Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 3 down vote accepted

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 $$ h'(t_0)=\int_a^b\frac{\partial\phi}{\partial t}(s,t_0)ds. $$

share|improve this answer
    
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 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.

\begin{equation} \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} \end{equation}

This is the Leibniz integral rule.

Look at this and this.

share|improve this answer
1  
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
2  
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 $[a,b]\times[c,b]$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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