How to show that functions of this type are strictly decreasing Let $f:[0,\infty)\to \mathbf{R}$ be defined by $$ f(x) = \frac{1}{x+1} \int_x^\infty  g(r,x) dr,$$ where $g(r,x)$ is a "nice" function and all of this makes sense.
Suppose that I want to show that $f(x)$ is strictly decreasing. In my particular problem, this should be the case and I'm trying to prove it. 
One way would be to show that the derivative is negative.
How to derive $f$?
Are there other ways to prove that   $f$ is strictly decreasing (assuming it is)?

There are three ways in which the function $f$ depends on $x$, and the derivative contains one term for each of them:
$$
\begin{eqnarray}
f'(x)
&=&
-\frac1{(x+1)^2}\int_x^\infty g(r,x)\mathrm dr
\\
&&+\frac1{x+1}g(x,x)
\\
&&
+\frac1{x+1}\int_x^\infty\frac{\partial}{\partial x}g(r,x)\mathrm dr\;.
\end{eqnarray}
$$
[Edit in response to the comment:]
I'll assume that you meant $g(r,x) = \exp(-r^2) (r-x)^{-1/2}$, since the version with a $t$ in the denominator wouldn't cause problems at $r=x$.
In such a case, you could obtain a result by replacing the lower bound of the integral by $x+\epsilon$; then two of the terms would go to infinity as $\epsilon\to0$, and you could cancel them before taking that limit. However, a simpler approach would be to substitute:
$$\frac1{x+1}\int_x^\infty \frac{\mathrm e^{-r^2}}{\sqrt{r-x}}\mathrm dr=\frac1{x+1}\int_0^\infty\frac{\mathrm e^{-(u+x)^2}}{\sqrt u}\mathrm du\;.$$
Now the bound doesn't depend on $x$, and the integral of the derivative of the integrand with respect to $x$ is well-defined. Of course you can always make this substitution, but unless $g(r,x)$ contains $r-x$, it just rearranges the terms without reducing the work.
