# IE2

Approximate

One method: Expand both and in
Taylor series; in powers of , and
in powers of . We can save
ourselves some effort if we note from the integral equation that
; and from repeatedly differentiating the integral
equation, that (so the coefficient of the
term in the expansion is 1), and that
for (because we get
the sine in the integrand for those terms). Thus, the Taylor series expansion
for (about ) has the form
.

The expansion for is
.

Inserting these in the integral equation, we get

or after some expanding and rearrangement (and subtracting the
term from both sides)

.

All of the integrals on the right side are readily evaluated.
For example, retaining only terms through , we get

Equating coefficients for each power of on the left and
right sides gives a system of equations for the coefficients
Upon solving that system, we get
. Repeating
the calculation, this time including terms through , gives
and leaves the lower-order coefficients essentially
unchanged, evidence that the above approximation for is quite good
for .