Let f be continuous over the cylinder produced by the disk of s cross the interval [a,b]. Furthermore, for each x in [a,b], f(s,x) is analytic across s.
Let g(s) be the integral of fs(x) from a to b. Since fs is continuous, this is well defined. Is g differentiable, and what is its derivative?
Given an integer n, subdivide [a,b] into n equal subintervals, let x step through these subintervals, find n functions of s corresponding to the n values of x, add up these functions, divide by n, and call the result hn(s). This is a riemann sum, hence hn(s) approaches g(s). Also, each hn is analytic.
Since f is continuous over the entire cylinder, which is closed and bounded, f is uniformly continuous. Given ε, choose n sufficiently large, so that hn(s) is within ε of g(s). This can be done in a manner that does not depend on s. Therefore hn approaches g uniformly, g is analytic, and g′ is the limit of hn′.
The derivative of hn is the sum of the derivatives of f, divided by n. In the limit, this becomes the integral of the derivatives. Therefore the derivative of the integrals equals the integral of the derivatives. We have swapped integration and differentiation.
If f is continuous over a region in the complex plane cross [a,b], with fx analytic, the above holds for each small disk in the region. We can swap integration and differentiation across the entire region.
Define f at a and b and apply the above theorem. Make sure the riemann net always uses the center of each subinterval, rather than the start or end, so we don't run into a or b. After all, we don't know that f(s) is analytic at a or b.
Integration and differentiation are once again interchangeable, and we're almost done. We only need a technical theorem that says the definite integral on [a,b] equals the indefinite integral on (a,b) when the function is bounded, and the image of our closed bounded cylinder is indeed closed and bounded in the complex plane, so we're all right.
This seems rather pointless until you apply it to functions that run from 0 to infinity, dropping quickly to 0. For each s in a region in the complex plane, let fs be a function of x from 0 to infinity that is integrable, And after a while, fs(x)is dominated by a constant over x3. (Any real exponent above 2 is fine; x3 is easy to read.) The earlier assumptions about continuity and analytic cross sections apply.
Replace x with u/(1-u). As u runs from 0 to 1, x runs from 0 to infinity. Integration by substitution brings in a factor of 1/(1-u)2. Let e = f/(1-u)2. Now the integral of e on [0,1) equals the integral of f on [0,∞). Furthermore, e is continuous, and analytic for each u. If we can define e at 1, we can apply the earlier result for the half open interval [0,1).
As u approaches 1, f is bounded by a constant times(1-u)3/u3. This in turn is divided by (1-u)2 to build e(u). The result is (1-u)/u3, which approaches 0. Furthermore, the approach is uniform, and does not depend on s. This is important, because e needs to be continuous over the entire cylinder, not just along each function es. Setting e(1) = 0 does the trick. Integration and differentiation can now be swapped. When differentiating with respect to s, the extra factor 1/(1-u)2 is merely a constant, and doesn't change a thing. so we can take the results for e and carry them back to f, where the integrals run to infinity. As before, integration and differentiation are interchangeable.
Similar results hold for functions that run from -∞ to +∞, provided both tails drop faster than c/x3.