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From: OsherD on 31 Dec 2009 10:02
From Osher Doctorow

The Laplace transform integrates an argument which is a function times
exp(-s), and its inverse integrates a function times exp(+s), so the
latter is closer to the idea of P(X > x), asymptotic E(X-->Y), etc.
since it doesn't decay for s real, unlike exp(-s), asymptotically.

We already see some arguably interesting results with standard
integrals:

1) I[x exp(ax)]dx = [1/a^2][ax - 1]exp(ax) + k, k constant of
integration, I...dx indefinite integral.
2) I[exp(ax)sin(bx)]dx = [1/(a^2 + b^2)]exp(ax)[a sin bx - b cos bx] +
k
3) Analogously to (2) with sine and cosine interchanged and - on the
right replaced by +.
4) I[x^n exp(x)]dx = x^n exp(x) - n I[x^(n-1) exp(x)]dx (integration
by parts, n positive integer)

Osher Doctorow

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