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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.

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  • $\begingroup$ What is referenced in the link is not the same as what you mentioned here. :) $\endgroup$ – H. R. Jan 7 '16 at 19:01
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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 $\epsilon\to0.\quad$ :-$)$

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

Therefore,

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

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