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Math412 Notes Select Topics in Mathematical Analysis of Functions of One or Several Variables

Zheng-Chao Han

Section 5.7 Differentiation of Integrals

Suppose that \(\phi(x, t)\) is differentiable with respect to \(t\) when each \(x\) is held fixed. We are interested in knowing under what conditions the following holds
\begin{equation} D_{t}\int_{a}^{b }\phi(x, t)\; dx = \int_{a}^{b }D_{t}\phi(x, t)\; dx.\tag{5.7.1} \end{equation}
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The question is really a case of Theorem 2.2.7, or our extension Theorem 2.3.2. For it is really asking whether
\begin{equation*} \lim_{h_{n}\to 0} \int_{a}^{b }\frac{\phi(x, t+h_{n})-\phi(x, t)}{h_{n}}\; dx =\int_{a}^{b }D_{t}\phi(x, t)\; dx. \end{equation*}
The condition in Theorem 2.2.7 requires that \([a, b]\) be a finite interval and \(\frac{\phi(x, t+h_{n})-\phi(x, t)}{h_{n}} \to D_{t}\phi(x, t)\) uniformly over \(x\in [a, b]\) as \(n\to \infty\text{.}\) Since the mean value theorem gives us \(\frac{\phi(x, t+h_{n})-\phi(x, t)}{h_{n}}=D_{t}\phi(x, t+\theta h_{n})\) for some \(0 < \theta < 1\) depending on \(x, t, h_{n}\text{,}\) a sufficient condition for this uniform convergence is the following condition: For any \(\epsilon >0\) there exists a \(\delta>0\) such that for all \(x\in [a, b], s \in (t-\delta, t+\delta)\)
\begin{equation*} |D_{t}\phi(x, s)- D_{t}\phi(x, t)| < \epsilon. \end{equation*}
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When the uniform integrability fails, differentiation under the integral may not hold.
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Example 5.7.1. Differentiability of the integral \(\int_{0}^{1}\frac{t^{2}x}{(t+x)^{3}}\,dx\).

The integrand \(f(x, t)=\frac{t^{2}x}{(t+x)^{3}}\) is continuously differentiable in the first quadrant. For any \(0 < x \le 1\text{,}\) \(f(x, t)\to 0\) as \(t\to 0+\text{,}\) so we may extend \(f(x, 0)=0\) to make the resulting function continuous for \(x>0, t\ge 0\) (is it continuous at \((x ,t)=(0, 0)\text{?}\)). We can consider either \(\frac{f(x, t)-f(x, 0)}{t}\) as \(t\to 0+\text{,}\) or the limit of \(D_{t}f(x, t)=\frac{2tx (t+x)-3t^{2}x}{(t+x)^{4}}\) as \(t\to 0+\text{.}\)
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In the particular example here,
\begin{equation*} \int_{0}^{1} \frac{f(x, t)-f(x, 0)}{t} \, dx = \int_{0}^{1} \frac{t x}{(t+x)^{3}}\, dx =\int_{0}^{1/t}\frac{y}{(1+y)^{3}}\, dy \end{equation*}
which tends to \(\int_{0}^{\infty}\frac{y}{(1+y)^{3}}\, dy>0\) as \(t\to 0+\text{.}\) On the other hand, \(\frac{f(x, t)-f(x, 0)}{t} \to 0\) for every \(0 < x \le 1\) as \(t\to 0+\text{.}\) It is instructive to examine how the uniform integrability has failed here.
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This example is constructed based on an earlier example when examining
\begin{equation*} \lim_{n\to \infty} \int_{0}^{1}\frac{n^{2} x}{(1+nx)^{3}}\, dx= \lim_{n\to \infty} \int_{0}^{1}\frac{\frac 1n x}{(\frac 1n + x)^{2}}\, dx. \end{equation*}
We simply replace \(\frac 1n\) by \(t>0\) to get this example.
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When the interval of integration is unbounded, even if we can make \(| \frac{\phi(x,t+h)-\phi(x,t)}{h} - \phi_t(x,t) |< \epsilon\text{,}\) uniformly for all \(x\) in the interval of integration and \(h\) with \(|h|< \delta\text{,}\) this may not guarantee that \(\left| \int_a^b \left( \frac{\phi(x,t+h)-\phi(x,t)}{h} - \phi_t(x,t) \right) dx \right|\) is small when \(a\) or \(b\) is infinite.
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For definiteness of exposition, let’s assume \(a\) to be finite and \(b=\infty\text{.}\) We will also use \(f(x, t)\) in place of \(\phi(x, t)\text{.}\) Our strategy is to break the integral into an integral on a finite integral, which we can handle by the earlier method, and the tail part, which will be made small by some reasonable assumption. More specifically, we assume that the improper integral \(\int_a^{\infty} D_t f(x,s)\,dx\) converges uniformly with respect to \(s\) in some neighborhood \(I\) of \(t\text{,}\) namely,
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For any \(\epsilon \gt 0\text{,}\) there exists \(L\) such that for any \(N>M\ge L\text{,}\) \(| \int_M^{N} D_t f(x,s)\,dx | < \epsilon/4\) for all \(s\in I\text{.}\)
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Proof.

