# Derivate of generalized function

1. Apr 25, 2008

### greisen

Hi,

I have to show that if the derivate f'(x) of a generalized function f(x) is defined by the sequence f'_n(x) where f(x) is defined

$$f_n(x)[\tex] then [tex]\int_{-\infty}^{\infty}f'(x)F(x) dx = - \int_{-\infty}^{\infty}f(x)F'(x) dx$$

I use the limits for generalized functions and get

$$lim_{n \to \infty} \int_{-\infty}^{\infty}f'_n(x)F(x) dx = - \int_{-\infty}^{\infty}f_n(x)F'(x) dx$$

which should show the above - I am a liltte confused where the minus sign comes from?
$$- \int_{-\infty}^{\infty}f(x)F'(x) dx$$

Any help appreciated - thanks in advance

2. Apr 25, 2008

### lurflurf

it is from integration by parts
let f,F be functions
d(fF)=(fF)'dx=f'Fdx+fF'dx
f'Fdx=d(fF)-fF'dx
integrating gives (assume f,F->0)
ʃf'Fdx=ʃd(fF)-ʃfF'dx
ʃd(fF)=0 so
ʃf'Fdx=-ʃfF'dx
now extent this to generalized functions by limits