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What is the closed-form of this integral? $$\int_0^\infty{J_0(\lambda a)\sin(\lambda a)d\lambda}$$where $J_0 $ is the Bessel function of the first kind of order zero, and $a$ is real. Based on the flow of the paper I am reading, this integral should have a closed form expression, but I couldn't find it everywhere I searched for.

Edit: Additional information about the origin of this integral, and why I think it should converge to some value:

I am working with the following function, written in cylindrical coordinate:$$\mathrm V(r,z)=-\frac{c}{2\pi a\sigma}\int_0^\infty\frac{\mathrm u(z, \lambda)}{\mathrm u'(0, \lambda)} \mathrm J_0(\lambda r)\sin(\lambda a)d\lambda$$, where $\mathrm u$ is of exponential form and monotonically decreasing as a function of $z$; and $a,\sigma$ are constants.

The paper says that this function has been constructed so that:$$\frac{\partial V}{\partial z}|_{z=0}=\begin{cases} 0, & \text{if $r>a$} \\ \ne 0, & \text{if $r\leq a$} \end{cases}$$

I couldn't understand why this is the case, given this construction. Here is my attempt: taking the derivative of $\mathrm V$ with respect to $z$:$$\frac{\partial V}{\partial z}=-\frac{c}{2\pi a\sigma}\int_0^\infty\frac{\mathrm u'(z, \lambda)}{\mathrm u'(0, \lambda)} \mathrm J_0(\lambda r)\sin(\lambda a)d\lambda$$

At $z=0$, this reduces to:$$\frac{\partial V}{\partial z}|_{z=0}=-\frac{c}{2\pi a\sigma}\int_0^\infty\mathrm J_0(\lambda r)\sin(\lambda a)d\lambda$$From here, I don't know what else to do to see why $\mathrm V$ satisfies the above conditions.

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  • $\begingroup$ Any conditions on the value of $a$? $\endgroup$ – Chappers Jan 22 '18 at 3:39
  • $\begingroup$ I don't think there is any condition on $a$. However, I added more information. Please have a look at the question again. Thank you. $\endgroup$ – A Slow Learner Jan 22 '18 at 5:46
  • $\begingroup$ @Geophysics $r,a$ in your equation corresponds to $a, b$ in my answer. If $r > a$ then the formula I quoted from $DLMF$ does say the integral is $0$. $\endgroup$ – achille hui Jan 22 '18 at 8:48
  • $\begingroup$ The $a$-parameter is irrelevant since it can be removed through a substitution and the integral $\int_{0}^{+\infty}J_0(x)\sin(x)\,dx$ is not convergent since $J_0(x)\approx \frac{\cos(x)+\sin(x)}{\sqrt{\pi x}}$ for large values of $x$. Indeed $$\int_{0}^{+\infty}\frac{\sin^2(x)}{\sqrt{x}}\,dx $$ is divergent. $\endgroup$ – Jack D'Aurizio Jan 22 '18 at 12:31
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Quoting DLMF 10.22.59, When $\Re \mu > -1$,

$$\int_0^\infty e^{ibt}J_\mu(at) dt = \begin{cases} \frac{\exp(i\mu \sin^{-1}(b/a))}{\sqrt{a^2-b^2}},& 0 \le b < a\\ \frac{ia^\mu \exp\left(\frac12\mu\pi i\right)}{\sqrt{b^2-a^2}\left(b + \sqrt{b^2-a^2}\right)^\mu}, & 0 < a < b\end{cases}$$

Setting $\mu$ to $0$ and taking imaginary part, one get $$\int_0^\infty J_0(at)\sin(bt)dt = \begin{cases} 0,& 0 \le b < a\\ \frac{1}{\sqrt{b^2-a^2}},& b > a > 0 \end{cases}$$

Your integral corresponds to the case $a = b$. As one can see, the integral tends to different value ( $0$ and $\infty$ ) as $b$ approaches $a$ from the left and from the right. There is a big chance your integral diverges.

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  • $\begingroup$ I added more information. Please help me. Thank you. $\endgroup$ – A Slow Learner Jan 22 '18 at 5:45
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For any $\lambda>0$

$$ \int_{0}^{+\infty}J_0(x)\,\sin(x)\,e^{-\lambda x}\,dx = \frac{2}{\pi}\int_{0}^{+\infty}\int_{0}^{\pi/2}\cos(x\cos u)\sin(x)\,e^{-\lambda x}\,du\,dx$$ can be written (through Fubini's theorem) as $$ \frac{2}{\pi}\int_{0}^{\pi/2}\frac{1+\lambda^2-\cos^2 u}{(1+\lambda^2+\cos^2 u)^2-4\cos^2 u}\,du \\= \frac{2}{\pi}\int_{0}^{1}\frac{t^2+\lambda^2 \left(1+t^2\right)}{t^4+\lambda^4 \left(1+t^2\right)^2+2 \lambda^2 \left(2+3 t^2+t^4\right)}\,dt $$ and as $\lambda\to 0^+$, the last integrand function tends to $\frac{1}{t^2}\not\in L^1(0,1)$.
It follows that the original integral is simply divergent.

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