DLMF:10.24.E3 (Q3479)

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DLMF:10.24.E3
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    test
    Γ ⁡ ( 1 + i ⁢ ν ) = ( π ⁢ ν sinh ⁡ ( π ⁢ ν ) ) 1 2 ⁢ e i ⁢ γ ν , Euler-Gamma 1 𝑖 𝜈 superscript 𝜋 𝜈 𝜋 𝜈 1 2 superscript 𝑒 𝑖 subscript 𝛾 𝜈 {\displaystyle{\displaystyle\Gamma\left(1+i\nu\right)=\left(\frac{\pi\nu}{% \sinh\left(\pi\nu\right)}\right)^{\frac{1}{2}}e^{i\gamma_{\nu}},}}
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    DLMF:10.24.E3
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