# Prove that $h'(t)=\int_{a}^{b}\frac{\partial\phi}{\partial t}ds$.

Let $\phi:[a,b]\times[c,b]\to \mathbb{R}$ be continuous. Define $h:[c,d]\to \mathbb{R},h(t)=\int_{a}^{b}\phi(s,t)ds$. Assume that $\frac{\partial\phi}{\partial t}$ exists and is continuous on $[a, b]\times[c, b]$. Prove that $h'(t)=\int_{a}^{b}\frac{\partial\phi}{\partial t}ds$.

I have tried to prove it by first differentiating $h(t)$ w.r.t $t$ but it shouldn't be that easy. Is it okay to just differentiate on $h(t)$ w.r.t $t$ directly? Or how to prove it?

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Although one can use measure theory (e.g. the dominated convergence theorem) to prove this, there is however a direct approach since we are dealing with a continuous function. Let $t_0 \in [c,d]$ and $\varepsilon>0$. Since $\displaystyle\frac{\partial\phi}{\partial t}(s,t_0)$ exists, then for each $s \in [a,b]$ there is a $\delta(t_0,s,\varepsilon)>0$ such that $$\left|\frac{\partial\phi}{\partial t}(s,t_0)-\frac{\phi(s,t)-\phi(s,t_0)}{t-t_0}\right|< \frac{\varepsilon}{b-a} \quad \forall (s,t) \in [a,b]\times[c,d],\ |t_0-t|<\delta(t_0,s,\varepsilon).$$ Since $$[a,b]\times[c,d] \subset \bigcup_{(s,t) \in [a,b]\times[c,d]}(s-\varepsilon,s+\varepsilon)\times(t-\delta(t,s,\varepsilon),t+\delta(t,s,\varepsilon))$$ and $[a,b]\times[c,d]$ is compact, therefore there exist $$(s_1,t_1),\ldots, (s_n,t_n) \in [a,b]\times[c,d]$$ such that $$[a,b]\times[c,d] \subset \bigcup_{i=1}^n(s_i-\varepsilon,s_i+\varepsilon)\times(t_i-\delta(t_i,s_i,\varepsilon),t_i+\delta(t_i,s_i,\varepsilon)).$$ Choosing $$\delta_0(\varepsilon)=\min_{1 \le i \le n}\delta(t_i,s_i,\varepsilon),$$ we have for every $t \in [c,d]$ with $|t_0-t|< \delta_0(\varepsilon)$ \begin{eqnarray} \left|\int_a^b\frac{\partial\phi}{\partial t}(s,t_0)ds-\frac{h(t)-h(t_0)}{t-t_0}\right|&=&\left|\int_a^b\left[\frac{\partial\phi}{\partial t}(s,t_0)-\frac{\phi(s,t)-\phi(s,t_0)}{t-t_0}\right]ds\right|\\ &\le&\int_a^b\left|\frac{\partial\phi}{\partial t}(s,t_0)-\frac{\phi(s,t)-\phi(s,t_0)}{t-t_0}\right|ds\\ &\le&\int_a^b\frac{\varepsilon}{b-a}ds=\varepsilon, \end{eqnarray} i.e. $h$ is differentiable at $t_0 \in [c,d]$ and $$h'(t_0)=\int_a^b\frac{\partial\phi}{\partial t}(s,t_0)ds.$$

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I have a question, is the last part somewhat of definition of a differentiable function or what? Why would you find an $\delta_o$ to prove $|...|\le\epsilon$ – Mathematics Dec 10 '12 at 13:09
Yes it is! See en.wikipedia.org/wiki/Limit_of_a_function for the $\varepsilon-\delta$ definition of the limit. – Mercy King Dec 10 '12 at 14:12
Oh i see. I knew that definition of limit but didn't apply it to differentiablity, forgot that it is equivalent saying $\lim_{t\to t_0}\frac{h(t)-h(t_0)}{t-t_0}=f'(t_0)$. – Mathematics Dec 10 '12 at 14:19

Since $\phi$ is continuous and smooth (its derivative is also continuous), and since integration domain is fixed, then you can exchange integral with derivative operator.

$$\nonumber \begin{array}{lcl} h'(t) & = & \frac{\partial h}{\partial t} = \\ & = & \frac{\partial}{\partial t} \int_{a}^{b}\phi(s,t)ds = \\ & = & \int_{a}^{b} \frac{\partial \phi(s,t)}{\partial t}ds & \end{array}$$

This is the Leibniz integral rule.

Look at this and this.

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Is there any theorem or proof support this statement? – Mathematics Dec 10 '12 at 10:30
The Leibniz integral rule is a theorem itself. The second link shows a proof of it. – the_candyman Dec 10 '12 at 10:35
Not to be too pedantic, but I wouldn't call $\phi$ smooth exactly: the continuity of $\frac{\partial \phi}{\partial t}$ does not imply (for example) the continuity of $\frac{\partial \phi}{\partial s}$. – Jesse Madnick Dec 10 '12 at 10:54
that's right, I'd better to say that it is "smooth" with respect to variable $t$. – the_candyman Dec 11 '12 at 10:36

Actually,it's a corollary of the Lebesgue dominated convergence theorem.The exchange is availbale here because $\frac{\partial \phi(s,t)}{\partial t}$ is dominated by a integrable function i.e. it's upper-bound in $[a,b]\times[c,b]$.

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