Note that the left side of Equation 15 is the integrand of Sn(a), shown in Equation 6. Now, we can replace Jn(u) in the left bracketed term by the right side of Equation 9b. This gives: Note that both terms in square brackets in Equation 16 contain a function times its derivative. This is equal to one-half of the derivative of the square of the function. Thus, Substituting the right side of Equation 17 into Equation 6 gives: The left bracketed term is simply the integral of a derivative. The right bracketed term can be integrated by parts. This gives:
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