Goodwin–Staton integral: Difference between revisions

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In mathematics the Goodwin–Staton integral is defined as :^[1]

G(z)=\int _{0}^{\infty }{\frac {e^{-t^{2}}}{t+z}}\,dt

It satisfies the following third-order nonlinear differential equation：

4w(z)+8\,z{\frac {d}{dz}}w(z)+(2+2\,z^{2}){\frac {d^{2}}{dz^{2}}}w(z)+z{\frac {d^{3}}{dz^{3}}}w\left(z\right)=0

$G(-z)={\sqrt {2}}\sin \left({\frac {\pi }{4}}+\pi {\frac {z^{2}}{4}}\right)-G(z)$

$G(z)=e^{-z^{2}}+\operatorname {Ei} (1,-z^{2})e^{-z^{2}}+e^{-z^{2}}\operatorname {erf} (iz)$ $G(z)=e^{-z^{2}}+\operatorname {Ei} (1,-z^{2})e^{-z^{2}}+e^{-z^{2}}\operatorname {erf} (iz)$
- $G(z)=e^{-z^{2}}+\operatorname {Ei} (1,-z^{2})e^{-z^{2}}+{\frac {ze^{-z^{2}}\left(-i\operatorname {erfc} \left({\sqrt {-z^{2}}}\right)+i\right)}{\sqrt {-z^{2}}}}$

$G(z)={\frac {1}{2}}\,{\frac {G_{2,3}^{3,2}\left({z}^{2}\,{\Big \vert }\,_{1/2,0,0}^{0,1/2}\right)}{\pi }}$

$G(z)=e^{-z^{2}}+{\rm {U}}(1,\,1,\,-z^{2})e^{z^{2}}e^{-z^{2}}+{\frac {2ie^{-z^{2}}z{\rm {M}}(1/2,3/2,z^{2})}{\sqrt {\pi }}}$

$G(z)=e^{-z^{2}}+\operatorname {Ei} (1,-z^{2})e^{-z^{2}}+{\frac {2ie^{-z^{2}}z\operatorname {HeunB} \left(1,0,1,0,{\sqrt {z^{2}}}\right)}{\sqrt {\pi }}}$

$G(z)=e^{-z^{2}}+{\it {Ei}}(1,-z^{2})e^{-z^{2}}+i{e^{-z^{2}}}{\sqrt {\pi }}z\operatorname {LaguerreL} (-1/2,1/2,z^{2})$

$G(z)=10z^{-1}-50z^{-2}-{\frac {1000}{3}}\,{\frac {z^{2}-1}{z^{3}}}+2500\,{\frac {z^{2}-1}{z^{4}}}+10000\,{\frac {2-2z^{2}+z^{4}}{z^{5}}}-{\frac {250000}{3}}\,{\frac {2-2z^{2}+z^{4}}{z^{6}}}-{\frac {5000000}{21}}\,{\frac {-6+6z^{2}-3z^{4}+z^{6}}{z^{7}}}+{\frac {6250000}{3}}\,{\frac {-6+6z^{2}-3z^{4}+z^{6}}{z^{8}}}+{\frac {125000000}{27}}\,{\frac {24-24z^{2}+12z^{4}-4z^{6}+z^{8}}{z^{9}}}-{\frac {125000000}{3}}\,{\frac {24-24z^{2}+12z^{4}-4z^{6}+z^{8}}{z^{10}}}$

$G(z)=\left(1-\gamma -\ln(z^{2})-i\operatorname {csgn} (iz^{2})\pi +{\frac {2\,i}{\sqrt {\pi }}}z+(-2+\gamma +\ln(z^{2})+i\operatorname {csgn} (iz^{2})\pi \right)z^{2}+{\frac {-4/3\,i}{\sqrt {\pi }}}z^{3}+\left({\frac {5}{4}}-1/2\,\gamma -1/2\,\ln(z^{2})-1/2\,i\operatorname {csgn} (iz^{2})\pi \right)z^{4}+O(z^{5}))$

^ Frank Oliver, NIST Handbook of Mathematical Functions, p160,Cambridge University Press 2010

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  </math>
-;[[Laguerrel polynomial]]
+;[[Laguerrel polynomials]]
 *<math>G(z) =e^{-z^2} +{\it Ei}( 1,-z^2) e^{-z^2}+i{e^{-z^2}}\sqrt {\pi }z \operatorname{LaguerreL} (-1/2,1/2,z^2) </math>