DLMF:5.19#Ex4 (Q2205)

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DLMF:5.19#Ex4
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    test
    S = 2 ⁢ π 1 2 ⁢ n ⁢ r n - 1 Γ ⁡ ( 1 2 ⁢ n ) = n r ⁢ V . 𝑆 2 superscript 𝜋 1 2 𝑛 superscript 𝑟 𝑛 1 Euler-Gamma 1 2 𝑛 𝑛 𝑟 𝑉 {\displaystyle{\displaystyle S=\frac{2\pi^{\frac{1}{2}n}r^{n-1}}{\Gamma\left(% \frac{1}{2}n\right)}=\frac{n}{r}V.}}
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    DLMF:5.19#Ex4
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