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

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.

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.

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.

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.