Question Number 102089 by Ar Brandon last updated on 06/Jul/20
$$ \\ $$ $$\mathrm{The}\:\mathrm{Gamma}\:\mathrm{function}\:\Gamma\left(\alpha\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{follows}; \\ $$ $$\Gamma\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{y}^{\alpha−\mathrm{1}} \mathrm{e}^{−\mathrm{y}} \mathrm{dy}\:,\:\alpha>\mathrm{0} \\ $$ $$\mathrm{a}\backslash\:\mathrm{Show}\:\mathrm{that}\:\Gamma\left(\alpha+\mathrm{1}\right)=\alpha\Gamma\left(\alpha\right). \\ $$ $$\mathrm{b}\backslash\mathrm{Conclude}\:\mathrm{that}\:\Gamma\left(\mathrm{n}\right)=\left(\mathrm{n}−\mathrm{1}\right)!\:,\:\mathrm{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:... \\ $$ $$\mathrm{c}\backslash\mathrm{Determine}\:\Gamma\left(\mathrm{55}\right). \\ $$
Answered by Ar Brandon last updated on 06/Jul/20
![a\Γ(α+1)=∫_0 ^∞ y^α e^(−y) dy=[y^α ∫e^(−y) dy]_0 ^∞ −∫_0 ^∞ {(dy^α /dy)∙∫e^(−y) dy}dy =[−y^α e^(−y) ]_0 ^∞ +∫_0 ^∞ αy^(α−1) e^(−y) dy=α∫_0 ^∞ y^(α−1) e^(−y) dy =αΓ(α)](Q102090.png)
$$\mathrm{a}\backslash\Gamma\left(\alpha+\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{y}^{\alpha} \mathrm{e}^{−\mathrm{y}} \mathrm{dy}=\left[\mathrm{y}^{\alpha} \int\mathrm{e}^{−\mathrm{y}} \mathrm{dy}\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \left\{\frac{\mathrm{dy}^{\alpha} }{\mathrm{dy}}\centerdot\int\mathrm{e}^{−\mathrm{y}} \mathrm{dy}\right\}\mathrm{dy} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[−\mathrm{y}^{\alpha} \mathrm{e}^{−\mathrm{y}} \right]_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \alpha\mathrm{y}^{\alpha−\mathrm{1}} \mathrm{e}^{−\mathrm{y}} \mathrm{dy}=\alpha\int_{\mathrm{0}} ^{\infty} \mathrm{y}^{\alpha−\mathrm{1}} \mathrm{e}^{−\mathrm{y}} \mathrm{dy} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\alpha\Gamma\left(\alpha\right) \\ $$
Answered by floor(10²Eta[1]) last updated on 06/Jul/20
$$\left.\Gamma\left(\mathrm{55}\right)=\mathrm{54}!\:\:\:\:\:\:\::\right) \\ $$