Chapter 2, Problem 6

[dstefan]

Chapter 2, Problem 6

Let h(x) be a continuous function such that
\int_{-\infty}^{\infty} |x h(x)| dx
exists.
Define g(t) by
g(t) ~=~ \int_{t}^{\infty} (x-t) h(x) ~dx
and show that g''(t) = h(t).


Solution:

Recall that
\frac{d}{dt} \left( ~\int_{a(t)}^{b(t)} f(x,t) ~dx ~\right) = \int_{a(t)}^{b(t)} \frac{\partial f}{\partial t}(x,t) ~dx + f(b(t),t) b'(t) - f(a(t),t) a'(t),
if a(t) and b(t) are differentiable functions and if f(x,t) is a continuous function such that \frac{\partial f}{\partial t}(x,t) exists and is continuous.

A similar result can be derived for improper integrals, i.e.,
\frac{d}{dt} \left( ~\int_{a(t)}^{\infty} f(x,t) ~dx ~\right) ~=~ \int_{a(t)}^{\infty} \frac{\partial f}{\partial t}(x,t) ~dx ~-~ f(a(t),t) a'(t).

For our problem, a(t) = t and f(x,t) = (x-t) h(x), where h(x) is a continuous function. Then, \frac{\partial f}{\partial t}(x,t) = - h(x) is continuous. Note that f(a(t),t) = f(t,t) = 0.
We conclude that
g'(t) ~=~ - \int_{t}^{\infty} h(x) ~dx

Since
\frac{d}{dt} \left( ~\int_{a(t)}^{\infty} f(x) ~dx ~\right) ~=~ - f(a(t)) ~a'(t),
it follows that
g''(t) = h(t).