How to show that functions of this type are strictly decreasing
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 $g(r,x) = \exp(r^2) (rx)^{1/2}$, 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:
$$\frac1{x+1}\int_x^\infty \frac{\mathrm e^{r^2}}{\sqrt{rx}}\mathrm dr=\frac1{x+1}\int_0^\infty\frac{\mathrm e^{(u+x)^2}}{\sqrt u}\mathrm du\;.$$
Now the bound doesn't depend on $x$, and the integral of the derivative of the integrand with respect to $x$ is welldefined. Of course you can always make this substitution, but unless $g(r,x)$ contains $rx$, it just rearranges the terms without reducing the work.
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seporhau
Updated on March 12, 2020Comments

seporhau over 3 years
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)?

Admin over 11 yearsDerive using Fundamental theorem of Calculus.

Admin over 11 yearsSadly, 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 $f$.

Admin over 11 yearsAs for your actual question, I believe you will need to differentiate under the integral sign.

Davide Giraudo over 11 yearsWhat is $g$ in your particular problem?

seporhau over 11 yearsHey 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 over 11 yearsWhat would you mean by $g(x,x)$ if $g(r,x) = \exp(r^2) (tx)^{1/2}$ or something like that? The integral from $x$ to $\infty$ is welldefined in this case, but the value $g(x,x)$ is not welldefined. What to do?

joriki over 11 years@seporhau: I edited the answer in response to your comment.