0
$\begingroup$

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.

MY ANSWER

(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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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