I need advice on how to solve the following integral:

$$\int_0^\infty J_0(bx) \sin(ax) dx$$

I've seen it referenced, e.g. here on MathSE, so I know the solution is $(a^2-b^2)^{-1/2}$ for $a>b$ and $0$ for $b>a$, but I don't know how to get there.

I have tried to solve it by using the integral representation of the Bessel function and switching the integrals, resulting in $$ \frac{1}{\pi}\int_0^\pi \int_0^\infty \sin(ax)\cos(bx\sin(\theta))dx d\theta. $$ Doing the dx-integration, I get $$ =\frac{1}{\pi}\int_0^\pi \frac{2a}{a^2-b^2\sin^2(\theta)}\left(1-\lim_{x\to\infty}\cos(ax)\cos(bx\sin(\theta)\right)d\theta $$ and have no idea how to proceed from there.

Is there anything wrong with my calculations? Should I use a totally different approach? Any help appreciated.

  • $\begingroup$ What is referenced in the link is not the same as what you mentioned here. :) $\endgroup$ – Hosein Rahnama Jan 7 '16 at 19:01

I need advice on how to solve the following integral

Then you shall receive it ! ;-$)$

Use Euler's formula in conjunction with the series expansion of the Bessel function. This will

require switching the order of summation and integration, and recognizing the binomial series

of $~\dfrac1{\sqrt{b^2\color{red}+c^2}}~=~\displaystyle\int_0^\infty J_0(bx)~e^{cx}~dx,~$ where the latter converges for $~c<0.~$ Now let

$c=\epsilon+ia,~$ where $\epsilon\to0.\quad$ :-$)$


We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \, b >0 )\tag{1}. $$

(See this question for a derivation of $(1)$ using contour integration.)

First let $s=p+ia$, where $p,a >0$.

A slight modification of the answer here shows that $\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx $ converges uniformly for all $p \in [0, \infty$).

This allows us to conclude that $$\begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \lim_{p \downarrow 0}\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx \\ &= \lim_{p \downarrow 0} \frac{\left(\sqrt{(-p+ia)^2+b^{2}}-p-ia\right)^{n}}{b^{n}\sqrt{(p+ia)^2+b^{2}}} \\ &= \frac{\left(\sqrt{b^{2}-a^{2}}-ia\right)^{n}}{b^{n}\sqrt{b^{2}-a^{2}}}. \end{align}$$

So if $ a < b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(\sqrt{b^{2}-a^{2}+a^{2}} e^{-i \arcsin \left(\frac{a}{b}\right)}\right)^{n}}{b^{n} \sqrt{b^{2}-a^{2}}} \\ &= \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}} .\end{align}$$

And if $a >b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(i\sqrt{a^{2}-b^{2}}-ia \right)^{n}}{b^{n}i \sqrt{a^{2}-b^{2}}} \\ &= \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}. \end{align}$$


$$\int_{0}^{\infty} J_{n}(bx) \sin(ax) \, dx = \begin{cases} \frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \, & \quad 0 < a < b \\ \frac{\cos \left(\frac{\pi n}{2} \right) \left(\sqrt{a^{2}-b^{2}} -a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}} & \quad a > b >0 \end{cases} $$


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.