My professor gave me to prove this statement:
Let $A(t,x) \in C^1([0,1]\times\Bbb R, \Bbb R).$
Then $\displaystyle \frac{\partial}{\partial x}\int_0^1A(t,x)\ dt = \int_0^1\frac{\partial A}{\partial x}(t,x) \ dt$
This is my proof, I'm not sure if it is completely correct.
$\displaystyle \frac{\int_0^1A(t,x+h) \ dt - \int_0^1A(t,x) \ dt}{h} =\int_0^1 \frac{A(t,x+h)-A(t,x)}{h} \ dt \stackrel{Lagrange}{=} \int_0^1 \frac{\partial A}{\partial x}(t,\xi_{t,h}) \ dt \quad $ where $\ \xi_{t,h} \in [x,x+h]$ and depends on $t$.
I want to prove that $\forall \ x \in \Bbb R \quad \forall \ \epsilon>0 \quad \exists \ h>0 \ : \displaystyle \left| \frac{\partial A}{\partial x}(t,\xi_{t,h})-\frac{\partial A}{\partial x}(t,x) \right| < \epsilon \quad \quad(1)$
Let $r>0$ and $f_t(x) \stackrel{def}{=} \displaystyle \frac{\partial A}{\partial x}(t,x)$. Then $f_t$ is uniformously continuous in $\overline B(x,r)$ so $\exists \ \delta > 0 : |x_1-x_2| < \delta \implies |f_t(x_1)-f_t(x_2)|<\epsilon$. If we pick $h<\delta$ we have (1) so $$\displaystyle \left|\int_0^1\frac{\partial A}{\partial x}(t,\xi_{t,h}) \ dx - \int_0^1\frac{\partial A}{\partial x}(t,x) \ dx \right| =\left|\int_0^1\left[\frac{\partial A}{\partial x}(t,\xi_{t,h})-\frac{\partial A}{\partial x}(t,x) \right] \ dx \right| < \epsilon$$ that is the thesis.
I'm not sure if my proof is correct since i was a bit confused by the use of quantificator. Thank you in advance
I have a question about derivative under the integral sign, perhaps someone could help me with this issue. The problem is the following: Let us consider two intervals $I_{x} \in (0, +\infty)$ and $I_{y} \in (0, +\infty)$ and a function $f(x,y)$ that is smooth enough over $I_{x}\times I_{y}$ in the sense that is integrable and one can take derivatives as well. I would like to calculate the following derivative: \begin{equation} \frac{\partial}{\partial x} \int^{\infty}_{x} f(x,y)dy \end{equation}
I was using the Leibniz's rule but I think that I am doing some wrong because the results that I obtained It does not seem to be right. Does anyone know how to do such computation?
Sincerely,
Martin.