Exact smoothness condition necessary for differentiation under integration sign to hold.

Question

Theorem. Let $f(x, t)$ be a function such that both $f(x, t)$ and its partial derivative $f_x(x, t)$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x), x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then for $x_0 ≤ x ≤ x_1$:

$$\frac{\mathrm{d}}{\mathrm{d}x}\int_{a(x)} ^{b(x)} f(x,t)\,\mathrm{d}t =f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)} ^{b(x)}\frac{\partial f}{\partial x}(x,t)\,\mathrm{d}t$$

This comes from wikipedia, but on

http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign gives the basic result, but as the it is remarked at the top of the the wiki page.

it is noted that

the specific problem is: vague conditions for applicability; insufficiently rigorous proof that appears to be separated vaguely into possibly non-exhaustive cases;generally unclear presentation;reference to a proof of the fundamental theorem of calculus without specifying which proof.

In particular, the $f_x(x, t)$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane bit is confusing. Does this mean $f_x$ needs to be jointly continuous in $(x,t)$ or does it need to be component-wise continuous?

I would like an authentic source, e.g. a textbook for a proof for the statement and proof of this result.

Answer 1

The problem is to find $$ \frac{d}{dx}F(a(x),b(x),x), $$ where $$ F(u,v,x) = \int_{u}^{v}f(x,t)\,dt. $$ Fix $x$ and assume that $f(x,t)$ is Riemann integrable on $[u-\delta,v+\delta]$ for some $\delta > 0$ and is continuous at $t=u$, $t=v$. Then, by the Fundamental Theorem of Calculus, the following exist: $$ \begin{align} \frac{\partial}{\partial u}F(u,v,x) & =-f(x,u),\\ \frac{\partial}{\partial v}F(u,v,x) & =f(x,v).\\ \end{align} $$ Fix $u$ and $v$, and assume that $f_{x}(x,t)$ is jointly-continuous in $x$, $t$ for $(x,t)\in(x-\delta,x+\delta)\times (u,v)$ for some $\delta > 0$. Then, for $0 < |h| < \delta$, the following steps are justified: $$ \begin{align} \frac{1}{h}\{F(u,v,x+h)-F(u,v,x)\}& =\frac{1}{h}\int_{u}^{v}f(x+h,t)-f(x,t)\,dt \\ & = \frac{1}{h}\int_{u}^{v}\int_{x}^{x+h}f_{x}(x',t)\,dx'\,dt \\ & = \frac{1}{h}\int_{x}^{x+h}\int_{u}^{v}f_{x}(x',t)\,dt\,dx'. \end{align} $$ The above assumptions also guarantee that the function $x'\mapsto \int_{u}^{v}(x',t)\,dt$ is continuous in $x'$ for $x'\in(x-\delta,x+\delta)$. Therefore, the Fundamental Theorem of Calculus and the above combine to give the existence of the partial derivative of $F$ with respect to $x$: $$ \frac{\partial}{\partial x}F(u,v,x)=\int_{u}^{v}f_{x}(x,t)\,dt. $$ Finally: Combine all of the above conditions by assuming that $f(x,t)$ is jointly continuous in $x$, $t$ in an open region $\Omega$ containing $\{ (x,t) : a(x) \le t \le b(x),\;\; x_{0} \le x \le x_{1}\}$. Further assume that $f_{x}$ is continuous on $\Omega$. Then $F(u,v,x)$ is continuously differentiable on $\Omega$ because all of its partial derivatives are continuous on $\Omega$. Finally, assume that $a$ and $b$ are continuously differentiable on $[x_{0},x_{1}]$. The map $x\mapsto F(a(x),b(x),x)$ is then continuously differentiable on $[x_{0},x_{1}]$ because $x\mapsto (a(x),b(x),x)$ is continuously differentiable into $\Omega$, and $F(u,v,x)$ is continuously differentiable on $\Omega$. By the chain rule $$ \begin{align} \frac{d}{dx}F(a(x),b(x),x) & =F_{u}(a(x),b(x),x)a'(x)+F_{v}(a(x),b(x),x)b'(x)+F_{x}(a(x),b(x),x) \\ & = -f(a,b,x)a'(x)+f(a,b,x)b'(x)+\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}(x,t)\,dt. \end{align} $$ The joint continuity of $f$, $f_{x}$ on $\Omega$, and the continuous differentiability of $a$, $b$ on $[x_{0},x_{1}]$ are sufficient to do what you want.

Relaxed Conditions: It is difficult to relax the continuous differentiability of $F$ on $\Omega$, or the continuous differentiability of $a$, $b$ on $[x_{0},x_{1}]$, except to deal in one-sided derivatives at endpoints; it's hard to do without $F$ being differentiable on the closed region, or to relax conditions on $a$, $b$. Joint continuity of the partial derivatives of $F$ are about the only easily-verified conditions which ensure $F$ is differentiable, and that is hard to prove without the requirement that $f$, $f_{x}$ be jointly continuous.

Answer 2

According to

• Kaplan: Advanced Calculus, Chpt. 4.9, p.254ff

The conditions are for the theorem to hold for $t\in[t_1,t_2]$ are

• $a,b$ are $C^1$ on $[t_1,t_2]$
• $f$ is $C^1$ on $\{(x,t): t\in [t_1,t_2], x\in[a(t),b(t)]\}$

A proof is also included.

