# §10.38 Derivatives with Respect to Order

 10.38.1 $\frac{\partial\mathop{I_{\nu}\/}\nolimits\!\left(z\right)}{\partial\nu}=% \mathop{I_{\nu}\/}\nolimits\!\left(z\right)\mathop{\ln\/}\nolimits\!\left(% \tfrac{1}{2}z\right)-(\tfrac{1}{2}z)^{\nu}\sum_{k=0}^{\infty}\frac{\mathop{% \psi\/}\nolimits\!\left(\nu+k+1\right)}{\mathop{\Gamma\/}\nolimits\!\left(\nu+% k+1\right)}\frac{(\frac{1}{4}z^{2})^{k}}{k!},$
 10.38.2 $\frac{\partial\mathop{K_{\nu}\/}\nolimits\!\left(z\right)}{\partial\nu}=\tfrac% {1}{2}\pi\mathop{\csc\/}\nolimits\!\left(\nu\pi\right)\*\left(\frac{\partial% \mathop{I_{-\nu}\/}\nolimits\!\left(z\right)}{\partial\nu}-\frac{\partial% \mathop{I_{\nu}\/}\nolimits\!\left(z\right)}{\partial\nu}\right)-\pi\mathop{% \cot\/}\nolimits\!\left(\nu\pi\right)\mathop{K_{\nu}\/}\nolimits\!\left(z% \right),$ $\nu\notin\Integer$.

## Integer Values of $\nu$

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

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

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

## Half-Integer Values of $\nu$

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

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

For further results see Brychkov and Geddes (2005).