San José State University

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The Asymptotic Sine Integral of Functions

Consider integrals of the form

J(λ) = ∫_{L} ^{H} F(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)=x^{n} for n as a nonnegative integer.

First of all, by letting λx=z

I_{n} (λ) = ∫_{L} ^{H} x^{n} sin(λx)dx
can be transformed into
(1/λ^{n+1} )∫_{λL} ^{λH} z^{n} sin(z)dz
When Integration-by-Parts (IbP) is applied to ∫z^{n} sin(z)dz the result is

∫z^{n} sin(z)dz = −z^{n} ·cos(z) + nz^{n-1} ·sin(z) + n(n-1)∫z^{n-2} sin(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/λ^{n} )∫_{L} ^{H} x^{n} sin(λx)dx = (1/λ)[−H^{n} cos(H)+L^{n} cos(L)]
+(n/λ²)[ H^{n-1} sin(H) − L^{n-1} sin(L)] + (n(n-1)/λ^{n+1} )∫_{λL} ^{λH} z^{n-2} sin(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 K_{0} =0 and K_{1} =0. Therefore K_{n} =0 for all nonnegative n.

Thus for any function defined by a polynomial series

F(x) = c_{0} +c_{1} x + c_{2} x^{2} +... +c_{n} x^{n}
the integral J(λ) = ∫_{L} ^{H} F(x)sin(λx)dx has an
asymptotic limit of zero as λ increases without limit.