# Integral of sine multiplied by Bessel function with complicated argument

I need a help with integral below, $$\int_0^\infty \sin(ax)\ J_0\left(b\sqrt{1+x^2}\right)\ \mathrm{d}x,$$ where $a,b > 0$ and real, $J_0(x)$ is the zeroth-order of Bessel function of the first kind.

I found some integrals similar to the integral above, but I don't have any idea on how to apply it. Here are some integrals that might help. $$\int_0^\infty \cos(ax)\ J_0\left(b\sqrt{1+x^2}\right)\ \mathrm{d}x = \frac{\cos\sqrt{b^2-a^2}}{\sqrt{b^2-a^2}}; \mathrm{~~for~0 < a < b}$$

$$\int_0^\infty \sin(ax)\ J_0(bx)\ \mathrm{d}x = \frac{1}{\sqrt{a^2-b^2}}; \mathrm{~~for~0 < b < a}$$

The proof of the first integral can be seen here.

By exploiting the integral representation for the $J_0$ function: $$J_0(z) = \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\cos t}\,dt \tag{1}$$ we have: $$\int_{0}^{+\infty}\sin(ax)\, J_0(b\sqrt{1+x^2})\,dx =\frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{2\pi}\sin(ax)e^{ib\sqrt{x^2+1}\cos t}\,dt\,dx \tag{2}$$ where: $$\sqrt{x^2+1}\cos t = \cos(t-t_0)-x\sin(t-t_0),\qquad x=\tan t_0 \tag{3}$$ so: $$\begin{eqnarray*} I &=& \frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{2\pi}\sin(ax)\,e^{ib\cos t-ibx\sin t}\,dt\,dx\\&=&\frac{1}{2\pi}\int_{0}^{2\pi}\frac{a}{a^2-b^2\sin^2 t}e^{ib\cos t}\,dt\\&=&\frac{1}{2\pi}\int_{0}^{2\pi}\frac{a}{a^2-b^2\sin^2 t}\,\cos(b\cos t)\,dt\tag{4} \end{eqnarray*}$$ but since: $$\frac{1}{2\pi}\int_{0}^{2\pi}\sin^{2n}(t)\cos(b\cos t)\,dt = \frac{ (2n-1)!!}{b^n}\, J_n(b) \tag{5}$$ we have: $$\begin{eqnarray*} I &=& \frac{1}{a}\sum_{n\geq 0}\left(\frac{b}{a}\right)^{2n} \frac{1}{2\pi}\int_{0}^{2\pi}\sin^{2n}(t)\cos(b\cos t)\,dt \\&=&\frac{1}{a}\sum_{n\geq 0}\frac{(2n)!}{n!}\left(\frac{b}{2a^2}\right)^n J_n(b)\\&=&\frac{1}{a}\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!}\left(\frac{b}{2}\right)^m\sum_{n\geq 0}\left(\frac{b}{2a}\right)^{2n}\frac{(2n)!}{n!(n+m)!}\\&=&\frac{1}{\sqrt{a^2-b^2}}+\frac{1}{a}\int_{0}^{1}\sum_{m\geq 1}\frac{(-1)^m}{m!(m-1)!}\left(\frac{b}{2}\right)^m(1-u)^{m-1}\sum_{n\geq 0}\binom{2n}{n}\left(\frac{b^2}{4a^2}\right)^{n}u^n\,du\\&=& \frac{1}{\sqrt{a^2-b^2}}+\int_{0}^{1}\sum_{m\geq 1}\left(\frac{b}{2}\right)^m \frac{(-1)^m}{m!(m-1)!}\frac{(1-u)^{m-1}}{\sqrt{a^2-b^2 u}}\,du\\&=& \frac{1}{\sqrt{a^2-b^2}}-\sqrt{\frac{b}{2}}\int_{0}^{1}\frac{J_1(\sqrt{2b(1-u)})}{\sqrt{(a^2-b^2 u)(1-u)}}\,du\\&=&\frac{1}{\sqrt{a^2-b^2}}-\sqrt{2b}\int_{0}^{1}\frac{J_1(t\sqrt{2b})}{\sqrt{(a^2-b^2)+b^2 t^2}}\,dt.\tag{6}\end{eqnarray*}$$
• i'm wondering why this result is so more difficult then in the case where $sin$ is replaced by $cos$. The only thing i can think about is the missing inversion symmetry in the present case.... (+1) – tired Feb 24 '15 at 10:06
• Yes, the difference between sin and cos for this case is the symmetry (even and odd function). That's why I included the second integral in my question. The proof of the second integral follows similar steps as the first integral, except that we use $J_0(x) = \frac{2}{\pi} \int_0^\infty \sin\left(x\cosh(t)\right) \mathrm{d}t$. – Firman Feb 24 '15 at 10:21