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I need some help with solving this integral involving Bessel function:

$\hspace{2in}\displaystyle\int_0^\infty$$J_0(x)\ $$\text{sinc}(\pi\,x)\ $$e^{-x}\,\mathrm dx.$

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Prudnikov-Brychkov-Marychev (Vol. II, formula 2.12.8.5) tells us that $$\int_0^{\infty}\frac{1-e^{-px}}{x}J_0(x)dx=\ln\left(p+\sqrt{p^2+1}\right)$$ Taking the difference of this formula with $p=1-i\pi$ and $p=1+i\pi$, one obtains \begin{align}\int_0^{\infty}J_0(x)\mathrm{sinc}(\pi x)e^{-x}dx&=\frac{1}{2\pi i}\ln\frac{1+i\pi+\sqrt{(1+i\pi)^2+1}}{1-i\pi+\sqrt{(1-i\pi)^2+1}}=\\ &=\frac{1}{\pi}\arctan\frac{\pi\sqrt{2}+\sqrt{\pi^2-2+\sqrt{\pi^4+4}}}{ \sqrt{2}+\sqrt{2-\pi^2+\sqrt{\pi^4+4}}}. \end{align}

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    $\begingroup$ Amazing answer! $\endgroup$ – nbubis May 20 '13 at 20:42
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    $\begingroup$ @nbubis: I wouldn't say so. One should just know the right book. $\endgroup$ – Start wearing purple May 20 '13 at 20:45
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    $\begingroup$ Great answer! It is a shame that Mathematica cannot solve it, especially when they claim "It can do almost any integral that can be done in terms of standard mathematical functions." reference.wolfram.com/mathematica/howto/DoAnIntegral.html $\endgroup$ – Zakharia Stanley May 21 '13 at 3:01
  • $\begingroup$ @Zakharia, you did notice the word "almost", no? ;) $\endgroup$ – J. M. is a poor mathematician May 24 '13 at 16:23
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O.L.'s answer is absolutely correct. I just want to generalize the formula a little: $$\forall a>0,\hspace{5mm}\int_0^\infty J_0(x)\ \text{sinc}(a\,x)\ e^{-x}\,\mathrm dx=\frac1a\arctan\sqrt{\frac{2}{\sqrt{a^4+4}-a^2}-1}.$$

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Here's an evaluation of the integral used by O.L..

$$\begin{align} \int_{0}^{\infty} \frac{1-e^{-px}}{x} J_{0}(x) \ dx &= \int_{0}^{\infty} \int_{0}^{\infty} (1-e^{-px})J_{0}(x) e^{-xt} \ dt \ dx \\ &= \int_{0}^{\infty} \left( \int_{0}^{\infty} J_{0}(x) e^{-tx} \ dx \ - \int_{0}^{\infty} J_{0}(x) e^{-(p+t)x} \ dx \right) \ dt \\ &= \int_{0}^{\infty} \Bigg( \frac{1}{\sqrt{1+t^{2}}} - \frac{1}{\sqrt{1+(p+t)^{2}}} \Bigg)\ dt \tag{1} \\ &= \text{arcsinh}(t) -\text{arcsinh}(p+t) \Bigg|^{\infty}_{0} \\ &= \text{arcsinh}(p) \\ &= \ln \left(p+\sqrt{p^{2}+1} \right) \end{align}$$

$(1)$ http://mathworld.wolfram.com/LaplaceTransform.html

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