The FunctionAdvisor(specialize, math_function_1) command attempts to specialize this function in terms of every other mathematical function. When the function is given with parameters, as in math_function_1(a, b, ..., z), only specializations valid for those values of the parameters or particular cases of them are returned.

The FunctionAdvisor(specialize, math_function_1, math_function_2) command attempts to specialize the first function, math_function_1 in terms of the other one using Maple algorithms. When the first function is given with parameters, as in math_function_1(a, b, ..., z), only specializations valid for those values of the parameters or particular cases of them are returned. For more information, see convert/to_special_function.

Note the specialization operation typically requires additional constraints for the function parameters involved. These constraints are returned as a boolean function. If no constraints are required, this is explicitly indicated.

$\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\,\sin \,\mathrm{hypergeom}\right)$

$\left[\sin \left(z\right)=z\mathrm{hypergeom}\left(\left[\right]\,\left[\frac{3}{2}\right]\,-\frac{z^2}{4}\right)\,\mathrm{with\; no\; restrictions\; on}\left(z\right)\right]$

$\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\,\mathrm{hypergeom}\,\sin \right)$

$\left[\mathrm{hypergeom}\left(\left[\right]\,\left[a\right]\,z\right)=\frac{\sin \left(2\sqrt{-z}\right)}{2\sqrt{-z}}\,a=\frac{3}{2}\right]$

$\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\,\mathrm{hypergeom}\,\mathrm{dilog}\right)$

$\left[\mathrm{hypergeom}\left(\left[a\,b\,c\right]\,\left[d\,e\right]\,z\right)=\frac{\mathrm{dilog}\left(1-z\right)}{z}\,a=1\wedge b=1\wedge c=1\wedge d=2\wedge e=2\right]$

$\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\,\mathrm{hypergeom}\left(\left[1\,1\,a\right]\,\left[b\,c\right]\,1\right)\,\mathrm{dilog}\right)$

$\left[\mathrm{hypergeom}\left(\left[1\,1\,a\right]\,\left[b\,c\right]\,1\right)=\mathrm{dilog}\left(0\right)\,a=1\wedge b=2\wedge c=2\right]$

$\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\,\mathrm{HermiteH}\,\mathrm{KummerU}\right)$

$\left[\mathrm{HermiteH}\left(a\&comma;z\right)=2^a\mathrm{KummerU}\left(-\frac{a}{2}\&comma;\frac{1}{2}\&comma;z^2\right)\&comma;0<\mathrm{\Re }\left(z\right)\vee \left(\mathrm{\Re }\left(z\right)=0\wedge 0<\mathrm{\Im }\left(z\right)\right)\right]$

$\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\&comma;\mathrm{KummerU}\&comma;\mathrm{HermiteH}\right)$

$\left[\mathrm{KummerU}\left(a\&comma;b\&comma;z\right)=\mathrm{HermiteH}\left(-2a\&comma;\sqrt{z}\right)2^{2a}\&comma;b=\frac{1}{2}\right],\left[\mathrm{KummerU}\left(a\&comma;b\&comma;z\right)=\frac{\mathrm{HermiteH}\left(-2a+1\&comma;\sqrt{z}\right)2^{2a}}{2\sqrt{z}}\&comma;b=\frac{3}{2}\right]$

$\mathrm{convert}\left(2\sin \left(z\right)+\mathrm{KummerU}\left(a\&comma;b\&comma;z\right)\&comma;\mathrm{HermiteH}\right)\mathbf{assuming}b=\frac{3}{2}$

$2\sin \left(z\right)+\frac{\mathrm{HermiteH}\left(-2a+1\&comma;\sqrt{z}\right)2^{2a}}{2\sqrt{z}}$

