Good day all, here is a question I have concerning Leibniz's rule for differentiation under the integral sign. Let \begin{align} G(x)=\int^{\beta(x)}_{\alpha(x)}g(x,t)dt\end{align} (a.) Place conditions on $\alpha,\;\beta$ and $g$ so that $G'(x)$ exists for $0<x<1.$

(b.) Give the formula for computing $G'(x)$ and prove its correctness.


(a.) $\alpha(x)$ and $\beta(x)$ must be continuous and must have continuous derivatives for $0<x<1.$ Also, $g(x,t)$ must be a function such that $g_x(x,t)$ are continuous in $t$ and $x$ in some region of $(x,t)-$plane including $\alpha(x)\leq x \leq \beta(x),\;0<x<1.$

(b.) \begin{align} G'(x)&=\dfrac{d}{dx}\left(\int^{\beta(x)}_{\alpha(x)}g(x,t)dt\right)\\&=g(x,\beta(x))\dfrac{d}{dx}\beta(x)-g(x,\alpha(x))\dfrac{d}{dx}\alpha(x)+\dfrac{\partial}{\partial x}\int^{\beta(x)}_{\alpha(x)}g(x,t)dt\end{align}

MY QUESTION IS: How do I prove its correctness? Can anyone help me out? Proofs and references will be highly appreciated. Thanks


Hint: Assume there exists a function $H$ so that $\frac{\partial H}{\partial t} = g$. Then $\int_I g(x,t) dt = H(x,t)]_I$ then expand on the interval H, then differentiate the two terms you'll get.


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