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In calculus, the Leibniz integral rule for differentiation under the integral sign, named after Gottfried Wilhelm Leibniz, states that for an integral of the form () (,), where < (), < and the integrands are functions dependent on , the derivative of this integral is expressible as (() (,)) = (, ()) (, ()) + () (,) where the partial derivative indicates that inside the integral, only the ...
The proof of the general Leibniz rule [2]: 68–69 proceeds by induction. Let f {\displaystyle f} and g {\displaystyle g} be n {\displaystyle n} -times differentiable functions. The base case when n = 1 {\displaystyle n=1} claims that: ( f g ) ′ = f ′ g + f g ′ , {\displaystyle (fg)'=f'g+fg',} which is the usual product rule and is known ...
The fundamental theorem of calculus is a theorem that links the concept of differentiating a function (calculating its slopes, or rate of change at each point in time) with the concept of integrating a function (calculating the area under its graph, or the cumulative effect of small contributions). Roughly speaking, the two operations can be ...
The problem of the differentiation of integrals is much harder in an infinite-dimensional setting. Consider a separable Hilbert space (H, , ) equipped with a Gaussian measure γ. As stated in the article on the Vitali covering theorem, the Vitali covering theorem fails for Gaussian measures on infinite-dimensional Hilbert spaces. Two results of ...
Here all the integrands are continuous on the circle S(z, r), which justifies differentiation under the integral sign. In particular, if f is once complex differentiable on the open set U , then it is actually infinitely many times complex differentiable on U .
This visualization also explains why integration by parts may help find the integral of an inverse function f −1 (x) when the integral of the function f(x) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx.
It's possible to prove the result using elementary calculus by applying the differentiation under the integral sign technique to an integral due to Freitas: [10] = (+ +). While the primitive function of the integrand cannot be expressed in terms of elementary functions, by differentiating with respect to α {\displaystyle \alpha } we arrive at
(The proof shows it is not necessary to take to be the sup over ... Thus, after the differentiation under the integral sign, the claimed estimate follows.