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This integral is from (6.616.2) in Gradshteyn and Ryzhik. $$ \int_1^\infty e^{-\alpha x}J_0(\beta\sqrt{x^2-1})\mathrm{d}x \,=\frac{1}{\sqrt{\alpha^2+\beta^2}}e^{-\sqrt{\alpha^2+\beta^2}} $$

I want to know how to do this integral and the restriction of $\alpha$ and $\beta$. The integral table doesn't mention it.

I doubt the results must have some restriction on $\alpha$ and $\beta$, because:

  1. it seems that when $\mathrm{Re}\,\alpha <0$, the integral diverges.

  2. also when $\alpha$ is purely imaginary, and for $\beta$ real, the result should be complex conjugate when $\alpha$ takes conjugate purely imaginary pairs, $\pm i$ for example, however the results given depends only on $\alpha^2$.

  3. I also found by Mathematica numerical integration that when $\alpha$ is purely imaginary, the integration also seems troublesome.

Edit:

The answer by @Fabian give a general condition that when $\mathrm{Re}\alpha>\mathrm{Im}\beta$, the integral converges. However, what about $\mathrm{Re}\alpha=\mathrm{Im}\beta$. For the simplest case when $\alpha$ is purely imaginary and $\beta$ is real, Mathematica can give the sensible result when $|\alpha|>|\beta|$, while seems diverges when $|\alpha|<|\beta|$:

\[Alpha]=-3I ;\[Beta]=2;
NIntegrate[Exp[-\[Alpha] x]BesselJ[0,\[Beta] Sqrt[x^2-1]],{x,1,Infinity}]
f[a_,b_]:=Exp[-Sqrt[a^2+b^2]]/Sqrt[a^2+b^2]
f[\[Alpha],\[Beta]]//N
-0.351845-0.276053 I
-0.351845+0.276053 I
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  • $\begingroup$ @havakok fixed. $\endgroup$ – an offer can't refuse Aug 30 '16 at 9:34
  • $\begingroup$ Are you a physicist? $\endgroup$ – Santiago Aug 30 '16 at 9:38
  • $\begingroup$ @Santiago yeah. $\endgroup$ – an offer can't refuse Aug 30 '16 at 9:41
  • $\begingroup$ The left hand side is analytical in $\alpha$ and $\beta$. So is the right hand side. So the result holds (by analytical continuation) as long as the integral on the left converges, e.g., for $\text{Re} \alpha >0$ and $\beta \in \mathbb{R}$. $\endgroup$ – Fabian Aug 30 '16 at 12:05
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The two expressions are equal by analytical continuation whenever the left hand side exists. The only problem for the convergence of the integral is at $x\to \infty$. We have the asymptotic expansion $(|\arg z| < \pi)$ $$ J_0(z) \sim \sqrt{\frac{2}{\pi z}} \cos(z-\pi/4).$$

Thus, for $x\to \infty$, we have that the integrand behaves as $$ e^{-\alpha x}J_0(\beta\sqrt{x^2-1}) \sim e^{-\alpha x} \sqrt{\frac{2}{\pi \beta x}} \cos(\beta x-\pi/4). $$

Let us investigate first the case $\mathop{\rm Im} \beta>0$, we then have that $$ e^{-\alpha x}J_0(\beta\sqrt{x^2-1}) \sim \sqrt{\frac{i}{2\pi \beta x}} e^{-(\alpha+i \beta) x};$$ the integral thus converges for $\mathop{\rm Re}\alpha >\mathop{\rm Im} \beta$.

The case $\mathop{\rm Im} \beta<0$ can be treated similarly with the result that the integral converges whenever $$ \mathop{\rm Re} \alpha >|\mathop{\rm Im} \beta|.\tag{1}$$ The case $\mathop{\rm Re} \alpha =|\mathop{\rm Im} \beta|$ needs special attention as on this line the convergence is conditional.

Regarding your questions:

1) is incorrect, see (1) above. $\text{ }$

2) this is exactly covered by analytical continuation. If you continue $\alpha$ from the real line to the upper imaginary axis, then you obtain one branch of the square root. Continuing it to the lower part of the imaginary axis you get the other branch. In formula, we have that $$ \sqrt{\alpha^2 + \beta^2} \stackrel{\alpha \to\pm i a}\mapsto \pm i \sqrt{a^2-\beta^2}.$$

3) when $\alpha=i a$ is purely imaginary then the integral is troublesome as it is oscillating very fast. It is normal that in this case numerical routines run into troubles. So you should rely in this case to analytical continuation instead.

