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Let $\phi=\phi(x,t)\colon\,[a,b]\times [0,+\infty) \rightarrow R$ be a continuous function on $[a,b]\times [0,+\infty)$. Suppose also that$\frac{\partial}{\partial t}\phi=\phi_{t}$ exists and is continuous on $[a,b]\times [0,+\infty)$.

I have to show that the function $$\psi\colon [0,+\infty)\longmapsto \mathbb{R},\,\,\,\,\,\,\,\,\,\psi(t):=\int_{a}^{b}\phi(x,t)\,dx$$ is derivable on $[0,+\infty)$ with derivative $$\psi'(t):=\int_{a}^{b}\phi_{t}(x,t)\,dx$$ for all $t \in [0,+\infty)$.

Here's my attempt of proof:

Fix any $t_0\in [0,+\infty)$ and $\epsilon>0$. Consider the restriction of $\psi$

$$\psi_0\colon [t_0,t_0+\epsilon]\longmapsto \mathbb{R},\,\,\,\,\,\,\,\,\,\psi_0(t):=\int_{a}^{b}\phi(x,t)\,dx\,\,\,\,\,\,\,\,\,\,\,t\in[t_0,t_0+\epsilon] $$

By hypothesis, we have that the function $\phi=\phi(x,t):\,[a,b]\times [t_0,t_0+\epsilon] \longmapsto R$ is continuous on $[a,b]\times [t_0,t_0+\epsilon]$ and also $\frac{\partial}{\partial t}\phi=\phi_{t}$ exists and is continuous on $[a,b]\times [t_0,t_0+\epsilon]$.

Now, by Leibniz integration rule, $\psi_0$ is a continuously differentiable function of $t$ on $[t_0,t_0+\epsilon]$ with derivative $$\psi'_{0}(t):=\int_{a}^{b}\phi_{t}(x,t)\,dx\,\,\,\,\,\,\,\,\,\,\,t\in[t_0,t_0+\epsilon].$$ In particular, there exists, and is finite $$\psi'_{0}(t_0)=\lim_{h \to 0^+}{\frac{\psi_{0}(t_0+h)-\psi_{0}(t_0)}{h}}=\int_{a}^{b}\phi_{t}(x,t_0)\,dx.$$

On the other hand $$\lim_{h \to 0^+}{\frac{\psi_{0}(t_0+h)-\psi_{0}(t_0)}{h}}=\lim_{h \to 0^+}{\frac{\psi(t_0+h)-\psi(t_0)}{h}},$$ so, by definition, $\psi$ is derivable at $t_0$ and $\psi'(t_0)=\int_{a}^{b}\phi_{t}(x,t_0)\,dx$ .

Since $t_0 \in [0,+\infty)$ is arbitrary, then we've done.

Does it work for you? Thanks in advance!

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  • $\begingroup$ You should explain what you mean under Leibniz rule. For, according to Wikipedia, Leibniz rule is a statement more general than yours, so there is essentially nothing to prove. $\endgroup$ – user539887 Jun 7 '18 at 19:06
  • $\begingroup$ Incidentally, it is wrongly tagged as differential-equations. $\endgroup$ – user539887 Jun 7 '18 at 19:08

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