I would like to prove the following identity which I found implemented in some code given to me: $$\int_0^\infty k e^{-k^2} J_0(kx)Y_0(kx)~\mathrm{d}k = -\frac{1}{2\pi} e^{-\frac{1}{2}x^2} K_0\left(\frac{1}{2}x^2\right)$$ where $J_0$ and $Y_0$ are the Bessel functions of the first and second kind, respectively, and $K_0$ is the modified Bessel function of the second kind.

I am a physicist and not particularly familiar with Bessel functions or with their usual identities. Any particular ideas on how to prove this identity?

I did a quick simulation in Python and the two sides of the equation indeed seem equal.

  • 1
    $\begingroup$ Maybe this is useless, but I saw a recent big question on similar integrals, here. I hope you can find something useful there. $\endgroup$ Commented Nov 27, 2019 at 14:37
  • $\begingroup$ Thank you for your answer ! The other question is solved using Laplace transforms and the related Bessel identities. I don't think this can apply here due to the square in the exponentials, unless I am missing something. $\endgroup$
    – Ronan
    Commented Nov 27, 2019 at 15:15

1 Answer 1


We may use the Hankel transform representation of the Gaussian term (DLMF): \begin{equation} \int_{0}^{\infty}J_{\nu}\left(bt\right)\exp\left(-p^{2}t^{2}\right)t^{\nu+1}% \mathrm{d}t=\frac{b^{\nu}}{(2p^{2})^{\nu+1}}\exp\left(-\frac{b^{2}}{4p^{2}}% \right) \end{equation} with $\nu=0,p=1/2$ to express (by changing the order of integration) \begin{align} I(x)&=\int_0^\infty k e^{-k^2} J_0(kx)Y_0(kx)~\mathrm{d}k \\ &=\frac{1}{2}\int_0^\infty te^{-t^2/4}\,dt\int_0^\infty k J_0(kt) J_0(kx)Y_0(kx)~\mathrm{d}k \end{align} The integral of the triple product of Bessel function is (for a proof see Watson, "Theory of Bessel functions", §13.46): \begin{equation} \int_{0}^{\infty}Y_{\nu}\left(ak\right)J_{\nu}\left(bk\right)J_{\nu}\left(ck% \right)k^{1+\nu}\mathrm{d}k=\begin{cases}-\dfrac{(abc)^{\nu}(-A)^{-\nu-\frac{1% }{2}}}{\pi^{\frac{1}{2}}2^{\nu+1}\Gamma\left(\frac{1}{2}-\nu\right)},&0<a<|b-c% |,\\ 0,&|b-c|<a<b+c,\\ \dfrac{(abc)^{\nu}(-A)^{-\nu-\frac{1}{2}}}{\pi^{\frac{1}{2}}2^{\nu+1}\Gamma% \left(\frac{1}{2}-\nu\right)},&a>b+c.\end{cases} \end{equation} where $A=s(s-a)(s-b)(s-c), s=(a+b+c)/2$. Here, we choose $\nu=0,a=x,b=t,c=x$, and thus $A=-\left( t^2/4-x^2 \right)t^2/4$. This integral is non zero iff $t>2x$. In this case, \begin{equation} \int_{0}^{\infty}Y_{\nu}\left(ak\right)J_{\nu}\left(bk\right)J_{\nu}\left(ck% \right)k^{1+\nu}\mathrm{d}k=-\frac{t^{-1}\left( t^2/4 -x^2\right)^{-1/2}}{\pi} \end{equation} Using simple changes of variables and an integral representation of the modified Bessel function (DLMF), we have then \begin{align} I(x)&=-\frac{1}{2\pi}\int_{2x}^\infty \frac{e^{-t^2/4}}{\sqrt{t^2/4-x^2}}\,dt\\ &=\frac{-1}{\pi}\int_0^\infty e^{-x^2\cosh^2u}du\\ &=-\frac{e^{-x^2/2}}{\pi}\int_0^\infty e^{-\frac{x^2}{2}\cosh 2u}\,du\\ &=-\frac{e^{-x^2/2}}{2\pi}\int_0^\infty e^{-\frac{x^2}{2}\cosh v}\,dv\\ &=-\frac{e^{-x^2/2}}{2\pi}K_0\left( \frac{x^2}{2} \right) \end{align} as expected.


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