# Definite Integral of bessel function of first kind of order one.

How to prove $\int\limits_0^\infty J_1(x)~dx=1$ ?

I got $\int\limits_0^\infty J_1(x)dx=-[J_0(x)]_0^\infty$ .

Please help.

## 2 Answers

Using the generating function of the Bessel functions, sometimes designated as G(x,t): \large e^{\frac{x}{2}[t-t^{-1}]}=\sum_{n=-\infty}^{\infty}J_n(x)t^n\\[3em] \xrightarrow{x=0} 1 = \sum_{n=-\infty}^{\infty}J_n(0)t^n = \begin{align}&J_0(0) + \\&J_1(0) t + J_2(0) t^2 + ...\\&J_{-1}(0)t^{-1} + J_{-2}(0)t^{-2}+ ...\end{align}\\ now by equating coefficients we get: $$J_0(0)=1\; \wedge \;J_{n\ne 0}(0)=0\\ \rightarrow J_0(\infty)=0$$

$\because\lim\limits_{x\to\infty}J_0(x)=0$

$\therefore\int\limits_0^\infty J_1(x)dx=-[J_0(x)]_0^\infty=1$

• I couldn't prove $\lim\limits_{x\to\infty}(x)=0$ – Susmita Agarwal Apr 2 '14 at 4:47