Como encontrar

Como posso resolver isso? Eu preciso de equações intermediárias. Talvez a resposta seja $-tf(x)$ .

$$\frac{d}{dt} \left [\int_t^\infty xf(x)\,dx \right ]$$

$f(x)$ é a função de densidade de probabilidade.

Ou seja, $\lim\limits_{x \to \infty} f(x) = 0$ e $\lim\limits_{x \to \infty} F(x) = 1$

fonte: http://www.actuaries.jp/lib/collection/books/H22/H22A.pdf p.40

Tentando equações intermediárias abaixo:

$$\frac{d}{dt} \left [\int_t^\infty xf(x)\,dx \right ] = \frac{d}{dt} \left [\left [xF(x) \right ]_t^\infty - \int_t^\infty F(x)\,dx \right ]??$$

$$\frac{d}{dt} \int_t^a f(x)\,dx = -\frac{d}{dt} \int_a^t f(x)\,dx = -\frac{d}{dt} (F(t)-F(a))=F'(t)=f(t)$$

Por definição, a derivada ( se existir ) é o limite do quociente de diferença

$$\frac{1}{h}\left(\int_{t+h}^\infty x f(x) dx - \int_t^\infty x f(x) dx\right) = -\frac{1}{h}\int_t^{t+h} x f(x) dx$$

como h → 0 .$h\to 0$

Assumindo que f é contínuo dentro de um intervalo [ t , t + h ) para h > 0 suficientemente pequeno , x f também será contínuo ao longo deste intervalo. Então o Teorema do Valor Médio afirma que existe algum h ∗ entre 0 e $f$$[t, t+h)$$h\gt 0$$x f$$h^{*}$$0$$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.

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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.

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!!!