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))