Why Improper Integrals Need Local Bounds
An integral over an unbounded interval depends on infinitely much of the integrand, so pointwise differentiability in a parameter does not by itself justify moving a derivative inside the integral. The useful condition is local in the parameter: near the value of interest, both the integrand and its parameter derivative must be bounded by functions whose integrals are finite. The derivative bound controls difference quotients, which is what permits the Dominated Convergence Theorem to apply.
We use Lebesgue measure \(m\) on \([0,\infty)\) or \(\mathbb{R}\), with the Lebesgue sigma-algebra. These spaces are sigma-finite and complete, but not finite. The functions in the examples are real-valued. An improper integral below means the Lebesgue integral over the indicated unbounded set; it is finite whenever the absolute value of its integrand is integrable.
A Local Differentiation Criterion
Proof. Since \(|f(t,x)|\le g(x)\) and \(\int_X g\,d\mu<\infty\), each \(f(t,\cdot)\) is integrable, so \(F(t)\) is finite. The function \(x\mapsto\partial_t f(t_0,x)\) is measurable: it is the pointwise limit, as \(n\to\infty\), of the measurable difference quotients \(n(f(t_0+1/n,x)-f(t_0,x))\), after taking \(n\) large enough that \(t_0+1/n\) lies in the neighborhood. Its absolute value is bounded by \(h\), so it too is integrable. Take any sequence \(s_n\ne t_0\) with \(s_n\to t_0\). For all sufficiently large \(n\), \(s_n\in[t_0-\delta,t_0+\delta]\). The pointwise mean value theorem, applied to \(t\mapsto f(t,x)\) between \(t_0\) and \(s_n\), gives \[ \left|\frac{f(s_n,x)-f(t_0,x)}{s_n-t_0}\right|\le h(x). \] The quotient converges pointwise to \(\partial_t f(t_0,x)\). The Dominated Convergence Theorem applies because the quotients are measurable and bounded in absolute value by the integrable function \(h\). Thus their integrals converge to \(\int_X\partial_t f(t_0,x)\,d\mu(x)\). Linearity of the integral for integrable functions identifies each such integral with \((F(s_n)-F(t_0))/(s_n-t_0)\). The sequence criterion for limits now gives the derivative formula: if the difference quotient did not converge to that value as \(s\to t_0\), a sequence \(s_n\to t_0\) violating convergence could be chosen, contradicting what was just proved. \(\square\)
For improper integrals, the central task is often to find the common bound \(h\). A bound valid only at \(t=t_0\) is not enough: the difference quotient uses nearby parameter values. Exponential decay supplies a standard method. If \(t\ge a>0\), then \(e^{-tx}\le e^{-ax}\) for \(x\ge0\); polynomial factors in \(x\) remain integrable against this exponential. The lower bound \(a\) must hold throughout the chosen parameter neighborhood.
Worked Exponential Integrals
Worked Example: An Exponential with a Linear Weight
For \(t>0\), define \(F(t)=\int_0^\infty (1+x)e^{-tx}\,dx\). Differentiate \(F\) by verifying local bounds around an arbitrary \(t_0>0\).
Set \(a=t_0/2>0\) and restrict \(t\) to \([a,3a]\), a neighborhood of \(t_0=2a\). For \(x\ge0\), \[ |(1+x)e^{-tx}|\le(1+x)e^{-ax}, \qquad |\partial_t((1+x)e^{-tx})|=x(1+x)e^{-tx}\le x(1+x)e^{-ax}. \] Both bounds are integrable. In fact, for \(a>0\), integration by parts gives \(\int_0^\infty e^{-ax}\,dx=1/a\), \(\int_0^\infty xe^{-ax}\,dx=1/a^2\), and \(\int_0^\infty x^2e^{-ax}\,dx=2/a^3\). The boundary terms vanish because \(x^k e^{-ax}\to0\) as \(x\to\infty\). The theorem therefore gives \[ F'(t)=-\int_0^\infty x(1+x)e^{-tx}\,dx =-\frac1{t^2}-\frac2{t^3}. \] To check, compute the original integral directly: \[ F(t)=\int_0^\infty e^{-tx}\,dx+\int_0^\infty xe^{-tx}\,dx =\frac1t+\frac1{t^2}. \] Differentiating this explicit expression gives \(-1/t^2-2/t^3\), agreeing with the integral formula. For example, at \(t=2\), the value is \(1/2+1/4=3/4\), and its derivative is \(-1/4-2/8=-1/2\).
Worked Example: A Gaussian Integral
For \(t>0\), let \(G(t)=\int_{\mathbb{R}}e^{-t x^2}\,dm(x)\). Verify the local bounds and find \(G'(t)\).
