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I have been trying to calculate how an author of a book gets RHS from LHS. He states that he does a integration by parts. He states that when $x \rightarrow \pm~\infty$ function $f(x) \rightarrow 0$. $f(x)^*$ is conjugate of $f(x)$ but i don't think it plays a significant role here.

$$ \int\limits_{-\infty}^{\infty} \left( \frac{d\,f(x)}{dx} {f(x)}^* - \frac{d\,{f(x)}^*}{dx} f(x) \right) dx = 2 \int\limits_{-\infty}^{\infty} \frac{d\, f(x)}{dx} f(x)^* \, dx $$

I have tried this and all i have managed to do was to 1st write separate integrals for the difference in brackets and 2nd try to calculate last integral by parts:

$$ \int\limits_{-\infty}^{\infty} \left( \frac{d\,f(x)}{dx} {f(x)}^* - \frac{d\,{f(x)}^*}{dx} f(x) \right) dx = \int\limits_{-\infty}^{\infty} \frac{d\,f(x)}{dx} {f(x)}^* \, dx - \underbrace{\int\limits_{-\infty}^{\infty} \frac{d\,{f(x)}^*}{dx} f(x) \, dx}_{\text{by parts}} = \dots $$


$$ \int\limits_{-\infty}^{\infty} \underbrace{f(x)}_{u} \,\underbrace{\frac{d\,{f(x)}^*}{dx} \, dx}_{dv} = \underbrace{u\cdot v\Bigg|^{\infty}_{-\infty} - \int\limits_{-\infty}^{\infty} v\, du}_{\text{i used standard by parts formula}} = \underbrace{f(x)\cdot \frac{d\, f(x)^*}{dx} \Bigg|^{\infty}_{-\infty}}_{=0 ~ ???} - \int\limits_{-\infty}^{\infty} \frac{d \, f(x)^*}{dx} \, d f(x) $$

This gets weird especialy the last part ( i mean $df(x)$). I am not sure if i calculated $v$ and $du$ corectly... Could someone fix me i am sure i did something completely wrong.

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It's all right except for the very last term (you probably misplaced the asterisk). Remember, that $v\, du = v(x)\, u'(x)\, dx$. Thus $$\int\limits_{-\infty}^{\infty} v\, du = \int\limits_{-\infty}^{\infty} f^*(x)\, \frac{df(x)}{dx}\ dx$$ which is exactly what you want.

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  • $\begingroup$ Where did u get this from : $v\, du = v(x)\, u'(x)\, dx$ ??? $\endgroup$
    – 71GA
    Mar 21, 2013 at 16:47
  • $\begingroup$ I mean iit was never crystal clear to me how i can calculate $v$ and $du$ if i know $u$ and $dv$ in a formula $\int u \, dv = uv - \int v \,du$. PLEASE someone explain it to me on this specific case. $\endgroup$
    – 71GA
    Mar 21, 2013 at 16:53
  • $\begingroup$ I checked some videos and i figured out that i didnt even properly knew the formula for integration by parts. So this is now solved :) $\endgroup$
    – 71GA
    Mar 21, 2013 at 17:46
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You forgot change the $dv$ to $v$ in the last step.

$$ \int\limits_{-\infty}^{\infty} \underbrace{f(x)}_{u} \,\underbrace{\frac{d\,{f(x)}^*}{dx} \, dx}_{dv} = 0-\int\limits_{-\infty}^{\infty} \underbrace{\frac{d\,f(x)}{dx}dx}_{du} \,\underbrace{{f(x)}^*}_{v} $$

Other than that, you just need to use this (notational) identity: $d\,f(x)=\frac{d\,f(x)}{dx}dx$

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  • $\begingroup$ Where can i read more about diferential notation stuff like this: $d\,f(x)=\frac{d\,f(x)}{dx}dx$ $\endgroup$
    – 71GA
    Mar 21, 2013 at 16:49
  • $\begingroup$ $\int f(x) d g(x)$ comes from Stieltjes integrals, and it's not always equal to the Rienmann integral $\int f(x) g'(x) d x$, but people tend to use them interchangeably in a slight abuse of notation. $\endgroup$
    – ebsddd
    Mar 22, 2013 at 1:35

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