How to find integral of this abstract function?

$\int_{-\infty}^{u}n(n-1)[F(u)-F(v)]^{n-2}f(u)f(v)dv$

Question

How to find integral of this abstract function?

$\int_{-\infty}^{u}n(n-1)[F(u)-F(v)]^{n-2}f(u)f(v)dv$

kirbykirby · Answer

I got as far as pulling out the constants:

$n(n-1)f(u)\int_{-\infty}^{u}[F(u)-F(v)]^{n-2}f(v)dv$

kirbykirby · Answer

And also, F is the antiderivative of f

kirbykirby · Answer

Apparently the answer is $nf(u)[F(u)]^{n-1}$

abb0t · Answer

$$\int\limits a^n =  \frac{ a^{n+1} }{ n+1 }+k$$

abb0t · Answer

$$n-2+1 = n-1$$

abb0t · Answer

once you have your answer, evaluate from 0 to u and then
take the limit as it approaches ∞

kirbykirby · Answer

I'm not sure how to deal with the f(v) though :S

kirbykirby · Answer

since it's being multiplied

anonymous · Answer

$$n(n-1)\int_{-\infty}^u\left[F(u)-F(v)\right]^{n-2}f(u)f(v)\;dv$$
Assuming F(u) and F(v) are antiderivatives of f(u) and f(v), respectively, make the substitution
$$t=F(u)-F(v)\
dt=-f(v)\;dv$$
That should help you get the form @abb0t was suggesting.

kirbykirby · Answer

:o Oh I see what you mean.. that's a good idea :)