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Can someone explain to me the difference between the results of $ A$ and $B$, where $$A=\frac{d}{dc} \int_{-\infty}^c xf(x) dx $$ $$B= \frac{d}{dc} \int_c^{+\infty} xf(x) dx $$

You can image $f(x)$ to be the probability density function of X random variable. So that $F(c) = \int_{-\infty}^c f(x)$

Is it correct to write A as follows $$ A=\frac{d}{dc}\Big[x \int_{-\infty}^c f(x)dx -\int_{-\infty}^c \int_{-\infty}^c f(x)dx \Big]\\ =\frac{d}{dc}x \int_{-\infty}^c f(x) dx -\frac{d}{dc}\int_{-\infty}^c \int_{-\infty}^c f(x)dx\\ =xf(c)-F(c) $$

and B as follows $$ B= \frac{d}{dc}\Big[x \int_c^{+\infty} f(x)dx -\int_c^{+\infty} \int_c^{+\infty} f(x)dx \Big]\\ =\frac{d}{dc}x \int_c^{+\infty} f(x) dx -\frac{d}{dc}\int_c^{+\infty} \int_c^{+\infty} f(x)dx\\ =\frac{d}{dc}x [1-F(c)] - [1-F(c)]\\ =0+xf(c)-1+F(c) $$

user77005
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Your derivation makes little sense. There is no way for the variable $x$ to leave the integrals like that and those double integrals are not syntactically correct.

Under the assumption that $f$ is a continuous pdf for a random variable $X$ with finite expectation (hence the integrals converge) you can directly apply the Fundamental Theorem of Calculus (which does apply to convergent improper integrals, see this question) to get that:

$$A = cf(c)$$

and

$$B = -cf(c).$$

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John Coleman
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  • I was trying to apply a integral of multiple functions like $\int u(x)v(x)$. So I am trying to understand the link you gave me. Just for clarification generally $F(c) = P(X<=c)= \int_{-\infty}^cf(x)$ right? Is this understanding of mine correct? – user77005 Jul 27 '16 at 01:05
  • Yes, your understanding is correct. The cdf is the antiderivative of the pdf – John Coleman Jul 27 '16 at 20:26