# §10.38 Derivatives with Respect to Order

 10.38.1 $\frac{\partial I_{\pm\nu}\left(z\right)}{\partial\nu}=\pm I_{\pm\nu}\left(z% \right)\ln\left(\tfrac{1}{2}z\right)\mp(\tfrac{1}{2}z)^{\pm\nu}\sum_{k=0}^{% \infty}\frac{\psi\left(k+1\pm\nu\right)}{\Gamma\left(k+1\pm\nu\right)}\frac{(% \frac{1}{4}z^{2})^{k}}{k!},$ ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, $\psi\left(\NVar{z}\right)$: psi (or digamma) function, $!$: factorial (as in $n!$), $I_{\NVar{\nu}}\left(\NVar{z}\right)$: modified Bessel function of the first kind, $\ln\NVar{z}$: principal branch of logarithm function, $\frac{\partial\NVar{f}}{\partial\NVar{x}}$: partial derivative of $f$ with respect to $x$, $\,\partial\NVar{x}$: partial differential of $x$, $k$: nonnegative integer, $z$: complex variable and $\nu$: complex parameter A&S Ref: 9.6.42 Referenced by: §10.38, §10.38, Erratum (V1.0.17) for Equations (10.15.1), (10.38.1) Permalink: http://dlmf.nist.gov/10.38.E1 Encodings: TeX, pMML, png Addition (effective with 1.0.17): This equation has been generalized to include the additional case of $\ifrac{\partial I_{-\nu}\left(z\right)}{\partial\nu}$ because it will help the user who wants to combine it with (10.38.2). See also: Annotations for §10.38 and Ch.10
 10.38.2 $\frac{\partial K_{\nu}\left(z\right)}{\partial\nu}=\tfrac{1}{2}\pi\csc\left(% \nu\pi\right)\*\left(\frac{\partial I_{-\nu}\left(z\right)}{\partial\nu}-\frac% {\partial I_{\nu}\left(z\right)}{\partial\nu}\right)-\pi\cot\left(\nu\pi\right% )K_{\nu}\left(z\right),$ $\nu\notin\mathbb{Z}$.

## Integer Values of $\nu$

 10.38.3 $\left.(-1)^{n}\frac{\partial I_{\nu}\left(z\right)}{\partial\nu}\right|_{\nu=n% }=-K_{n}\left(z\right)+\frac{n!}{2(\frac{1}{2}z)^{n}}\sum_{k=0}^{n-1}(-1)^{k}% \frac{(\frac{1}{2}z)^{k}I_{k}\left(z\right)}{k!(n-k)},$

For $\ifrac{\partial I_{\nu}\left(z\right)}{\partial\nu}$ at $\nu=-n$ combine (10.38.1), (10.38.2), and (10.38.4).

 10.38.4 $\left.\frac{\partial K_{\nu}\left(z\right)}{\partial\nu}\right|_{\nu=n}=\frac{% n!}{2(\frac{1}{2}z)^{n}}\sum_{k=0}^{n-1}\frac{(\frac{1}{2}z)^{k}K_{k}\left(z% \right)}{k!(n-k)}.$
 10.38.5 $\displaystyle\left.\frac{\partial I_{\nu}\left(z\right)}{\partial\nu}\right|_{% \nu=0}$ $\displaystyle=-K_{0}\left(z\right),$ $\displaystyle\left.\frac{\partial K_{\nu}\left(z\right)}{\partial\nu}\right|_{% \nu=0}$ $\displaystyle=0.$

## Half-Integer Values of $\nu$

For the notations $E_{1}$ and $\operatorname{Ei}$ see §6.2(i). When $x>0$,

 10.38.6 $\left.\frac{\partial I_{\nu}\left(x\right)}{\partial\nu}\right|_{\nu=\pm\frac{% 1}{2}}=-\frac{1}{\sqrt{2\pi x}}\left(E_{1}\left(2x\right)e^{x}\pm\operatorname% {Ei}\left(2x\right)e^{-x}\right),$
 10.38.7 $\left.\frac{\partial K_{\nu}\left(x\right)}{\partial\nu}\right|_{\nu=\pm\frac{% 1}{2}}=\pm\sqrt{\frac{\pi}{2x}}E_{1}\left(2x\right)e^{x}.$

For further results see Brychkov and Geddes (2005).