Consider the integral I ( x ) = <msubsup> &#x222B;<!-- ∫ --> <mrow class="MJX-

Sonia Gay

Sonia Gay

Answered question

2022-06-22

Consider the integral
I ( x ) = 0 exp ( s 2 / 2 ) cos ( x s + λ s 3 ) d s ,

Answer & Explanation

kejohananws

kejohananws

Beginner2022-06-23Added 19 answers

This is an elaboration on Maxim's comment. Your integral is
I ( x ) = 1 2 + exp ( 1 2 s 2 + i λ s 3 + i x s ) d s .
With the change of variables s = t ( 3 λ ) 1 / 3 + 1 6 i λ , we find
I ( x ) = 1 2 1 ( 3 λ ) 1 / 3 exp ( x 6 λ + 1 108 λ 2 ) + i / ( 6 λ ) + + i / ( 6 λ ) exp ( i ( t 3 3 + 1 ( 3 λ ) 1 / 3 ( x + 1 12 λ ) t ) ) d t .
Pushing the contour downwards and using the know integral representation of the Airy function, we deduce
I ( x ) = 1 2 1 ( 3 λ ) 1 / 3 exp ( x 6 λ + 1 108 λ 2 ) + exp ( i ( t 3 3 + 1 ( 3 λ ) 1 / 3 ( x + 1 12 λ ) t ) ) d t = 1 ( 3 λ ) 1 / 3 exp ( x 6 λ + 1 108 λ 2 ) + cos ( t 3 3 + 1 ( 3 λ ) 1 / 3 ( x + 1 12 λ ) t ) d t = π ( 3 λ ) 1 / 3 exp ( x 6 λ + 1 108 λ 2 ) Ai ( 1 ( 3 λ ) 1 / 3 ( x + 1 12 λ ) ) .
Now we know that
Ai ( z ) = 1 2 π z 1 / 4 exp ( 2 z 3 / 2 3 ) ( 1 + O ( 1 z 3 / 2 ) )
as z in the sector | arg z | π δ). Since
2 3 1 3 λ ( x + 1 12 λ ) 3 / 2 = 2 x 3 / 2 3 3 λ + x 1 / 2 12 3 λ 3 / 2 + 1 576 3 λ 5 / 2 x 1 / 2
and
( 3 λ ) 1 / 12 ( x + 1 12 λ ) 1 / 4 = ( 3 λ ) 1 / 12 x 1 / 4 ( 1 1 48 λ x + ) ,
we obtain
I ( x ) = π 2 ( 3 λ x ) 1 / 4 exp ( 2 x 3 / 2 3 3 λ + x 6 λ x 1 / 2 12 3 λ 3 / 2 + 1 108 λ 2 ) ( 1 + O ( 1 x 1 / 2 ) )
as x in the sector | arg x | π δ). More terms in the expansion can be obtained if necessary.

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