First we break up the integrals into parts and estimate them separately
\begin{align*} \amp \left| \int_a^{\infty} \left( \frac{f(x,t+h)-f(x,t)}{h} - D_t f(x,t) \right) dx \right|\\ \le \amp \left| \int_a^{L} \left( \frac{f(x,t+h)-f(x,t)}{h} - D_t f(x,t) \right) dx \right| \\ \amp + \left| \int_L^{\infty} \frac{f(x,t+h)-f(x,t)}{h} \,dx \right| + \left| \int_L^{\infty} D_t f(x,t) \, dx \right|. \end{align*}
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Next, we apply the fundamental theorem of calculus to \(f(x,t+h)-f(x,t)\) to get
\begin{equation*} f(x,t+h)-f(x,t)=\int_t^{t+h} D_t f(x,s)\,ds, \end{equation*}
so using the uniform integrability of \(|D_t f(x,s)|\) we get, for \(N>L\text{,}\)
\begin{align*} \left| \int_L^{N} \frac{f(x,t+h)-f(x,t)}{h} \,dx \right| \amp= \left| \int_L^{N} h^{-1} \int_t^{t+h} D_t f(x,s)\,ds dx \right| \\ \amp \le |h|^{-1} | \int_t^{t+h} \int_L^{N} D_t f(x,s) dx ds |\\ \amp \le \epsilon/4. \end{align*}
Therefore,
\begin{equation*} \left| \int_L^{\infty} \frac{f(x,t+h)-f(x,t)}{h} \,dx \right| = \left| \lim_{N\to \infty} \int_L^{N} \frac{f(x,t+h)-f(x,t)}{h} \,dx \right| \le \epsilon/4. \end{equation*}
We also have
\begin{equation*} \left| \int_L^{\infty} D_t f(x,t) \, dx \right| < \epsilon/4. \end{equation*}
For \(\left| \int_a^{L} \left( \frac{f(x,t+h)-f(x,t)}{h} - D_t f(x,t) \right) dx \right|\text{,}\) we apply our earlier argument to find a \(\delta >0\) such that when \(0 < |h| < \delta\text{,}\) we have
\begin{equation*} \left| \int_a^{L} \left( \frac{f(x,t+h)-f(x,t)}{h} - D_t f(x,t) \right) dx \right| < \epsilon/2. \end{equation*}
This concludes our proof.
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Example 5.7.3. Differentiation of a Poisson Integral.

