Bessel to integral form

Question

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

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

Answer 1

$\newcommand{\dif}{\mathrm d}$ $$\begin{align*} \int_0^1 \frac{\cos (xt)}{\sqrt{1-t^2}} \dif t &= \int_0^1 \sum_{j=0}^\infty \frac{(-1)^j (xt)^{2j}}{(2j)!\sqrt{1-t^2}} \dif t \\ &= \sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}} \end{align*}$$ 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 $\mathrm B(x,y) \Gamma(x+y)=\Gamma(x) \Gamma(y)$, $$\begin{align*} \int_0^1 \frac{t^{2j}}{\sqrt{1-t^2}} \dif t &= \int_0^1 \frac{1}{2s^{1/2}}\frac{s^j}{\sqrt{1-s}} \dif s \\ &= \frac12 \int_0^1 s^{j-1/2} (1-s)^{-1/2} \dif s \\ &= \frac12\mathrm B(j+1/2,1/2) \\ &= \frac{\sqrt\pi \Gamma(j+1/2)}{2\Gamma(j+1)} \end{align*}$$

Answer 2

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.