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  • $\begingroup$ Could you please elaborate your second point to my question, either in this answer or in this post: math.stackexchange.com/questions/1912132/… . $\endgroup$ – an offer can't refuse Sep 3 '16 at 1:33
  • $\begingroup$ And please elaborate how the convergence is conditional... How to judge whether it is convergent or not in the case when $\text{Re}\alpha=\text{Im}\beta$ $\endgroup$ – an offer can't refuse Sep 3 '16 at 1:35
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For the purpose of evaluation, I'm just going to consider the case where $\alpha>0$ and $\beta \in \mathbb{R}$.

From the answers here, we know that $$\int_{0}^{\infty} \exp \left(-ax^{2}-\frac{b}{x^{2}} \right) \, dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} e^{-2 \sqrt{ab}}, \quad a, b>0. \tag{1}$$

Using this formula, we get $$ \begin{align} \int_{1}^{\infty} e^{- \alpha x} J_{0}(\beta \sqrt{x^{2}-1}) \, dx &= \int_{0}^{\infty} \frac{u e^{-\alpha \sqrt{1+u^{2}}} J_{0}(\beta u)}{\sqrt{1+u^{2}}} \, du \\ &= \frac{2}{\sqrt{\pi}}\int_{0}^{\infty} u J_{0}(\beta u) \int_{0}^{\infty}\exp \left(-(1+u^{2})t^{2} -\frac{\alpha^{2}}{4t^{2}}\right) \, dt \, du \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \exp \left(- t^{2}- \frac{\alpha^{2}}{4t^{2}} \right)\int_{0}^{\infty} uJ_{0}(\beta u) \exp \left(-t^{2}u^{2} \right) \, du \, dt, \end{align}$$

where

$$ \begin{align}\int_{0}^{\infty} u J_{0}(\beta u) e^{-t^{2}u^{2}} \, du &= \frac{1}{2}\int_{0}^{\infty} J_{0}(\beta \sqrt{w}) e^{-t^{2}w} \, dw \\ &= \frac{1}{2} \int_{0}^{\infty} e^{-t^{2}w} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k!)^{2}} \left(\frac{\beta^{2}w}{4} \right)^{k} \, dw\\ &= \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k!)^{2}} \left(\frac{\beta^{2}}{4} \right)^{k} \int_{0}^{\infty} e^{-t^{2}w} w^{k} \, dw\\ &= \frac{1}{2t^{2}} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \left(\frac{\beta^{2}}{4t^{2}} \right)^{k} \\ &= \frac{1}{2t^{2}} \, \exp \left(-\frac{\beta^{2}}{4t^{2}} \right). \end{align} $$

Therefore, using $(1)$ again, we get $$\begin{align} \int_{1}^{\infty} e^{- \alpha x} J_{0}(\beta \sqrt{x^{2}-1}) \, dx &= \frac{2 }{\sqrt{\pi}} \int_{0}^{\infty} \exp \left(- t^{2}- \frac{\alpha^{2}}{4t^{2}} \right) \frac{1}{2t^{2}} \exp \left(- \frac{\beta^{2}}{4t^{2}}\right) \, dt \\ &=\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{1}{t^{2}} \, \exp \left(-t^{2} - \frac{\alpha^{2}+\beta^{2}}{4} \frac{1}{t^{2}}\right) \, dt \\ &= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \exp \left( - \frac{1}{v^{2}}-\frac{\alpha^{2}+\beta^{2}}{4} v^{2} \right) \, dv \\ &= \frac{1}{\sqrt{\pi}} \frac{1}{2} \sqrt{\frac{4 \pi}{\alpha^{2}+\beta^{2}}} \, e^{-\sqrt{\alpha^{2}+\beta^{2}}} \\&= \frac{1}{\sqrt{\alpha^{2}+\beta^{2}}} \, e^{-\sqrt{\alpha^{2}+\beta^{2}}}. \end{align} $$


A generalization of this integral, namely,

$$\small \int_{0}^{\infty} J_{\mu}(b t) \, \frac{K_{\nu}(a\sqrt{t^{2}+z^{2}})}{(t^{2}+z^{2})^{\nu/2}} \, t^{\mu+1} \, dt = \frac{b^{\mu}}{a^{\nu}} \left(\frac{\sqrt{a^{2}+b^{2}}}{z} \right)^{\nu-\mu-1}K_{\nu-\mu-1} \left(z \sqrt{a^{2}+b^{2}} \right) $$

is the second entry in section 47 of chapter 13 of the textbook A Treatise on the Theory of Bessel Functions.

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