DLMF:25.12#Ex2 (Q7736)

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DLMF:25.12#Ex2
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    test
    G s ⁢ ( x ) = Li s + 1 ⁡ ( e x ) . subscript 𝐺 𝑠 𝑥 polylogarithm 𝑠 1 superscript 𝑒 𝑥 {\displaystyle{\displaystyle G_{s}(x)=\mathrm{Li}_{s+1}\left(e^{x}\right).}}
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    DLMF:25.12#Ex2
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