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I'm trying to get the following result confirmed in Mathematica: $$ \left.\frac{\partial\mathop{K_{\nu}}\nolimits\!\left(x\right)}{\partial\nu}% \right|_{\nu=\pm\frac{1}{2}}=\pm\sqrt{\frac{\pi}{2x}}\mathop{E_{1}}\nolimits% \!\left(2x\right)e^{x}, $$ cf. http://dlmf.nist.gov/10.38.E7. Here $K_{\nu}(z)$ denotes the MacDonald function and $E_1$ denotes the exponential integral.

However, Mathematica seems to be unaware of the above result. All I get for

D[BesselK[a, z], {a, 1}] /. a -> 1/2

is

$$ \text{BesselK}^{(1,0)}\left(\frac{1}{2},z\right). $$

Can Mathematica be made more specific here?

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  • 2
    $\begingroup$ FullSimplify[FunctionExpand[Derivative[1, 0][BesselK][1/2, z]]] might be somewhat closer. $\endgroup$ – b.gates.you.know.what Mar 21 '16 at 20:47
  • $\begingroup$ ...and of course the result in b.gatessucks's comment can be massaged into the required form. $\endgroup$ – J. M. is away Mar 21 '16 at 22:10
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As noted by b.gatessucks, a preliminary application of FunctionExpand[] allows the confirmation of this identity:

FullSimplify[FunctionExpand[Derivative[1, 0][BesselK][1/2, z]] ==
             Sqrt[π/2] ExpIntegralE[1, 2 z] Exp[z]/Sqrt[z]]
   True

FullSimplify[FunctionExpand[Derivative[1, 0][BesselK][-1/2, z]] ==
             -Sqrt[π/2] ExpIntegralE[1, 2 z] Exp[z]/Sqrt[z]]
   True

It should be noted that the particular form of the result in Mathematica uses $\frac1{\sqrt{z}}$ instead of $\sqrt{\frac1z}$, which corresponds to a different choice of branch cut on the negative real axis.


Only recently has a formula for $\frac{\partial}{\partial\nu}K_\nu(z)$ for general values of $\nu$ been found; see e.g. Brychkov's article and this accompanying blog post.

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