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Question Number 148816 by ArielVyny last updated on 31/Jul/21

∫_0 ^∞ x^m e^(ix^n ) dx=??

$$\int_{\mathrm{0}} ^{\infty} {x}^{{m}} {e}^{{ix}^{{n}} } {dx}=?? \\ $$

Answered by mathmax by abdo last updated on 01/Aug/21

A_m =∫_0 ^∞ x^m e^(ix^n ) dx changement ix^n =−z give −x^n =−iz ⇒ x^n =iz ⇒x=i^(1/n) z^(1/n) ⇒A_m =∫_0 ^∞ i^(m/n) z^(m/n) e^(−z) (1/n)i^(1/n) z^((1/n)−1) dz =i^((m+1)/n) ∫_0 ^∞ z^(((m+1)/n)−1) e^(−z) dz but Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt (x>0) ⇒ A_m =(e^((iπ)/2) )^((m+1)/n) Γ(((m+1)/n)) =e^((iπ(m+1))/(2n)) ×Γ(((m+1)/n))

$$\mathrm{A}_{\mathrm{m}} =\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{m}} \:\:\mathrm{e}^{\mathrm{ix}^{\mathrm{n}} } \mathrm{dx}\:\:\mathrm{changement}\:\mathrm{ix}^{\mathrm{n}} \:=−\mathrm{z}\:\mathrm{give}\:−\mathrm{x}^{\mathrm{n}} \:=−\mathrm{iz}\:\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{n}} \:=\mathrm{iz}\:\Rightarrow\mathrm{x}=\mathrm{i}^{\frac{\mathrm{1}}{\mathrm{n}}} \:\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{n}}} \:\Rightarrow\mathrm{A}_{\mathrm{m}} =\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{i}^{\frac{\mathrm{m}}{\mathrm{n}}} \:\mathrm{z}^{\frac{\mathrm{m}}{\mathrm{n}}} \:\mathrm{e}^{−\mathrm{z}} \frac{\mathrm{1}}{\mathrm{n}}\mathrm{i}^{\frac{\mathrm{1}}{\mathrm{n}}} \:\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} \:\mathrm{dz} \\ $$$$=\mathrm{i}^{\frac{\mathrm{m}+\mathrm{1}}{\mathrm{n}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{z}^{\frac{\mathrm{m}+\mathrm{1}}{\mathrm{n}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{z}} \:\mathrm{dz}\:\:\:\mathrm{but}\:\Gamma\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{x}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\:\left(\mathrm{x}>\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{A}_{\mathrm{m}} =\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{m}+\mathrm{1}}{\mathrm{n}}} \Gamma\left(\frac{\mathrm{m}+\mathrm{1}}{\mathrm{n}}\right)\:=\mathrm{e}^{\frac{\mathrm{i}\pi\left(\mathrm{m}+\mathrm{1}\right)}{\mathrm{2n}}} \:×\Gamma\left(\frac{\mathrm{m}+\mathrm{1}}{\mathrm{n}}\right) \\ $$

Commented by ArielVyny last updated on 01/Aug/21

nice sir but you have forgot (1/n) if you look second line then the answer is ∫_0 ^∞ x^m e^(ix^n ) dx=(1/n)Γ(((m+1)/n))e^((iπ(m+1))/(2n))

$${nice}\:{sir}\:{but}\:{you}\:{have}\:{forgot}\:\frac{\mathrm{1}}{{n}}\:{if}\:{you}\:{look} \\ $$$${second}\:{line}\:{then}\:{the}\:{answer}\:{is} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{m}} {e}^{{ix}^{{n}} } {dx}=\frac{\mathrm{1}}{{n}}\Gamma\left(\frac{{m}+\mathrm{1}}{{n}}\right){e}^{\frac{{i}\pi\left({m}+\mathrm{1}\right)}{\mathrm{2}{n}}} \\ $$

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