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I want to show that the derivative of integral formulation of zeroth order Bessel function of second kind $K_{0}(z)$, with respect to argument $z$, is equal to minus the integral formulation of first order Bessel function of second kind $K_{1}(z)$, i.e.

\begin{equation} \frac{d}{dz} K_{0}(z) = -K_{1}(z) \end{equation}

I use that \begin{equation} K_{0}(z) = \int\limits_{0}^{\infty} \frac{\cos{(zt)}}{\sqrt{t^2 + 1}}\,dt \end{equation} and \begin{equation} K_{1}(z) = z\int\limits_{0}^{\infty} \frac{\cos{(t)}}{(t^2 + z^2)^{3/2}}\,dt \end{equation} from http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html.

So far I have only got that \begin{equation} \frac{d}{dz} K_{0}(z) = \int\limits_{0}^{\infty} \frac{-\sin{(zt)}}{\sqrt{t^2 + 1}}t\,dt = \int\limits_{0}^{\infty} \frac{zt\cos{(zt)}-\sin{(zt)}}{z^{2}(t^2 + 1)^{3/2}}t\,dt, \end{equation} but from here I do not know which path to pursue. Any guidance would be most welcome.

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    $\begingroup$ You're looking to get $\cos(t)$ inside the trig functions, so try a substitution $u=zt$ $\endgroup$ – B. Mehta May 19 '17 at 13:13
  • $\begingroup$ Also, where did that second integral come from? $\endgroup$ – B. Mehta May 19 '17 at 13:14
  • $\begingroup$ Via integrations by parts. The other part is \begin{equation} \frac{zt\cos{(zt)}-\sin{(zt)}}{z^2t(t^2+1)^{1/2}} \end{equation} which is goes to zero for t going to zero or infinity. $\endgroup$ – Raibyo May 19 '17 at 13:23
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If $s=zt$, the $K_0$ integral becomes $$ \int_0^{\infty} \frac{\cos{s}}{z\sqrt{1+s^2/z^2}} \, ds = \int_0^{\infty} \frac{\cos{s}}{\sqrt{z^2+s^2}} \, ds $$ for $z$ nonnegative (fine since we expect a branch cut anyway). Now you can just differentiate with respect to $z$ to get the $K_1$ formula, since $$ \frac{\partial}{\partial z} \frac{1}{\sqrt{z^2+s^2}} = -\frac{z}{(z^2+s^2)^{3/2}}. $$

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  • $\begingroup$ With that change of variables, don't I get $$\int_0^{\infty} \frac{\cos{s(z)}}{\sqrt{z^2+s(z)^2}} \, ds $$, which gives me another expression when differentiating with respect to z? $\endgroup$ – Raibyo May 19 '17 at 13:29
  • $\begingroup$ I don't think so. $t=s/z$, so $dt=ds/z$, and the original integral was $\int_0^{\infty} \frac{\cos{zt}}{\sqrt{1+t^2}} \, dt$? $\endgroup$ – Chappers May 19 '17 at 14:31
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Use this definition

$$ K_{\alpha }(x)=\int _{0}^{\infty }\exp(-x\cosh t)\cosh(\alpha t)\,dt $$

So that

$$ \frac{d}{dx}K_0(x) = \int _{0}^{\infty }\frac{d}{dt}\exp(-x\cosh t)\,dt = -\int _{0}^{\infty }\exp(-x\cosh t)\cosh(t)\,dt = -K_1(x) $$

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  • $\begingroup$ I guess you mean to differentiate w.r.t. x, but thanks! I did this earlier, so the problem is really to go between the expressions I have stated. $\endgroup$ – Raibyo May 19 '17 at 13:31

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