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How to see that $$\int_0^{\infty} \sqrt{a^2+b^2} \exp \left[ -\frac{a^2+b^2}{c} \right] da= cb^2K_2 \left[ \frac{b}{c} \right],$$ where $$ K_\alpha(x) = \int_0^\infty e^{-x\cosh t} \cosh \alpha t \,dt ,$$ is the modified Bessel functions of the second kind.


EDIT

I realized that I made a stupid mistake in previous calculations. Now I get the following integral for which I should understand how to get the final result, sorry for my ignorance.

$$\int_0^{\infty} a \sqrt{a^2-b^2} \exp \left[ -\frac{a}{c} \right] da= cb^2K_2 \left[ \frac{b}{c} \right],$$

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  • $\begingroup$ Are you sure about the $\sqrt{a^2-b^2}$ ? $\endgroup$ – Claude Leibovici Jun 27 '20 at 13:09
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$$I=\int_0^{\infty} \sqrt{a^2+b^2} \exp \left[ -\frac{a^2+b^2}{c} \right] da=\frac{c\sqrt{\pi } }{2} e^{-\frac{b^2}{c}} \, U\left(-\frac{1}{2},0,\frac{b^2}{c}\right)$$ where appears the Tricomi hypergeometric function.

This can be also expressed as $$I=\frac{b^2}{4} e^{-\frac{b^2}{2 c}} \left(K_0\left(\frac{b^2}{2 c}\right)+K_1\left(\frac{b^2}{2 c}\right) \right)$$ but this does not seem to be the rhs (except if I am mistaken).

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$$I=\int_{0}^{\infty} \sqrt{a^2+b^2} e^{-\frac{(a^2+b^2)}{c}} da$$ $a=b \sinh t \implies da= b \cosh t$ Then $$I=b^2\int_{0}^{\infty} \cosh^2 t ~e^{-(b^2\cosh^2 t)/c} dt$$ $$\implies I=b^2 \int_{0}^{\infty} \frac{(1+\cosh 2t)}{2} Exp\left[-b^2 \frac{(\cosh 2t+1)}{2c}\right] dt$$ Let $t=z/2$, then $$I=\frac{b^2e^{-b^2/(2c)}}{4}\left[\int_{0}^{\infty} \cosh z ~ e^{\frac{-b^2 \cosh z}{2c}}dz + \int_{0}^{\infty} e^{-\frac{b^2 \cosh z}{2c}} dz\right]$$ $$\implies I=\frac{b^2e^{-b^2/(2c)}}{4} [K_1(u)+K_0(u)], u=b^2/(2c).$$

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