I am having hard time solving this integral: $\int_0 ^\infty J_n(bx)dx,$ where $J_n(x)$ is the $n$-th order Bessel function of the first kind.

$\textbf{My attempt:}$ We know the Bessel integral: $$J_n(x)=\frac{1}{\pi}\int_{0}^{\pi}\cos(n\phi-x\sin\phi)d\phi.$$ Taking the hint from the comments, by the change of variable $bx=t,$ we can write $$\int_{0}^{\infty}J_n(bx)dx=\frac{1}{b}\int_{0}^{\infty}J_n(x)dx$$

So, we can now solve for $\int_{0}^{\infty}J_n(x)dx,$ which is simpler.

So, we have that $$\int_{0}^{\infty}J_n(x)dx=\frac{1}{\pi}\int_{0}^{\infty}\int_{0}^{\pi}\cos(n\phi-x\sin\phi)d\phi dx$$ By change of order of integration, $$\int_{0}^{\infty}J_n(x)dx=\frac{1}{\pi}\int_{0}^{\pi}\int_{0}^{\infty}\cos(n\phi-x\sin\phi)dxd\phi $$ So, $$\int_{0}^{\infty}J_n(x)dx=\frac{1}{\pi}\int_{0}^{\pi}\bigg[\frac{\sin(n\phi-x\sin\phi)}{-\sin\phi}\bigg]_{x=0}^{x=\infty}d\phi $$ The problem is how to evaluate the inner function at the $\textbf{upper and lower limits}$. When I try to evaluate the upper limit $x=\infty$ in the inner function, I get $$\lim_{x \rightarrow \infty}\frac{\sin(n\phi-x\sin\phi)}{-\sin\phi},$$ the place from where I am not able to go ahead. I am troubled here.

To simplify I shall take a substituiton after the change of order of integration as follows. We have $$\int_{0}^{\infty}J_n(x)dx=\frac{1}{\pi}\int_{0}^{\pi}\int_{0}^{\infty}\cos(n\phi-x\sin\phi)dxd\phi .$$ Take $$I(\phi)=\int_{0}^{\infty}\cos(n\phi-x\sin\phi)dx.$$ Put $t=n\phi-x\sin\phi.$ So, we have that $dt=-\sin\phi dx.$ Also $x=0 \implies t=n\phi$ and $x\rightarrow \infty \implies t \rightarrow - \infty.$ So, we have that $$I(\phi)=\int_{n\phi}^{-\infty}\frac{\cos t}{-\sin\phi} dt$$ $$I(\phi)=-\frac{1}{\sin \phi}\big[\sin t \big]_{n\phi}^{-\infty}$$ Now that the final answer is already given by @gpmath, that $\int_{0}^{\infty}J_n(x)dx=1.$ Suppose that I evaluate $I(\phi),$ the problem would be easier. Please help.

I am bit bad at manipulating symbols and stuck here for some time now. Please help. Thanks.

  • $\begingroup$ Start letting $bx=t$ to make the problem more than simple. $\endgroup$ Jul 31, 2023 at 12:08
  • 1
    $\begingroup$ $$\int_0^{\infty } J_n(b x) dx = \begin{cases} -\frac{e^{i \pi n}}{b},& b<0\land n>-1\land n\in \mathbb{R} \\[2ex] +\frac{1}{b},& b>0\land n>-1\land n\in \mathbb{R} \end{cases}$$ $\endgroup$
    – gpmath
    Jul 31, 2023 at 18:01
  • $\begingroup$ What is your assumption on $b$? $\endgroup$
    – Gary
    Aug 1, 2023 at 0:07
  • $\begingroup$ Assume that $b>0$ and $n \geq0.$ $\endgroup$ Aug 1, 2023 at 2:23

1 Answer 1


The integral $\color{blue}{\int_0^\infty J_\nu(x)\,dx=1}$ (for $\Re\nu>-1$) can be obtained from $\int_0^\infty e^{-ax}J_\nu(x)\,dx$ by taking $a\to 0$ (which needs some — not too hard — justification).

In turn, a way to get the last integral (for $a>0$) is to use Schläfli’s integrals $$J_\nu(x)=\frac1{2\pi i}\int_\lambda\exp\frac{x}2\left(z-\frac1z\right)\frac{dz}{z^{\nu+1}}=\frac1{2\pi i}\int_\Lambda e^{x\sinh t-\nu t}\,dt,$$ where the contour $\lambda$ encircles the negative real axis in the $z$-plane, and the contour $\Lambda$ may consist of three straight lines (coming from $+\infty-\pi i$ to $-\pi i$, then to $+\pi i$, and finally to $+\infty+\pi i$).

The first of these integrals is obtained using Hankel's $\int_\lambda z^{-s}e^z\,dz=\frac{2\pi i}{\Gamma(s)}$, the exponential power series for $e^{-x/2z}$, and the definition of $J_\nu$ as a power series; the second one results from the substitution $z=e^t$. Note also that the Bessel integral you know is a special case of this second integral, when $\nu$ is an integer: the integrals along the infinite parts of $\Lambda$ cancel each other.

So, using the second integral, we get (after exchanging the integrations) $$\int_0^\infty e^{-ax}J_\nu(x)\,dx=\frac1{2\pi i}\int_\Lambda e^{-\nu t}\int_0^\infty e^{(\sinh t-a)x}\,dx\,dt=\frac1{2\pi i}\int_\Lambda\frac{e^{-\nu t}\,dt}{a-\sinh t},$$ which equals minus the residue of the integrand at $t=\sinh^{-1}a$ (this is seen using a standard argument: let $\Lambda_T$ be the boundary of $[0,T]+[-\pi,\pi]i$; then for large $T>0$, the $\frac1{2\pi i}\int_{\Lambda_T}$ equals the residue, and we have $\lim\limits_{T\to+\infty}\int_{\Lambda_T}=-\int_\Lambda$). Evaluating the residue the usual way, we get $$\int_0^\infty e^{-ax}J_\nu(x)\,dx=\left.\frac{e^{-\nu t}}{\cosh t}\right|_{t=\log(a+\sqrt{1+a^2})}=\frac{(\sqrt{1+a^2}-a)^\nu}{\sqrt{1+a^2}}.$$


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