I need to do this integral:

$$\int_0^\infty dx\cdot x \sqrt{x^2+1}K_0(ax)$$

where K is the modified Bessel of second kind. I have seen that in Gradhsteyn 7th edition in 6.565.7 says that this integral is equal to:

$$\int_0^\infty dx\cdot x \sqrt{x^2+1}K_0(ax)=a^{-3/2}S_{1/2,3/2}(a)$$

where S is the Lommel function. My problem is that I need to know the limit of this Lommel function when a->0 and when a->$\infty$ but if I look for this function on the net or in Gradhsteyn I can only see a formula that is valid when the summ or the subtraction of the 2 parameters of this function are not a negative odd and here I have 1/2-3/2=-1. Does anyone know how is this function when you have my condition? Or maybe, could you see another solution for this integral?

The most complete information I have gotten is this web: http://dlmf.nist.gov/11.9


1 Answer 1


Check out the asymptotic expansion on the DLMF. It says there that if the sum or difference of the parameters is an integer then the sum on the right-hand side is finite. If you wanted you could just use the leading order term,

$$ S_{1/2,3/2}(a) \sim \frac{1}{\sqrt{a}} $$

as $a \to \infty$ to get

$$ \int_0^\infty dx\cdot x \sqrt{x^2+1}K_0(ax) \sim \frac{1}{a^2} $$ as $a \to \infty$.

The limiting form when $\mu-\nu$ is an odd negative integer can be found in Watson as formula $(3)$ in §10.73. The formula relevant to your case is

$$ S_{\nu-1,\nu}(z) = \frac{z^{\nu}}{4} \Gamma(\nu) \sum_{m=0}^{\infty} \frac{(-1)^m (z/2)^{2m}}{m! \Gamma(\nu+m+1)} \Bigl[2\log(z/2) - \psi(\nu+m+1) - \psi(m+1)\Bigr] - 2^{\nu-2} \pi \Gamma(\nu) Y_{\nu}(z). $$

The sum moonlights as an asymptotic series when $z \to 0$. If we just keep the first term we have

$$ \sum_{m=0}^{\infty} \frac{(-1)^m (z/2)^{2m}}{m! \Gamma(\nu+m+1)} \Bigl[2\log(z/2) - \psi(\nu+m+1) - \psi(m+1)\Bigr] = O(\log z) $$

as $z \to 0$, so that the first term in $S_{\nu-1,\nu}(z)$ is $O(z^\nu \log z)$ and thus tends to $0$ if $\nu > 0$. In other words,

$$ S_{\nu-1,\nu}(z) = - 2^{\nu-2} \pi \Gamma(\nu) Y_{\nu}(z) + O(z^\nu \log z) $$

as $z \to 0$. The asymptotic behavior of $Y_\nu(z)$ as $z \to 0^+$ can be found on Wikipedia, where it says that if $\nu > 0$ then

$$ Y_{\nu}(z) \sim -\frac{\Gamma(\nu)}{\pi} \left(\frac{2}{z}\right)^\nu. $$

It follows that

$$ S_{\nu-1,\nu}(z) \sim \frac{2^{2\nu-2}\Gamma(\nu)^2}{z^\nu} $$

as $z \to 0^+$. Setting $z = a$ and $\nu = 3/2$ we can conclude that

$$ \int_0^\infty dx\cdot x \sqrt{x^2+1}K_0(ax) \sim \frac{\pi}{2a^3} $$ as $a \to 0^+$.


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