6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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: $$\begin{eqnarray*} I &=& \frac{1}{a}\sum_{n\geq 0}\left(\frac{b}{a}\right)^{2n} \frac{1}{2\pi}\int_{0}^{2\pi}\sin^{2n}(t)\cos(b\cos t)\,dt \\&=&\frac{1}{a}\sum_{n\geq 0}\frac{(2n)!}{n!}\left(\frac{b}{2a^2}\right)^n J_n(b)\\&=&\frac{1}{a}\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!}\left(\frac{b}{2}\right)^m\sum_{n\geq 0}\left(\frac{b}{2a}\right)^{2n}\frac{(2n)!}{n!(n+m)!}\\&=&\frac{1}{\sqrt{a^2-b^2}}+\frac{1}{a}\int_{0}^{1}\sum_{m\geq 1}\frac{(-1)^m}{m!(m-1)!}\left(\frac{b}{2}\right)^m(1-u)^{m-1}\sum_{n\geq 0}\binom{2n}{n}\left(\frac{b^2}{4a^2}\right)^{n}u^n\,du\\&=& \frac{1}{\sqrt{a^2-b^2}}+\int_{0}^{1}\sum_{m\geq 1}\left(\frac{b}{2}\right)^m \frac{(-1)^m}{m!(m-1)!}\frac{(1-u)^{m-1}}{\sqrt{a^2-b^2 u}}\,du\\&=& \frac{1}{\sqrt{a^2-b^2}}-\sqrt{\frac{b}{2}}\int_{0}^{1}\frac{J_1(\sqrt{2b(1-u)})}{\sqrt{(a^2-b^2 u)(1-u)}}\,du\\&=&\frac{1}{\sqrt{a^2-b^2}}-\sqrt{2b}\int_{0}^{1}\frac{J_1(t\sqrt{2b})}{\sqrt{(a^2-b^2)+b^2 t^2}}\,dt.\tag{6}\end{eqnarray*}$$

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – tired Feb 24 '15 at 10:06
  • $\begingroup$ 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 $J_0(x) = \frac{2}{\pi} \int_0^\infty \sin\left(x\cosh(t)\right) \mathrm{d}t$. $\endgroup$ – Firman Feb 24 '15 at 10:21

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.