Question Number 62416 by mathmax by abdo last updated on 20/Jun/19
$${let}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} {e}^{−{t}} \:{dt}\:\:\:{with}\:{x}>\mathrm{1}\:{calculate}\:\Gamma^{\left({n}\right)} \left({x}\right)\:{for}\:{all}\:{integr}\:{n}. \\ $$
Commented bymathmax by abdo last updated on 23/Jun/19
![the function Γ is C^∞ on ]0,+∞[ we have Γ(x) =∫_0 ^∞ e^((x−1)ln(t)) e^(−t) dt ⇒ Γ^((1)) (x) =∫_0 ^∞ ln(t) t^(x−1) e^(−t) dt and by recurence we get Γ^((n)) (x) =∫_0 ^∞ (ln(t))^n t^(x−1) e^(−t) dt ∀ n ≥1 .](Q62566.png)
$$\left.{the}\:{function}\:\Gamma\:{is}\:{C}^{\infty} \:\:{on}\:\right]\mathrm{0},+\infty\left[\:\:{we}\:{have}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{\left({x}−\mathrm{1}\right){ln}\left({t}\right)} \:{e}^{−{t}} \:{dt}\:\Rightarrow\right. \\ $$ $$\Gamma^{\left(\mathrm{1}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:{ln}\left({t}\right)\:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:\:{dt}\:\:\:\:{and}\:{by}\:{recurence}\:{we}\:{get} \\ $$ $$\Gamma^{\left({n}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\left({ln}\left({t}\right)\right)^{{n}} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:\:{dt}\:\:\:\forall\:{n}\:\geqslant\mathrm{1}\:. \\ $$