Consider integrals of the form

J(λ) = ∫_L^HF(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

I_n(λ) = ∫_L^Hxⁿsin(λx)dx
can be transformed into
(1/λⁿ⁺¹)∫_λL^λHzⁿsin(z)dz

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

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

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

(1/λⁿ)∫_L^Hxⁿsin(λx)dx = (1/λ)[−Hⁿcos(H)+Lⁿcos(L)]
+(n/λ²)[ H^n-1sin(H) − L^n-1sin(L)]
+ (n(n-1)/λⁿ⁺¹)∫_λL^λHz^n-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 K_n as

K_n = lim_λ→∞[I_n(λ)]

Then from the previous result

K_n = lim_λ→∞[I_n-2(λ)/λ²]
= lim_λ→∞[K_n-2/λ²]

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

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

Thus for any function defined by a polynomial series

F(x) = c₀ +c₁x + c₂x² +... +c_nxⁿ

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