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I'm a beginner in integrals and derivatives, please let me know if my understandings are incorrect.

If I write $$g(x)=\int f(x)dx$$ that means $g(x)$ is the anti-derivative or indefinite integral of $f(x)$, and the derivative $\dfrac{dg(x)}{dx}=f(x)$.

If I write $$h=\int^\infty_{-\infty} f(x)dx$$ then $h$ is a constant if $f(x)$ is given, so $\dfrac{dh}{dx}$ is not defined.

If we think of $h$ as a functional of $f$, what is the functional derivative $\dfrac{dh}{df}$?

If $f_\theta$ is governed by parameter $\theta$, then $h$ can be thought of as a function of $\theta$, how should I compute $\dfrac{dh(\theta)}{d\theta}$? does it involve $\dfrac{dh}{df}$?

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If $h(\theta) = \int_{-\infty}^\infty f(x,\theta) dx$, then, if $f$ is nice enough (i.e., the integral always exists, and $f$ is smooth enough...), then $$\frac{dh}{d\theta} = \int_{-\infty}^\infty \frac{df}{d\theta}(x,\theta) dx$$

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  • $\begingroup$ thanks a lot, could you give a link or something where I can learn the derivatives of this kind $\frac{d}{d\theta}\int_{-\infty}^\infty f(x,\theta) dx$ in general? because when I google "integral derivatives" it's mostly about something like $\frac{d}{d\theta}\int_a^\theta f(x) dx$? $\endgroup$ – dontloo Feb 14 '17 at 7:56
  • $\begingroup$ @dontloo I would start here: en.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$ – 5xum Feb 14 '17 at 8:02
  • $\begingroup$ looks great, thanks $\endgroup$ – dontloo Feb 14 '17 at 8:09

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