A.3)
Suppose that $ y>0, \ x \in (\frac{1}{2}, \frac{3}{2}) $ and $ 0<h<\frac{1}{4} $, say.
Then$ -\frac{e^{-(x+h)y}-e^{-xy}}{h} \frac{1}{y^3+1} =- \frac{d}{dx} [e^{-xy}]_a \frac{1}{y^3+1} = \frac{ye^{-ay}}{y^3+1} $ by MVT for some $ 0<a<2. $
$ =\frac{ye^{-ay}}{y^3+1} \leq \frac{y}{y^3+1} $, which is integrable on $ (0, \infty). $
So,
$ F'(x) = -lim_{h \rightarrow 0} - \frac{ \int_0^\infty \frac{e^{-(x+h)y}}{y^3+1} dy - \int_0^\infty \frac{e^{-(x)y}}{y^3+1} dy}{h} $
$ = - lim_{h \rightarrow 0} \int_0^\infty - \frac{e^{-(x+h)y}-e^{-xy}}{h}\frac{1}{y^3+1} dy $ $ = - \int_0^\infty lim_{h \rightarrow 0} -\frac{e^{-(x+h)y}-e^{-xy}}{h}\frac{1}{y^3+1} dy $ (see * Below) $ = \int_0^\infty lim_{h \rightarrow 0} \frac{e^{-(x+h)y}-e^{-xy}}{h}\frac{1}{y^3+1} dy $ $ = \int_0^\infty \frac{d}{dx}[e^{-xy}]\frac{1}{y^3+1} dy $ $ = \int_0^\infty - \frac{ye^{-xy}}{y^3+1} dy $
So $ F'(1)= - \int_0^\infty \frac{ye^{-y}}{y^3+1} dy, $ and since the integrand is positive and integrable (since $ \frac{y}{y^3+1} $ is integrable)
$ F'(1) $ exists and is negative.
(*)
We can put the limit inside because the integrand is nonnegative and is dominated by $ \frac{y}{y^3+1} $ which is integrable on $ (0,\infty ) $ as shown above.
--Wardbc 13:30, 24 July 2008 (EDT)
