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Question Number 122137 by sdfg last updated on 14/Nov/20

Answered by Bird last updated on 14/Nov/20

Γ(x)=∫_0 ^∞  t^(x−1)  e^(−t) dt  ∫_(−∞) ^(+∞)  e^(xt−e^t ) dt  =_(e^t =u)    ∫_0 ^∞  e^(−u)  u^x   (du/u)  =∫_0 ^∞  u^(x−1)  e^(−u)  du =Γ(x)  with x>0

$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} {dt} \\ $$$$\int_{−\infty} ^{+\infty} \:{e}^{{xt}−{e}^{{t}} } {dt}\:\:=_{{e}^{{t}} ={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}} \:{u}^{{x}} \:\:\frac{{du}}{{u}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{u}^{{x}−\mathrm{1}} \:{e}^{−{u}} \:{du}\:=\Gamma\left({x}\right)\:\:{with}\:{x}>\mathrm{0} \\ $$

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