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How does one show the following integral (which was obtained in Maple)

$\int_0^{\infty } x \exp \left(-a x^2\right) Y_0(b x) \, dx=\frac{1}{2\pi a}\exp\left(-\frac{b^2}{4 a}\right) \text{Ei}\left(\frac{b^2}{4 a}\right)$,

where $Y_0$ is the Bessel function of the 2nd kind, and Ei is the exponential integral function?

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There may well be a simpler method, but the following should work:

We start by writing $$\phi \colon (0,\infty)^2 \to \mathbb{R} \, , \, \phi(a,b) = \int \limits_0^\infty x \, \mathrm{e}^{-a x^2} \operatorname{Y}_0 (b x) \, \mathrm{d} x \, . $$ Then we can let $\sqrt{a} x = \xi$, use an integral representation of the Bessel function and justify changing the order of integration to obtain $$ \phi(a,b) = - \frac{2}{\pi a} \int \limits_0^\infty \int \limits_0^\infty \xi \, \mathrm{e}^{-\xi^2} \cos \left(\frac{b}{\sqrt{a}} \cosh(t) \xi\right) \, \mathrm{d} \xi \, \mathrm{d} t \equiv \frac{1}{2\pi a} f\left(\frac{b^2}{4 a}\right) . $$ Here we have defined $f \colon (0,\infty) \to \mathbb{R}$ by \begin{align} f(u) &= - 4 \int \limits_0^\infty \int \limits_0^\infty \xi \, \mathrm{e}^{-\xi^2} \cos \left(2 \sqrt{u} \cosh(t) \xi \right) \, \mathrm{d} \xi \, \mathrm{d} t \\ &= 2 \int \limits_0^\infty \left[\int \limits_0^\infty 2 \sqrt{u} \cosh(t) \sin(2 \sqrt{u} \cosh(t) \xi) \mathrm{e}^{-\xi^2} \, \mathrm{d} \xi - 1 \right] \mathrm{d} t \\ &= 2 \int \limits_0^\infty \left[2 \sqrt{u} \cosh(t) \operatorname{F}(\sqrt{u} \cosh(t)) - 1\right] \mathrm{d} t \\ &\hspace{-17pt}\stackrel{s = \sqrt{u} \sinh(t)}{=} 2 \int \limits_0^\infty \left[2 \operatorname{F}(\sqrt{u + s^2}) - \frac{1}{\sqrt{u + s^2}}\right] \mathrm{d} s \, . \end{align} The integral over $\xi$ can be calculated using $\sin(y) = \mathrm{e}^{\mathrm{i} y}$ for $y \in \mathbb{R}$ and $\operatorname{F}$ is Dawson's integral. We note that $f$ is differentiable and compute $$ f'(u) = - f(u) + \int \limits_0^\infty \frac{\mathrm{d}s}{(u + s^2)^{3/2}} = - f(u) + \frac{1}{u} \, , \, u > 0 \, . $$ The usual variation of parameters method allows us to conclude that there is a constant $C \in \mathbb{R}$ such that $$ f(u) = \mathrm{e}^{-u} [\operatorname{Ei}(u) + C] \, , \, u > 0 \, . $$ Therefore, $$ \phi(a,b) = \frac{1}{2\pi a} \mathrm{e}^{-b^2/4a} \left[\operatorname{Ei}\left(\frac{b^2}{4a}\right) + C \right] \, , \, a,b > 0 \, . $$ For $a > 0$ and small $\varepsilon > 0$ we can use the series expansion of the exponential integral to obtain $$ 2 \pi a \, \phi(a,2\sqrt{a}\varepsilon) \stackrel{\varepsilon \to 0^+}{\sim} 2 \log(\varepsilon) + \gamma + C + \mathcal{O}[\varepsilon^2 \log(\varepsilon)] \, . $$ On the other hand, the expansion of the Bessel function yields \begin{align} 2 \pi a \, \phi(a,2\sqrt{a}\varepsilon) &\stackrel{\varepsilon \to 0^+}{\sim} 4 a \int \limits_0^\infty x \, \mathrm{e}^{-a x^2} \left[\log(\sqrt{a} \varepsilon x) + \gamma\right] \mathrm{d} x + \mathcal{O}[\varepsilon^2 \log(\varepsilon)] \\ &\hspace{-4pt}\stackrel{\sqrt{a} x^2 = y}{=} \int \limits_0^\infty \mathrm{e}^{-y} [2 \log(\varepsilon) + 2\gamma + \log(y)] \, \mathrm{d} y + \mathcal{O}[\varepsilon^2 \log(\varepsilon)] \\ &\hspace{4pt}= 2 \log(\varepsilon) + \gamma + \mathcal{O}[\varepsilon^2 \log(\varepsilon)] \, . \end{align} Thus we find $C = 0$ by comparison and the desired result follows.

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This answer is based on the fact that $Y_\nu(2\sqrt{u})$ is a solution to the confluent hypergeometric limit equation $x y''(x) + (\nu+1)y'(x) + y(x) = 0$. Since this has only linear coefficients, it can be used to create a first-order differential equation for the integral.

Let $z = b^2/4a$. Then the integral is equivalent to $$ \int_0^\infty xe^{-ax^2}Y_0(bx)dx = \frac{2}{b^2}\int_0^\infty Y_0(2\sqrt{u})e^{-u/z}du = \frac{2}{b^2}I(z) $$ The integral $I(z)$ is now in the form of the Laplace transform of the function $f(u) = Y_0(2\sqrt{u})$ that satisfies the differential equation $$ (uf'(u))'+f(u) = 0. $$ Taking the Laplace transform of this equation gives $$ -s\frac{d}{ds}\left[sF(s)\right] - \lim_{u\rightarrow 0^+}uf'(u)+F(s) = 0 $$ The limit can be found from $uf'(u) = -\sqrt{u}Y_1(2\sqrt{u})$, and turns out to be $1/\pi$. So $F(s)$ satisfies $$ s^2F'(s) + (s-1)F(s) = -\frac{1}{\pi}, $$ and $I(z) = F(1/z)$ satisfies $$ zI'(z)+(z-1)I(z) = \frac{z}{\pi}\Longrightarrow \frac{d}{dz}\left[\frac{e^z}{z}i(z)\right] = \frac{e^z}{\pi z}\Longrightarrow i(z) = \frac{ze^{-z}}{\pi}\left[C + \mathrm{Ei}\left(z\right)\right] $$ Which gives for the original integral $$ \int_0^\infty xe^{-ax^2}Y_0(bx)dx = \frac{2}{b^2}I(z) = \frac{2ze^{-z}}{\pi b^2}\left[C + \mathrm{Ei}\left(z\right)\right] = \frac{e^{-b^2/(4a)}}{2\pi a}\left[C + \mathrm{Ei}\left(\frac{b^2}{4a}\right)\right]. $$

I'm not sure of an easy way to show $C = 0$ other than the asymptotic analysis ComplexButTrivial did.

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