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Exact(6)
We need to prove that the functions under the integral signs are measurable.
The two-variable unknown function F that must be determined appears inside and outside the integral signs.
The proofs of measurability of the functions that stay under integral signs (Lemmas 1-5) use ideas from [5] (Chapters 2-5) and [6] (Chapter 2).
It is important to point out that if the unknown function appears only inside the integral signs, the resulting equation is of first kind.
where we have used the divergence theorem along with ϕ=0 on ∂ Ω to arrive at the first and last equalities, and the uniform convergence of (tilde {boldsymbol {x}}^{i}) to interchange the limit and the integral signs in the first equality.
Because of the assumption of the unique existence of the maximizer, Here, the exchangeability of differential and integral signs is assured by A3.
Similar(53)
The functions under integral sign are all integrable, and regardless of the values of or, the left-hand side of (3.29) tends to zero as.
Define ℓ = d - 1 2. It then follows that f ∈ C ℓ ( 0, ∞ ) and we can write f ( t ) = ∫ 0 ∞ r d - 1 + ℓ Y ( r ) σ ^ ( r t ) d r. (8)In other words, we can differentiate under the integral sign ℓ times since the successive derivatives are integrable, due to (7) and (5).
Some of the resultant integrals are evaluated by using the Gauss Chebyshev integration rules after moving the series coefficients to the outside of the integral sign; others are evaluated exactly, including the modified hypersingular integral.
In addition, the integral can be differentiated with respect to x ∈ R under the integral sign.
That (iii) implies (i) follows from differentiating under the integral sign and then applying Theorem 1.12 in [5].
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