Derivate of generalized function

greisen
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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] <br /> <br /> then <br /> <br /> \int_{-\infty}^{\infty}f&amp;#039;(x)F(x) dx = - \int_{-\infty}^{\infty}f(x)F&amp;#039;(x) dx<br /> <br /> I use the limits for generalized functions and get<br /> <br /> lim_{n \to \infty} \int_{-\infty}^{\infty}f&amp;#039;_n(x)F(x) dx = - \int_{-\infty}^{\infty}f_n(x)F&amp;#039;(x) dx<br /> <br /> which should show the above - I am a liltte confused where the minus sign comes from?<br /> - \int_{-\infty}^{\infty}f(x)F&amp;#039;(x) dx<br /> <br /> Any help appreciated - thanks in advance
 
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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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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