Fix \(t_0>0\) and write \(a=t_0/2\). For \(t\in[a,3a]\), \[ 0\le e^{-t x^2}\le e^{-a x^2}, \qquad |\partial_t e^{-t x^2}|=x^2e^{-t x^2}\le x^2e^{-a x^2}. \] The right sides are integrable on \(\mathbb{R}\). Indeed, the Gaussian integral is \(\int_{\mathbb{R}}e^{-a x^2}\,dx=\sqrt{\pi/a}\), and integration by parts gives \(\int_{\mathbb{R}}x^2e^{-a x^2}\,dx=\sqrt{\pi}/(2a^{3/2})\). Thus the theorem applies and \[ G'(t)=-\int_{\mathbb{R}}x^2e^{-t x^2}\,dx. \] Substitute \(u=\sqrt{t}\,x\) into the Gaussian integral to obtain \(G(t)=\sqrt{\pi/t}\). The derivative of this explicit expression is \(-\sqrt{\pi}/(2t^{3/2})\). The integral formula agrees: substituting \(u=\sqrt t\,x\) in the integral of \(x^2e^{-t x^2}\) gives \(t^{-3/2}\int_{\mathbb{R}}u^2e^{-u^2}\,du=\sqrt{\pi}/(2t^{3/2})\). At \(t=1\), the integral is \(\sqrt{\pi}\) and its derivative is \(-\sqrt{\pi}/2\); both follow from the formulas just computed.
Worked Example: A Difference of Exponentials
For \(t>0\), set \(H(t)=\int_0^\infty(e^{-tx}-e^{-(t+1)x})\,dx\). Differentiate at an arbitrary \(t_0>0\), taking care that the domination is local.
Let \(a=t_0/2\) and take \(t\in[a,3a]\). The absolute value of the integrand is at most \(e^{-tx}+e^{-(t+1)x}\le2e^{-ax}\), an integrable bound. Its parameter derivative is \(-xe^{-tx}+xe^{-(t+1)x}\), whose absolute value is at most \(xe^{-tx}+xe^{-(t+1)x}\le2xe^{-ax}\), also integrable. The theorem yields \[ H'(t)=\int_0^\infty\bigl(-xe^{-tx}+xe^{-(t+1)x}\bigr)\,dx =-\frac1{t^2}+\frac1{(t+1)^2}. \] For an independent check, integrate each exponential to obtain \(H(t)=1/t-1/(t+1)\); differentiating gives exactly \(-1/t^2+1/(t+1)^2\). At \(t=1\), \(H(1)=1-1/2=1/2\), and \(H'(1)=-1+1/4=-3/4\). The integral formula gives the same derivative, since its two terms are \(-\int_0^\infty xe^{-x}\,dx+\int_0^\infty xe^{-2x}\,dx=-1+1/4=-3/4\).
What the Bounds Do—and Do Not—Say
The theorem is local, not a claim that one integrable bound works for every parameter in the whole domain. For example, the bounds in the first worked example use \(a=t_0/2\); as \(t_0\) varies toward zero, this particular envelope changes and its integral need not remain uniformly bounded. To differentiate at a given \(t_0>0\), however, a single integrable envelope on one neighborhood of \(t_0\) is sufficient.
A common mistake is to differentiate the integrand pointwise and immediately integrate the resulting derivative, without checking a neighborhood-wide integrable bound. Here is a concrete failure. On \([0,\infty)\) with Lebesgue measure, define \(q(t,x)=t/(1+t^2x^2)\) for real \(t\). For each fixed \(t\), \(q(t,\cdot)\) is integrable: at \(t=0\) it is zero, and for \(t\ne0\) the substitution \(u=|t|x\) gives \[ Q(t)=\int_0^\infty q(t,x)\,dx= \begin{cases} \pi/2,&t>0,\\ 0,&t=0,\\ -\pi/2,&t<0. \end{cases} \] Thus \(Q\) is not continuous at zero, and in particular is not differentiable there. But for every fixed \(x\), \(\partial_t q(0,x)=1\), whose integral over \([0,\infty)\) is infinite. The proposed interchange would not even produce a finite derivative. The obstruction is visible in the missing hypothesis: there is no integrable local bound for the parameter derivative. Indeed, \(\partial_tq(t,x)=(1-t^2x^2)/(1+t^2x^2)^2\), which at \(t=0\) equals \(1\) for every \(x\). This example shows why pointwise differentiability and integrability of each parameter slice do not replace local domination.
When applying the criterion to an unbounded interval, check the same two bounds for every parameter in a neighborhood, not just at the parameter where the derivative is wanted. In practice:
- Choose a closed neighborhood that stays inside the parameter domain.
- Use its lower decay rate to bound the integrand and its derivative for all \(x\) in the integration domain.
- Verify that both resulting functions have finite integrals.
- Only then apply the theorem to the difference quotients.
Check Your Understanding
For each question, identify the parameter neighborhood and the integrable bound before using the differentiation formula.
- For \(J(t)=\int_0^\infty e^{-tx}\,dx\), \(t>0\), find a local bound for the integrand and its parameter derivative at \(t_0>0\), then compute \(J'(t_0)\).
- In the Gaussian example, why is \(x^2e^{-a x^2}\) integrable for \(a>0\), and where is that fact used?
- For \(K(t)=\int_0^\infty (1+x+x^2)e^{-tx}\,dx\), propose a common integrable bound for both the integrand and its parameter derivative near \(t_0>0\).
- In the difference-of-exponentials example, why must the neighborhood be chosen so that its lower endpoint is positive?
- For \(q(t,x)=t/(1+t^2x^2)\), which assumption of the local differentiation criterion fails at \(t=0\), and what is \(Q(t)\) for \(t<0\)?