There is a little disambiguity, as the set on which $f$ has to be $C^1$ is not explicitly given. Here the text says let $f(x,t)$ be as above redirecting to a simpler version of the theorem with constant integral limits. However, if you look at the text and the proof, this is the only way to interpret it.

Answer 3

Here is a set of conditions that does not require $a$, $b$ or $f$ to be $\mathcal C^1$.

Assume for simplicity that $f$ is defined on $\mathbb R^2$ and that the functions $a,b$ are defined on $\mathbb R$. Then, the function $F$ defined by $$F(x)=\int_{a(x)}^{b(x)} f(x,t)\, dt $$ will be differentiable with derivative given by the above formula provided that

(1) the functions $a$ and $b$ are differentiable;

(2) $f(x,t)$ is continuous with respect to $t$;

(3) $\frac{\partial f}{\partial x}(x,t)$ exists at all points;

(4) for any compact intervals $I,J\subset\mathbb R$, there exists an integrable function $g:J\to\mathbb R^+$ such that $$\left\vert \frac{\partial f}{\partial x}(x,t)\right\vert\leq g(t)\quad {\rm on}\quad I\times J $$

As you know, the idea is to consider the function of 3 variables defined by $$\Phi(u,v,x)=\int_u^v f(x,t)dt\, .$$

By condition (2) and the fundamental theorem of calculus, $\frac{\partial\Phi}{\partial v}$ exists at all points and is given by $\frac{\partial\Phi}{\partial v}(u,v,x)=f(x,v)$. Likewise, $\frac{\partial\Phi}{\partial u}$ exists at all points and is given by $\frac{\partial\Phi}{\partial u}=-f(x,u)$. Finally, conditions (3) and (4) imply (by the "standard" theorem on differentiating under the integral sign) that $\frac{\partial\Phi}{\partial x}$ also exists at each point and is given by $\frac{\partial\Phi}{\partial x}(u,v,x)=\int_u^v \frac{\partial f}{\partial x}(x,t)\, dt$.

Now, let us show that in fact the function $\Phi$ is differentiable. If we can do this, then, we'll get the required result by (1) since $F(x)=\Phi(a(x),b(x),x)$.

Let us fix $(u,v,x)\in\mathbb R^3$. We have to check that $$\Phi(u+\delta u, v+\delta v, x+\delta x)-\Phi(u,v,x)=L(\delta u,\delta v,\delta x)+o(\Vert (\delta u,\delta v,\delta x)\Vert) $$ as $(\delta u,\delta v,\delta x)\to (0,0,0)$, where $$L(\delta u,\delta v,\delta x)=-f(x,u)\,\delta u +f(x,v)\, \delta v+\left(\int_u^v \frac{\partial f}{\partial x}(x,t)\, dt\right) \delta x\, .$$

Without loss of generality, we may assume that $u,v$ $u+\delta$, $v+\delta v$ belong to some fixed compact interval $J$, and that $x, x+\delta x$ belong to some fixed compact interval $I$. So we may use the integrable function $g$ from (4).

Write $$\alpha(\delta u,\delta v,\delta x):=\Phi(u+\delta u,v+\delta v,x+\delta x)-\Phi(u,v,x)-L(\delta u,\delta v,\delta x)\, .$$

We have \begin{eqnarray} \alpha(\delta u,\delta v,\delta x)&=&-\left(\int_u^{u+\delta u} f(x+\delta x, t)\, dt -f(x,u)\, \delta u\right)\\& &+\int_v^{v+\delta v} f(x+\delta x, t)\, dt -f(x,v)\, \delta v\\ & &+\int_u^v \left( f(x+\delta x,t)-f(x,t)-\delta x\,\frac{\partial f}{\partial x}(x,t)\right) dt\\ &:=& \varepsilon_1 +\varepsilon_2 +\eta\, . \end{eqnarray}

We may write \begin{eqnarray}-\varepsilon_1&=& \int_u^{u+\delta u} \left(f(x+\delta x,t)-f(x,t)\right)dt+\int_u^{u+\delta u}f(x,t)\, dt-f(x,u)\, \delta u\\ &=&\int_u^{u+\delta u} \delta x\, \frac{\partial f}{\partial x}(z(\delta x,t), t)\, dt+\int_u^{u+\delta u} f(x,t)\, dt -f(x,u)\, \delta u\, , \end{eqnarray} where $z(\delta x,t)$ lies between $u$ and $u+\delta u$, and hence belongs to $I$. Using (4), (1) and the fundamental theorem of calculus, it follows that $$ \varepsilon_1=O(\delta u\delta x)+o(\delta u)=o(\Vert(\delta u,\delta v, \delta x)\Vert)\, .$$

In the same way, we get $$\varepsilon_2=o(\Vert(\delta u,\delta v,\delta x)\Vert\, .$$

Finally, using (4) and the dominated convergence theorem we obtain that $$\eta =o(\delta x)=o(\Vert (\delta u,\delta v,\delta x)\Vert)\, .$$

Altogether, the map $\Phi$ is indeed differentiable, and the proof is complete.