Consider the Poisson integral
\begin{equation*} u(s,t)=\int_{\mathbb R} \frac {t \, u(x,0)}{(x-s)^2 + t^2} dx \end{equation*}
on the upper half plane \(\left\{(s,t): t>0\right\}\text{,}\) where \(u(x,0)\) is a given bounded function.
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To consider the differentiability of \(u(s, t)\) for \((s,t)\) near \((s_0,t_0)\) with \(t_0>0\text{,}\) note that the integrand \(f(x,s,t)= \frac {t \, u(x,0)}{(x-s)^2 + t^2}\) is continuous for \((x,s,t) \in \mathbb R \times D_{r}(s_0,t_0)\text{,}\) where \(D_{r}(s_0,t_0)\) is the disk of radius \(r\) centered at \((s_0,t_0)\) with \(0 < r < t_0/2\text{,}\) so that when \((s,t)\in D_{r}(s_0,t_0)\text{,}\) we have \(t>t_0/2\text{;}\) \(f(x,s,t)\) is also differentiable in \((s,t)\) for \((s,t)\in D_{r}(s_0,t_0)\text{.}\)
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To verify the uniform convergence of the improper integral \(\int_{\mathbb R} f_s(x,s,t) \, dx\text{,}\) we need to start with \(\epsilon >0\text{,}\) and look for \(L>0\) such that for any \(N>M>L\text{,}\)
\begin{equation*} \left| \int_{N>|x|>M} f_s(x,s,t) \, dx \right| < \epsilon\quad \text{for all } (s,t)\in D_{r}(s_0,t_0). \end{equation*}
In fact, to verify differentiability in \(s\) at \((s_0,t_0)\text{,}\) it suffices to check the above inequality for \(s\) near \(s_0\) and \(t=t_0\text{.}\) The main focus will be to make sure that \(f_s(x,s,t_0)\) tends to \(0\) at a sufficiently fast rate in \(x\) as \(x\to \infty\text{,}\) with uniform control on the coefficients for \(s\) near \(s_0\text{,}\) such that the above estimate holds; so we will watch out for how \(f_s(x,s,t_0)\) depends on \(x\) and \(s\text{.}\)
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Since
\begin{equation*} |f_s(x,s,t_0) | \le \frac{ 2 t_0 |s-x| |u(x,0)|}{[ (s-x)^2 +t_0^2]^2} \le \frac{ 2 t_0 U |s-x| }{[ (s-x)^2 +t_0^2]^2}, \end{equation*}
where \(U>| u(x,0)|\) for all \(x\in \mathbb R\text{.}\) We argue that when \(s\) is near \(s_0\text{,}\) and \(|x|>M\text{,}\) \(|s-x|\) will be large; more precisely, for \(s\) is sufficiently near \(s_0\text{,}\) we have \(|s|\le 2 |s_0|\text{,}\) and \(|s-x| \ge |x| -|s| \ge M-2|s_0|\ge M/2\text{,}\) if \(M\) is chosen so that \(M> 4|s_0|\text{,}\) then
\begin{align*} \left| \int_{N>|x|>M} f_s(x,s,t) \, dx \right| \amp \le \int_{|x-s| \ge M/2} \frac{ 2 t_0 U |s-x| }{[ (s-x)^2 +t_0^2]^2} dx \\ \amp \le \int_{M/2}^{\infty} \frac{4t_0U z }{[z^2 +t_0^2]^2} dz \end{align*}
which can be made smaller than \(\epsilon\) when \(M\) is sufficiently large, as the integral \(\int_{M/2}^{\infty} \frac{z }{[z^2 +t_0^2]^2} dz\) is convergent.
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Note that \(t_0\) is considered fixed in this argument; there will be difficulty to verify the uniform integrability if we allow \(t\to 0\text{.}\)
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Exercise 5.7.4. Differentiation of the integral \(F(t)=\int_{\mathbb R} \frac{dx}{(1+t^2x^2)(1+x^2)}\).

  • Check whether \(\int_{\mathbb R} D_t f(x,t)\, dx\) satisfies the uniform convergence criterion near \(t>0\) and \(t=0\text{,}\) where \(f(x, t)\) is the integrand.
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  • Determine whether \(F'(0)=\int_{\mathbb R} D_t f(x,0)\, dx\text{.}\)
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