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Question-124389




Question Number 124389 by Engr_Jidda last updated on 03/Dec/20
Answered by Dwaipayan Shikari last updated on 03/Dec/20
∫_0 ^∞ x^m e^(−ax^n ) dx       ax^n =u⇒anx^(n−1) =(du/dx)  =(1/(an))∫_0 ^∞ x^(m+1−n) e^(−u) du  =(1/(an))∫_0 ^∞ ((u/a))^((m+1−n)/n) e^(−u) du =(1/(a^((m+1)/n) n))∫_0 ^∞ u^((m+1−n)/n) e^(−u) du  =(1/(a^((m+1)/n) n))Γ(((m+1)/n))
$$\int_{\mathrm{0}} ^{\infty} {x}^{{m}} {e}^{−{ax}^{{n}} } {dx}\:\:\:\:\:\:\:{ax}^{{n}} ={u}\Rightarrow{anx}^{{n}−\mathrm{1}} =\frac{{du}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{{an}}\int_{\mathrm{0}} ^{\infty} {x}^{{m}+\mathrm{1}−{n}} {e}^{−{u}} {du} \\ $$$$=\frac{\mathrm{1}}{{an}}\int_{\mathrm{0}} ^{\infty} \left(\frac{{u}}{{a}}\right)^{\frac{{m}+\mathrm{1}−{n}}{{n}}} {e}^{−{u}} {du}\:=\frac{\mathrm{1}}{{a}^{\frac{{m}+\mathrm{1}}{{n}}} {n}}\int_{\mathrm{0}} ^{\infty} {u}^{\frac{{m}+\mathrm{1}−{n}}{{n}}} {e}^{−{u}} {du} \\ $$$$=\frac{\mathrm{1}}{{a}^{\frac{{m}+\mathrm{1}}{{n}}} {n}}\Gamma\left(\frac{{m}+\mathrm{1}}{{n}}\right) \\ $$
Commented by Engr_Jidda last updated on 03/Dec/20
thank you?sir
$$\mathrm{thank}\:\mathrm{you}?\mathrm{sir} \\ $$

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