How to find $ \frac{d}{dt} \left [\int_t^\infty xf(x)\,dx \right ]$ when $f$ is a probability density function? How can I solve this? I need intermediate equations. Maybe the answer is $-tf(x)$.
$$ \frac{d}{dt} \left [\int_t^\infty xf(x)\,dx \right ] $$
$f(x)$ is probability density function.
That is to say, $\lim\limits_{x \to \infty} f(x) = 0$ and $\lim\limits_{x \to \infty} F(x) = 1$
source: http://www.actuaries.jp/lib/collection/books/H22/H22A.pdf p.40
Trying intermediate equations below:
$$ \frac{d}{dt} \left [\int_t^\infty xf(x)\,dx \right ] = \frac{d}{dt} \left [\left [xF(x) \right ]_t^\infty - \int_t^\infty F(x)\,dx  \right ]??
$$
$$ \frac{d}{dt} \int_t^a f(x)\,dx = -\frac{d}{dt} \int_a^t f(x)\,dx = -\frac{d}{dt} (F(t)-F(a))=F'(t)=f(t)$$
 A: Solved...
$\displaystyle \dfrac{d}{dt} \left [\int_t^\infty xf(x)~dx \right ]$
$= \displaystyle \dfrac{d}{dt} \left [G(\infty)-G(t) \right ]$
$= \displaystyle \dfrac{d}{dt} \left [G(\infty) \right ] - \dfrac{d}{dt} \left [G(t) \right ]$
$= \displaystyle 0-tf(t)$
Thank you all!!!
A: By definition, the derivative (if it exists) is the limit of the difference quotient
$$\frac{1}{h}\left(\int_{t+h}^\infty x f(x) dx - \int_t^\infty x f(x) dx\right) = -\frac{1}{h}\int_t^{t+h} x f(x) dx$$
as $h\to 0$.
Assuming $f$ is continuous within an interval $[t, t+h)$ for sufficiently small $h\gt 0$, $x f$ will also be continuous throughout this interval.  Then the Mean Value Theorem asserts there is some $h^{*}$ between $0$ and $h$ for which
$$-(t+h^{*})f(t+h^{*}) = -\frac{1}{h}\int_t^{t+h} x f(x) dx.$$
As $h\to 0$, necessarily $h^{*}\to 0$, and the continuity of $f$ near $t$ then implies the left hand side has a limit equal to $-t f(t)$.
(It is nice to see that this analysis requires no reasoning about the existence of the original improper integral $\int_t^\infty x f(x) dx$.)
However, even when a distribution has a density $f$, that density does not have to be continuous.  At points of discontinuity, the difference quotient will have different left and right limits: the derivative does not exist there.

This is not a matter that can be dismissed as being some arcane mathematical "pathology" that practitioners can ignore.  The PDFs of many common and useful distributions have points of discontinuity.  For example, the Uniform$(a,b)$ distribution has discontinuous PDF at $a$ and $b$; a Gamma$(a,b)$ distribution has a discontinuous PDF at $0$ when $a\le  1$ (which includes the ubiquitous Exponential distribution and some of the $\chi^2$ distributions); and so on.  Therefore, it is important not to assert, without careful qualifications, that the answer is merely $-t f(t)$: that would be a mistake.
