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)?

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Derive using Fundamental theorem of Calculus. –  user21436 Mar 11 '12 at 12:57
Sadly, English is inconsistent even in mathematics. The operation of finding the derivative of a function is called differentiation, and you are asking how to differentiate $f$. –  Rahul Mar 11 '12 at 13:37
As for your actual question, I believe you will need to differentiate under the integral sign. –  Rahul Mar 11 '12 at 13:40
What is $g$ in your particular problem? –  Davide Giraudo Mar 11 '12 at 15:26
Hey Davide, $g$ is a complicated function. If I put it here, I'm afraid somebody will write the answer down for me. Nothing wrong with that of course, but I'd rather try to do it myself. I didn't know the "differentiate under the integral sign" so I'm understanding that now. –  seporhau Mar 11 '12 at 15:55

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.

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What would you mean by $g(x,x)$ if $g(r,x) = \exp(-r^2) (t-x)^{-1/2}$ or something like that? The integral from $x$ to $\infty$ is well-defined in this case, but the value $g(x,x)$ is not well-defined. What to do? –  seporhau Mar 11 '12 at 15:59
@seporhau: I edited the answer in response to your comment. –  joriki Mar 11 '12 at 16:18