# Opt3

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Given that the power spectrum and structure function are related by

${\displaystyle \phi _{n}(\kappa )={\frac {1}{4\pi ^{2}\kappa ^{2}}}\int _{0}^{\infty }{\frac {\sin(\kappa R)}{\kappa R}}{\frac {\partial }{\partial R}}\left[R^{2}{\frac {\partial D_{n}(R)}{\partial R}}\right]\,dR\,}$,

use the structure function ${\displaystyle D_{n}(R)=C_{n}^{2}R^{2/3}\,}$ and the integral formula

${\displaystyle \int _{0}^{\infty }x^{\alpha }\sin(x)\,dx=2^{\alpha }{\sqrt {\pi }}\,{\frac {\Gamma ({\frac {\alpha }{2}}+1)}{\Gamma ({\frac {1}{2}}-{\frac {\alpha }{2}})}}\,}$ to obtain the Kolmogorov power law spectrum:

${\displaystyle \phi _{n}(\kappa )=0.033\,C_{n}^{2}\kappa ^{-11/3},\,\,\,{\frac {1}{L_{0}}}<\!\!<\kappa <\!\!<{\frac {1}{l_{0}}}\,}$.

Substituting the structure function into the power spectrum above,

${\displaystyle \phi _{n}(\kappa )={\frac {1}{4\pi ^{2}\kappa ^{2}}}\int _{0}^{\infty }{\frac {\sin(\kappa R)}{\kappa R}}C_{n}^{2}{\frac {\partial }{\partial R}}\left[R^{2}{\frac {2}{3}}R^{-1/3}\right]\,dR\,}$

${\displaystyle ={\frac {1}{4\pi ^{2}\kappa ^{2}}}C_{n}^{2}{\frac {2}{3}}\int _{0}^{\infty }{\frac {\sin(\kappa R)}{\kappa R}}{\frac {5}{3}}R^{2/3}\,dR\,}$

${\displaystyle ={\frac {2}{3}}{\frac {5}{3}}{\frac {1}{4\pi ^{2}\kappa ^{2}}}\kappa ^{-1}C_{n}^{2}\int _{0}^{\infty }\sin(\kappa R)R^{-1/3}\,dR\,}$

Now to use the integral formula, let ${\displaystyle x=\kappa R,\,\,\,R={\frac {x}{\kappa }},\,\,\,dR={\frac {dx}{\kappa }}\,}$

${\displaystyle ={\frac {10}{36}}{\frac {1}{\pi ^{2}\kappa ^{3}}}C_{n}^{2}\int _{0}^{\infty }\sin(x)\left({\frac {x}{\kappa }}\right)^{-1/3}{\frac {dx}{\kappa }}\,}$

${\displaystyle ={\frac {10}{36}}{\frac {1}{\pi ^{2}\kappa ^{11/3}}}C_{n}^{2}\int _{0}^{\infty }x^{-1/3}\sin(x)\,dx\,}$

${\displaystyle ={\frac {10}{36}}{\frac {1}{\pi ^{2}\kappa ^{11/3}}}C_{n}^{2}2^{-1/3}{\sqrt {\pi }}\,{\frac {\Gamma ({\frac {-1}{6}}+1)}{\Gamma ({\frac {1}{2}}+{\frac {1}{6}})}}\,}$

${\displaystyle =.033\,C_{n}^{2}\kappa ^{-11/3}\,}$