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

∫_0 ^∞ 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))

Commented byArielVyny 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))