\sqrt{1-s}} \dif s \\ &= \frac12 \int_0^1 s^{j-1/2} (1-s)^{-1/2} \dif s \\ &= \frac12\mathrm B(j+1/2,1/2) \\ &= \frac{\sqrt\pi \Gamma(j+1/2)}{2\Gamma(j+1)} \end{align*}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Starting with the well-known integral representation for the Bessel function $J_n (x)$ of $$J_n (x) = \frac{1}{\pi} \int^\pi_0 \cos (n \theta - x\sin \theta) \, d \theta,$$ where $n$ is an integer, for the zeroth-order Bessel function we set $n = 0$ giving $$J_0 (x) = \frac{1}{\pi} \int^\pi_0 \cos (x \sin \theta) \, d \theta.$$

If we let $\theta = \pi/2 - u$ our integral for $J_0 (x)$ becomes $$J_0 (x) = \frac{1}{\pi} \int^{\pi/2}_{-\pi/2} \cos (x \cos u) \, du.$$ Since the integrand is an even function between symmetric limits we can write this as $$J_0 (x) = \frac{2}{\pi} \int^{\pi/2}_0 \cos (x \cos u) \, du.$$

Now setting $t = \cos u, dt = - \sin u \, du$. Thus $du = -\dfrac{dt}{\sqrt{1 - t^2}},$ while for the limits of integration we have $(0,\pi/2) \mapsto (1,0)$. Hence $$J_0 (x) = \frac{2}{\pi} \int^1_0 \frac{\cos (xt)}{\sqrt{1 - t^2}} \, dt,$$ as required.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.