2
$\begingroup$

This is a step in a proof I thought I understood.

Given that $\lim_{|x|\to \infty} |x\cdot f(x)| = 0$, show

\int_R x f(x) f'(x) \; dx = \left.x\cdot \frac{[f(x)]^2}{2}\right|_{-\infty}^\infty - \int_R \frac{[f(x)]^2}{2}\;dx = - \int_R \frac{[f(x)]^2}{2} \; dx

My best guess was to take

u = x f(x),\quad du = x f'(x)+ f(x),\quad dv = f'(x)dx,\quad v = f(x).

But then where does the factor of $1/2$ come from?

Thanks for hint(s).

  • 0
    Looks to be. I don't believe his answer was there when I posted though.2012-04-01

4 Answers 4

4

Let $I$ equal your original integral. Then when you substitute $u,v,$ and $dv$ you get a $-I$ on the right hand side! When you bring the $-I$ to the left hand side , we have $2I$. From there onwards you know what to do...

  • 0
    Both the answers are fine and I am accepting this because it's a technique I like...and I can't accept both. Thanks!2012-04-01
4

$\dfrac{f(x)^2}{2}$ is the antiderivative of f'(x)f(x).

If dv=f'(x)f(x)\;dx then $v= \dfrac{f(x)^2}{2}$.

  • 0
    I haven't noticed it happening.2012-04-01
2

$u=x\quad v=\frac{f(x)^2}{2}$

2

The problem with your idea is that
v \;du=f(x)\left(xf'(x)+f(x)\right)\;dx=\left(xf(x)f'(x)+f(x)^2\right)\;dx

Try the simpler $u=\dfrac {f(x)^2}2$ and $v=x$.

  • 0
    @daniel: no problem Daniel and Sri Pot's answer was the best concerning your specific change of variable (anyway my answer was the third only...). Cheers,2012-04-01