Differentiating Under the Integral Proof

Solution 1:

Isn't the proof sort of "follow your nose"? Let $\Delta x$ be nonzero, consider

$$ \phi(x+\Delta x)-\phi(x) = \int^{b}_{a}f(x+\Delta x,t)-f(x,t)\,\mathrm{d}t$$

Then construct the quotient

$$ \frac{\phi(x+\Delta x)-\phi(x)}{\Delta x} = \frac{\int^{b}_{a}f(x+\Delta x,t)-f(x,t)\,\mathrm{d}t}{\Delta x} $$

But because we do not integrate over $x$, we treat $x$ like a constant. So we can rewrite the integral as

$$ \frac{\phi(x+\Delta x)-\phi(x)}{\Delta x} = \int^{b}_{a}\frac{f(x+\Delta x,t)-f(x,t)}{\Delta x}\,\mathrm{d}t $$

Taking the limit as $\Delta x\to0$ gives us

$$ \frac{\mathrm{d}\phi(x)}{\mathrm{d} x} = \int^{b}_{a}\frac{\partial f(x,t)}{\partial x}\,\mathrm{d}t $$

precisely as desired? [Edit: We can take the limit under the integral sign, as Giuseppe Negro points out, if the function $f(x,t)$ is continuously differentiable in $x$.]

Addendum: Why, oh why, do we need $f(x,t)$ to be continuously differentiable in $x$?

Why can we take this limit? Well, there's a number of different arguments.

One is the Dominated convergence theorem, which states if we have a sequence of functions $f_{n}(t)\to F(t)$ which is "dominated" by some function $g(t)$, meaning $$ |f_{n}(t)|\leq g(t)\quad\mbox{for any }t $$ then we have $$ \lim_{n\to\infty}\int|f_{n}(t)-F(t)|\,\mathrm{d}t=0 $$ which implies $$ \lim_{n\to\infty}\int f_{n}(t)\,\mathrm{d}t=\int F(t)\,\mathrm{d}t. $$ Take $F(t)=\partial f(x,t)/\partial x$ and $f_{n}(t)$ to be $$ f_{n}(t) = \frac{f(x + \varepsilon_{n},t)-f(x,t)}{\varepsilon_{n}} $$ using any sequence $\varepsilon_{n}\to 0$.

Addendum 2: A second different way begins with the observation $$ \int^{b}_{a}\int^{x}_{0}\frac{\partial f(y,t)}{\partial y}\,\mathrm{d}y\,\mathrm{d}t = \phi(x)-\phi(0)$$ by the fundamental theorem of calculus. Fubini's theorem lets us switch the order of integration $$ \int^{x}_{0}\int^{b}_{a}\frac{\partial f(y,t)}{\partial y}\,\mathrm{d}t\,\mathrm{d}y = \phi(x)-\phi(0)$$ Then we can use Leibniz's rule differentiating both sides with respect to $x$. This gives us the desired result $$ \int^{b}_{a}\frac{\partial f(x,t)}{\partial x}\,\mathrm{d}t = \phi'(x).$$

Recall Leibniz's rule states if $G(x) = \int^{x}_{0}g(y)\,\mathrm{d}y$ then $$ G'(x) = g(x). $$ We can prove this quickly by $$ \frac{G(x+\Delta x)-G(x)}{\Delta x} = \frac{1}{\Delta x}\int^{x+\Delta x}_{x} g(y)\,\mathrm{d}y$$ and taking $\Delta x$ to be "sufficiently small", we can approximate the Riemann sum as $$ \int^{x+\Delta x}_{x} g(y)\,\mathrm{d}y\approx g(c)\Delta x$$ where $x\leq c\leq x+\Delta x$. Plugging this back in gives us $$ \frac{G(x+\Delta x)-G(x)}{\Delta x} = \frac{1}{\Delta x}\left(g(c)\Delta x\right) = g(c)$$ Taking $\Delta x\to 0$ gives us $c\to x$, and $$ G'(x) = g(x)$$ as desired.