$\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\&comma;\sin \right)$

$\left[\sin \left(z\right)=\frac{\sqrt{z}\sqrt{\mathrm{\pi }}\sqrt{2}\mathrm{BesselJ}\left(\frac{1}{2}\&comma;z\right)}{2}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\mathrm{ChebyshevT}\left(b\&comma;\cos \left(\frac{\mathrm{\pi }-2z}{2b}\right)\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\mathrm{ChebyshevU}\left(-\frac{-z+\arccos \left(c\right)}{\arccos \left(c\right)}\&comma;c\right)\sqrt{-c^2+1}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{-2z{\mathrm{Ei}}_0\left(-2\mathrm{I}z\right)+\mathrm{I}}{2{\&ExponentialE;}^{\mathrm{I}z}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{-\frac{\mathrm{I}}{2}\left(\mathrm{\Gamma }\left(1\&comma;-2\mathrm{I}z\right)-1\right)}{{\&ExponentialE;}^{\mathrm{I}z}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{z\mathrm{HeunB}\left(2\&comma;0\&comma;0\&comma;0\&comma;\sqrt{2}\sqrt{\mathrm{I}z}\right)}{{\&ExponentialE;}^{\mathrm{I}z}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{\left(2\mathrm{I}{z}^2+z\right)\mathrm{HeunC}\left(1\&comma;1\&comma;1\&comma;0\&comma;\frac{1}{2}\&comma;-2\mathrm{I}z\right)}{{\&ExponentialE;}^{\mathrm{I}z}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{z\mathrm{KummerM}\left(1\&comma;2\&comma;2\mathrm{I}z\right)}{{\&ExponentialE;}^{\mathrm{I}z}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{-\frac{\mathrm{I}}{2}\left(-1+{\&ExponentialE;}^{2\mathrm{I}z}\right)}{{\&ExponentialE;}^{\mathrm{I}z}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\sqrt{\mathrm{\pi }}\mathrm{MeijerG}\left(\left[\left[\right]\&comma;\left[\right]\right]\&comma;\left[\left[\frac{1}{2}\right]\&comma;\left[0\right]\right]\&comma;\frac{z^2}{4}\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{\sqrt{z}\mathrm{StruveH}\left(-\frac{1}{2}\&comma;z\right)\sqrt{\mathrm{\pi }}\sqrt{2}}{2}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=-\frac{\mathrm{I}}{2}\mathrm{WhittakerM}\left(0\&comma;\frac{1}{2}\&comma;2\mathrm{I}z\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=-\cos \left(z+\frac{\mathrm{\pi }}{2}\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=-\cosh \left(\mathrm{I}\left(z+\frac{\mathrm{\pi }}{2}\right)\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{2\cot \left(\frac{z}{2}\right)}{{\cot \left(\frac{z}{2}\right)}^2+1}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{-2\mathrm{I}\coth \left(\frac{\mathrm{I}}{2}z\right)}{{\coth \left(\frac{\mathrm{I}}{2}z\right)}^2-1}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{1}{\csc \left(z\right)}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{\mathrm{-I}}{\mathrm{csch}\left(\mathrm{I}z\right)}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=-\frac{\mathrm{I}}{2}\left({\&ExponentialE;}^{\mathrm{I}z}-\frac{1}{{\&ExponentialE;}^{\mathrm{I}z}}\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=z\mathrm{hypergeom}\left(\left[\right]\&comma;\left[\frac{3}{2}\right]\&comma;-\frac{z^2}{4}\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=-\frac{1}{\sec \left(z+\frac{\mathrm{\pi }}{2}\right)}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=-\frac{1}{\mathrm{sech}\left(\mathrm{I}\left(z+\frac{\mathrm{\pi }}{2}\right)\right)}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\mathrm{-I}\sinh \left(\mathrm{I}z\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{2\tan \left(\frac{z}{2}\right)}{1+{\tan \left(\frac{z}{2}\right)}^2}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right],\left[\sin \left(z\right)=\frac{2\mathrm{I}\tanh \left(\frac{\mathrm{I}}{2}z\right)}{{\tanh \left(\frac{\mathrm{I}}{2}z\right)}^2-1}\&comma;\mathrm{with\; no\; restrictions\; on}\left(z\right)\right]$

$\left[\mathrm{FunctionAdvisor}\left(\mathrm{specialize}\&comma;\mathrm{LegendreQ}\right)\right]$

