# Bessel integral

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.

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

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 $$\int_0^\infty e^{-ax}J_\nu(x)\,dx=\left.\frac{e^{-\nu t}}{\cosh t}\right|_{t=\log(a+\sqrt{1+a^2})}=\frac{(\sqrt{1+a^2}-a)^\nu}{\sqrt{1+a^2}}.$$