第5章 5.42

(1)

Γ(s+1)=∫0∞e−xxsdx=[−e−xxs]0∞+∫0∞e−x(s⋅xs−1)dx=s∫0∞e−xxs−1dx=sΓ(s)\Gamma_{(s+1)} = \int_0^\infty e^{-x} x^s dx = \left[-e^{-x} x^s\right]_0^\infty + \int_0^\infty e^{-x}(s \cdot x^{s-1})dx = s\int_0^\infty e^{-x} x^{s-1} dx = s\Gamma_{(s)}

(2)

Γ(5)=1⋅2⋅3⋅4=24\Gamma_{(5)} = 1 \cdot 2 \cdot 3 \cdot 4 = 24

解説: r31bn1z