San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

The Asymptotic Sine
Integral of Functions

Consider integrals of the form

J(λ) = ∫LHF(x)sin(λx)dx

What is sought is the condition on F(x) such that J(λ)→0 as λ→∞. The analysis will start with F(x)=xn for n as a nonnegative integer.

First of all, by letting λx=z

In(λ) = ∫LHxnsin(λx)dx
can be transformed into

When Integration-by-Parts (IbP) is applied to ∫znsin(z)dz the result is

∫znsin(z)dz = −zn·cos(z) + nzn-1·sin(z) + n(n-1)∫zn-2sin(z)dz

When this integral is evaluated at the limits of λL and λH and the results substituted into the transformed expression for In(λ) the result is

(1/λn)∫LHxnsin(λx)dx = (1/λ)[−Hncos(H)+Lncos(L)]
+(n/λ²)[ Hn-1sin(H) − Ln-1sin(L)]
+ (n(n-1)/λn+1)∫λLλHzn-2sin(z)dz

Both terms enclosed in square brackets on the RHS of the above are finite so the asymptotic limits of the first two terms on the RHS are zero. Now define Kn as

Kn = limλ→∞[In(λ)]

Then from the previous result

Kn = limλ→∞[In-2(λ)/λ²]
= limλ→∞[Kn-2/λ²]

Thus if Km = 0 then Km+2 = 0.

It is easily shown that K0=0 and K1=0. Therefore Kn=0 for all nonnegative n.

Thus for any function defined by a polynomial series

F(x) = c0 +c1x + c2x2 +... +cnxn

the integral J(λ) = ∫LHF(x)sin(λx)dx has an asymptotic limit of zero as λ increases without limit.

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