Newman's Tauberian Theorem
Let be a bounded function. Assume that its Laplace transform (which is well-defined by this formula for ) admits an analytic extension (which we'll denote by the same letter ) to some open domain containing the closed half-plane . Then, the integral converges and its value equals .
For every , let . The function is defined and analytic on the entire complex plane . The conclusion of the theorem is equivalent to the assertion . We choose some large , and some arbitrarily small such that is defined on the set Let be the counterclockwise contour on the boundary of this set. Let be the restriction of this contour to the half-plane . Let be the restriction of the contour to the set and let be the restriction to the set Let , as shown in the diagram below.
By the Cauchy integral formula, we have where We will estimate this integral separately in the left and right half-planes. In principle, could be arbitrary, but we have chosen to make it easier to estimate this integral.
We first estimate the difference for . Let be an upper bound for . In the the right half-plane , we note that
Thus, we should kill the denominator for the integral to converge. On the other hand, we can afford the kernel growth as in the right half-plane, which will allow us corresponding decay in the left half-plane. Hence our choice This is convenient because for , so that kills the unpleasant denominator on .
We then have
To estimate the integral over , we note that is analytic in the left half-plane, so we may change the integration path to the left semicircle of radius . Now, on , we have Then as before,
Now, let be an upper bound for the quantity on . Then for , and Therefore But as becomes arbitrarily large, the last term vanishes, so that We can make arbitrarily small, so that the second term vanishes. Then we pick an arbitrarily large , so that the first term vanishes, and the theorem follows.