calculate-0-e-x-n-dx-

Question Number 124920 by mathmax by abdo last updated on 07/Dec/20

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}^{\mathrm{n}} } \mathrm{dx}\: \\ $$

Answered by Dwaipayan Shikari last updated on 07/Dec/20

∫_0 ^∞ e^(−x^n ) dx x^n =u⇒nx^(n−1) =(du/dx) ⇒(1/n)∫_0 ^∞ x^(1−n) e^(−u) du =(1/n)∫_0 ^∞ u^((1−n)/n) e^(−u) du =(1/n)Γ((1/n))=Γ(((n+1)/n))

$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{{n}} } {dx}\:\:\:\:\:\:\:\:{x}^{{n}} ={u}\Rightarrow{nx}^{{n}−\mathrm{1}} =\frac{{du}}{{dx}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{1}−{n}} {e}^{−{u}} {du}\:=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{\mathrm{1}−{n}}{{n}}} {e}^{−{u}} {du}\:=\frac{\mathrm{1}}{{n}}\Gamma\left(\frac{\mathrm{1}}{{n}}\right)=\Gamma\left(\frac{{n}+\mathrm{1}}{{n}}\right) \\ $$

Answered by mathmax by abdo last updated on 07/Dec/20

A_n =∫_0 ^∞ e^(−x^n ) dx we do the changement x=t^(1/n) ⇒ A_n =∫_0 ^∞ e^(−t) ×(1/n)t^((1/n)−1) dt =(1/n)∫_0 ^∞ t^((1/n)−1) e^(−t) dt =(1/n)×Γ((1/n))

$$\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}^{\mathrm{n}} } \mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}} \:\Rightarrow \\ $$$$\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{t}} ×\frac{\mathrm{1}}{\mathrm{n}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{n}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}}×\Gamma\left(\frac{\mathrm{1}}{\mathrm{n}}\right) \\ $$

Facebook Tweet Pin

calculate-0-e-x-n-dx-

Leave a Reply Cancel reply