# Differentiating Under the Integral Proof

There are many variations of "differentiating under the integral sign" theorem; here is one:

If $U$ is an open subset of $\mathbb{R}^n$ and $f:U \times [a,b] \rightarrow \mathbb{R}$ is continuous with continuous partial derivatives $\partial_1 f, \dots \partial_n f$ then the function $$\phi(x) = \int^b_a f(x,t)dt$$ is continuously differentiable and $$\partial_i \phi (x) = \int^b_a \partial_i f(x,t)dt$$

Can anyone suggest a textbook that provides a proof of this version of the theorem?

• What are the tools you are allowed to use? For example, you can try a proof with dominated convergence theorem. – Davide Giraudo Jun 25 '12 at 16:03
• I am familiar with the statement of that theorem, but don't know it's proof. I would like a reference that proves a more self-contained proof, inasumuch as one is possible... – ItsNotObvious Jun 25 '12 at 16:05
• IMHO, you'll do faster by proving it yourself. Anyway, try looking at this Wikipedia page: en.wikipedia.org/wiki/Differentiation_under_the_integral_sign (which I find a bit too lengthy, though). – Giuseppe Negro Jun 25 '12 at 16:47

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.

• You need to justify the interchange of limit and integral in the last step. This can be done by exploiting the hypothesis of continuous differentiability of $f$. – Giuseppe Negro Jun 25 '12 at 16:45
• Good catch, @GiuseppeNegro, I'll update my post! – Alex Nelson Jun 25 '12 at 16:52
• Would you mind including the details as to why the fact that $f(x,t)$ is continuously differentiable implies you can swap the limit and the integral? My own feeling is that this is the real heart of the question. – Jason DeVito Jun 25 '12 at 16:58
• @AlexNelson It's unclear how you used the assumption of continuity to come up with your dominating function $g$ in your first proof. You may want to use mean value theorem together with the fact that continuous functions on compact sets are bounded. – Mark Jun 24 '14 at 6:36
• Also, I agree with Mark. In your DCT part, what is the dominating function $g$? This answer is incomplete without that information. – layman May 3 '15 at 15:44