1
$\begingroup$

enter image description hereenter image description here

I tried to understand the reasoning given in it but I couldn't understand it. It says that "as the gradient operation involves x and not the integration variable x', it can be taken outside the integral sign".

$\endgroup$

1 Answer 1

1
$\begingroup$

You are essentially using Leibinz integral rule

$$ {\displaystyle {\frac {{\rm d}}{{\rm d}x}}\left(\int _{a}^{b}f(x,x')\,{\rm d}x'\right)=\int _{a}^{b}{\frac {\partial }{\partial x}}f(x,x')\,{\rm d}x'.} $$

$\endgroup$
3
  • $\begingroup$ I know this but what does the statement mean that it does not involve integration variable x'? $\endgroup$ Commented Dec 27, 2018 at 12:31
  • $\begingroup$ @HaaranAjgaonkar It means that you are integrating w.r.t $x'$, but taking the derivative w.r.t $x$, so the rule above applies $\endgroup$
    – caverac
    Commented Dec 27, 2018 at 12:40
  • $\begingroup$ Thank you very much, I was a little bit confused by the language of the context. $\endgroup$ Commented Dec 27, 2018 at 12:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.