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Consider the following function that is defined by an infinite integral $$ I(r,t,a) = \int_0^\infty e^{-\lambda a} J_0(\lambda r) \sin(\lambda t) \, \mathrm{d}\lambda \, , $$ where $r, t,$ and $a$ are positive real numbers. Here, $J_0$ denotes the zeroth order Bessel function of the first kind.

Clearly, the integral above in convergent. i was wondering whether an explicit analytical expression for $I$ can be obtained analytically.

Any help or suggestion is highly appreciated.

Thank you

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    $\begingroup$ Please review the integral parameters (it should be $d\lambda$ I guess). Also what is the definition of $J_0$ ? $\endgroup$ – Zacky Oct 21 '19 at 11:00
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    $\begingroup$ I'm aware that it's Bessel function. What i mean iss what definition are you using, or anything can be used? $\endgroup$ – Zacky Oct 21 '19 at 11:59
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    $\begingroup$ I should have asked directly if one can use any integral representation, such as $J_0(x)=\frac{1}{\pi}\int_0^\pi \cos(x\cos z) dz=\int_{-\infty}^\infty \sin(x\cosh z)dz$. $\endgroup$ – Zacky Oct 21 '19 at 12:48
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    $\begingroup$ For $I_0$ write $\sin (\lambda t)=\Im e^{i\lambda t}$, $\Im$ denoting the imaginary part, then using the laplace transform of the Bessel function: $\int_0^\infty e^{-ax} J_0 (bx)dx=\frac{1}{\sqrt{a^2+b^2}}$ we obtain $I_0=\Im \frac{1}{\sqrt{r^2+(a-it)^2}}$. For $I_{-1}$ integrate once the above with respect to a. $\endgroup$ – Zacky Oct 21 '19 at 12:50
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    $\begingroup$ I'm not sure how to integrate that imaginary part, so maybe someone else can show it. $\endgroup$ – Zacky Oct 21 '19 at 13:51
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It's easier to start with Bessel's Integral and apply residue theorem. For simplicity make a change of variable to ensure $r = 1$. Then, we have

\begin{align*}I = \int_0^\infty e^{-(a-it)x} J_0(x)\,dx &= \frac{1}{2\pi}\int_0^{\infty} \int_{-\pi}^{\pi} e^{-(a-it)x} e^{ix\sin \tau}\,d\tau\,dx \\&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \int_0^{\infty} e^{-(a-i(t+\sin \tau))x} \,dx \,d\tau \\&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1}{a-i(t+\sin \tau)} \,d\tau \\&= \frac{1}{2\pi} \int_{|z| = 1} \frac{1}{a-i\left(t+\frac{1}{2i}\left(z - \frac{1}{z}\right)\right)} \,\frac{dz}{iz}\tag{$\ast$}\\&= \frac{i}{\pi} \int_{|z| = 1} \frac{1}{z^2 - 2(a-it)z - 1} \,dz\end{align*}

where, in step $(\ast)$ we made the change of variable $z = e^{i\tau}$.

The function has two simple poles at $\alpha^{\pm} = a-it \pm \sqrt{(a-it)^2 + 1}$ (where we are using the principle branch of square root since $a > 0$) and only $\alpha^{-}$ is inside the unit disk. Therefore, the integral is $$I = \frac{i}{\pi}\frac{2\pi i}{\alpha^- - \alpha^+} = \frac{1}{\sqrt{(a-it)^2 + 1}} = \frac{1}{\sqrt{a^2 - t^2 + 1 - 2iat}}.$$

Now, comparing imaginary parts on both sides should give the result.

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    $\begingroup$ $-1/\pi$ in front of the last integral should be $i/\pi$. $\endgroup$ – Maxim Oct 24 '19 at 13:40
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    $\begingroup$ Thanks .. You are right! Will fix it soon as I get near a pc. $\endgroup$ – r9m Oct 24 '19 at 13:48

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