$\left[\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(a+1\right)\mathrm{AppellF1}\left(1+\frac{a}{2}\&comma;0\&comma;\frac{a}{2}+\frac{1}{2}\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(a+b+1\right)\mathrm{AppellF1}\left(1+\frac{a}{2}+\frac{b}{2}\&comma;0\&comma;\frac{a}{2}+\frac{b}{2}+\frac{1}{2}\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+b+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(a+1\right)\mathrm{AppellF2}\left(1+\frac{a}{2}\&comma;0\&comma;\frac{a}{2}+\frac{1}{2}\&comma;1\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(a+b+1\right)\mathrm{AppellF2}\left(1+\frac{a}{2}+\frac{b}{2}\&comma;0\&comma;\frac{a}{2}+\frac{b}{2}+\frac{1}{2}\&comma;1\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+b+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(a+1\right)\mathrm{AppellF3}\left(0\&comma;1+\frac{a}{2}\&comma;0\&comma;\frac{a}{2}+\frac{1}{2}\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(a+b+1\right)\mathrm{AppellF3}\left(0\&comma;1+\frac{a}{2}+\frac{b}{2}\&comma;0\&comma;\frac{a}{2}+\frac{b}{2}+\frac{1}{2}\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+b+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(a+1\right)\mathrm{AppellF4}\left(1+\frac{a}{2}\&comma;\frac{a}{2}+\frac{1}{2}\&comma;1\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(a+b+1\right)\mathrm{AppellF4}\left(1+\frac{a}{2}+\frac{b}{2}\&comma;\frac{a}{2}+\frac{b}{2}+\frac{1}{2}\&comma;1\&comma;\frac{3}{2}+a\&comma;0\&comma;\frac{1}{z^2}\right)}{2z^{a+b+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{z^2{\left(\frac{z^2-1}{z^2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\sqrt{2}\mathrm{EllipticK}\left(\frac{\sqrt{2}\sqrt{\frac{z^2+\sqrt{z^4-z^2}-1}{z^2-1}}}{2}\right)}{\left(z^2-1\right)\sqrt{\frac{\sqrt{z^4-z^2}}{\sqrt{\frac{-z^2+\sqrt{z^4-z^2}+1}{z^2-1}}\left(z^2-1\right)\sqrt{\frac{z^2+\sqrt{z^4-z^2}-1}{z^2-1}}}}}\&comma;a=-\frac{1}{2}\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(a+1\right)\mathrm{HeunC}\left(0\&comma;\frac{1}{2}+a\&comma;-\frac{1}{2}\&comma;0\&comma;\frac{1}{4}a+\frac{3}{8}+\frac{1}{4}{a}^2\&comma;\frac{1}{z^2\left(\frac{1}{z^2}-1\right)}\right)}{2z^{a+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right){\left(1-\frac{1}{z^2}\right)}^{\frac{a}{2}+\frac{1}{2}}{2}^a}\&comma;\left(a+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(a+b+1\right)\mathrm{HeunC}\left(0\&comma;\frac{1}{2}+a\&comma;-\frac{1}{2}\&comma;0\&comma;\frac{1}{4}a+\frac{3}{8}-\frac{1}{4}{b}^2+\frac{1}{4}{a}^2\&comma;\frac{1}{z^2\left(\frac{1}{z^2}-1\right)}\right)}{2z^{a+b+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right){\left(1-\frac{1}{z^2}\right)}^{\frac{a}{2}+\frac{b}{2}+\frac{1}{2}}{2}^a}\&comma;\left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(a+1\right)\mathrm{HeunG}\left(0\&comma;0\&comma;1+\frac{a}{2}\&comma;\frac{a}{2}+\frac{1}{2}\&comma;0\&comma;1\&comma;\frac{1}{z^2}\right)}{2z^{a+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(a+b+1\right)\mathrm{HeunG}\left(0\&comma;0\&comma;1+\frac{a}{2}+\frac{b}{2}\&comma;\frac{a}{2}+\frac{b}{2}+\frac{1}{2}\&comma;0\&comma;1+b\&comma;\frac{1}{z^2}\right)}{2z^{a+b+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\mathrm{\pi }\mathrm{JacobiP}\left(-1-\frac{a}{2}\&comma;\frac{1}{2}+a\&comma;0\&comma;\frac{z^2-2}{z^2}\right)2^{1+a}}{4z^{1+a}{2}^a\sin \left(\frac{\mathrm{\pi }\left(2+a\right)}{2}\right)}\&comma;\mathrm{with\; no\; restrictions\; on}\left(a\&comma;z\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(1+b+a\right)\mathrm{\Gamma }\left(-\frac{b}{2}-\frac{a}{2}\right)\mathrm{JacobiP}\left(-1-\frac{a}{2}-\frac{b}{2}\&comma;\frac{1}{2}+a\&comma;b\&comma;\frac{z^2-2}{z^2}\right)}{2z^{1+b+a}\mathrm{\Gamma }\left(\frac{1}{2}+\frac{a}{2}-\frac{b}{2}\right)2^a}\&comma;\mathrm{with\; no\; restrictions\; on}\left(a\&comma;b\&comma;z\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{\mathrm{\pi }{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\csc \left(b\mathrm{\pi }\right)\left(\mathrm{LegendreP}\left(a\&comma;b\&comma;z\right)-\frac{\mathrm{\Gamma }\left(a+b+1\right)\mathrm{LegendreP}\left(a\&comma;-b\&comma;z\right)}{\mathrm{\Gamma }\left(a-b+1\right)}\right)}{2}\&comma;b::\left(\neg \mathbb{Z}\right)\wedge \left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;+\right)}\right)\wedge \left(a-b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;+\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\mathrm{MeijerG}\left(\left[\left[-\frac{a}{2}\&comma;\frac{1}{2}-\frac{a}{2}\right]\&comma;\left[\right]\right]\&comma;\left[\left[0\right]\&comma;\left[-\frac{1}{2}-a\right]\right]\&comma;-\frac{1}{z^2}\right)}{2z^{1+a}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(a\&comma;z\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{MeijerG}\left(\left[\left[-\frac{b}{2}-\frac{a}{2}\&comma;\frac{1}{2}-\frac{b}{2}-\frac{a}{2}\right]\&comma;\left[\right]\right]\&comma;\left[\left[0\right]\&comma;\left[-\frac{1}{2}-a\right]\right]\&comma;-\frac{1}{z^2}\right)2^b}{2z^{1+b+a}}\&comma;\mathrm{with\; no\; restrictions\; on}\left(a\&comma;b\&comma;z\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\mathrm{I}}{2}\left(-\mathrm{\pi }+2\mathrm{arccot}\left(\frac{\mathrm{I}}{z}\right)\right)\&comma;a=0\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\mathrm{\pi }\sqrt{-\frac{{\left(-1+z\right)}^2}{z^2}}z+2\mathrm{arccoth}\left(\frac{1}{z}\right)z-2\mathrm{arccoth}\left(\frac{1}{z}\right)}{2\left(-1+z\right)}\&comma;a=0\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\mathrm{arctanh}\left(\frac{1}{z}\right)\&comma;a=0\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;z\right)=\frac{\sqrt{\mathrm{\pi }}\mathrm{\Gamma }\left(a+1\right)\mathrm{hypergeom}\left(\left[1+\frac{a}{2}\&comma;\frac{a}{2}+\frac{1}{2}\right]\&comma;\left[\frac{3}{2}+a\right]\&comma;\frac{1}{z^2}\right)}{2z^{a+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\&comma;\left[\mathrm{LegendreQ}\left(a\&comma;b\&comma;z\right)=\frac{{\&ExponentialE;}^{\mathrm{I}b\mathrm{\pi }}\sqrt{\mathrm{\pi }}{\left(z+1\right)}^{\frac{b}{2}}{\left(z-1\right)}^{\frac{b}{2}}\mathrm{\Gamma }\left(a+b+1\right)\mathrm{hypergeom}\left(\left[1+\frac{a}{2}+\frac{b}{2}\&comma;\frac{a}{2}+\frac{b}{2}+\frac{1}{2}\right]\&comma;\left[\frac{3}{2}+a\right]\&comma;\frac{1}{z^2}\right)}{2z^{a+b+1}\mathrm{\Gamma }\left(\frac{3}{2}+a\right)2^a}\&comma;\left(a+b+1\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\wedge \left(\frac{3}{2}+a\right)::\left(\neg {\mathbb{Z}}^{\left(0\&comma;-\right)}\right)\right]\right]$