Finding Dominating Functions for Parameter Families

Choosing an Envelope Before Taking a Limit

For a parameter-dependent integral, the central question is often not how to compute each integral separately, but whether the family of integrands has one integrable bound near the parameter value of interest. Such a bound is called an integrable envelope. Finding one lets the Dominated Convergence Theorem justify a limit under the integral; a bound on parameter derivatives can similarly justify differentiating under the integral.

Throughout, the general statements concern a measure space \((X,\mathcal{M},\mu)\), with no assumption that \(\mu\) is finite or \(\sigma\)-finite. Functions in the limit test may be real-valued or complex-valued and are measurable; an envelope is a nonnegative measurable function \(G:X\to[0,\infty]\) with \(\int G\,d\mu<\infty\). The examples use Lebesgue measure \(m\) on the stated subsets of \(\mathbb{R}\). All pointwise bounds in the examples hold everywhere on their stated domains.

Envelope test: Let \(t_0\) be a real parameter and let \(f_t:X\to\mathbb{C}\) be measurable for \(t\) in a neighborhood of \(t_0\). Suppose \(f_t(x)\to f_{t_0}(x)\) for \(\mu\)-almost every \(x\) as \(t\to t_0\), and there is one \(G\in L^1(\mu)\) such that \(|f_t(x)|\le G(x)\) for all parameters in that neighborhood and for \(\mu\)-almost every \(x\). Then \(\int f_t\,d\mu\to\int f_{t_0}\,d\mu\). No finiteness assumption on the measure space is required.

Justification. For any sequence \(t_n\to t_0\), the functions \(f_{t_n}\) are measurable, converge to \(f_{t_0}\) almost everywhere, and are dominated by the same integrable \(G\). The Dominated Convergence Theorem therefore gives \(\int f_{t_n}\,d\mu\to\int f_{t_0}\,d\mu\). The sequential criterion for limits of real parameters gives the asserted limit. In particular, the bound must be uniform over the nearby parameters at the same \(x\); a different integrable bound for each \(t\) does not meet this test.

Power-Exponential Families

A common way to find an envelope is to restrict the parameter to a small interval and replace it in the integrand by the endpoint that gives the slowest decay. For example, if \(a\ge a_0>0\) and \(x\ge0\), then \(e^{-ax}\le e^{-a_0x}\). Polynomial factors remain integrable when paired with a sufficiently strong exponential decay. The choice of \(a_0\) must be positive: allowing \(a\) to approach zero can remove the decay that made the bound integrable.

Worked Example: A Power Times an Exponential

On \([0,\infty)\) with Lebesgue measure, consider \(f_a(x)=x^2e^{-ax}\) as \(a\to2\), restricting \(a\) to \([1,3]\). For every \(x\ge0\), \(a\ge1\), so \(0\le f_a(x)\le G(x)=x^2e^{-x}\). The function \(G\) is integrable: integration by parts twice gives

\int_0^\infty x^2e^{-x}\,dx =\left[-x^2e^{-x}\right]_0^\infty+2\int_0^\infty xe^{-x}\,dx =2\left(\left[-xe^{-x}\right]_0^\infty+\int_0^\infty e^{-x}\,dx\right)=2.

The boundary terms vanish because \(x^2e^{-x}\) and \(xe^{-x}\) tend to zero at infinity. For each fixed \(x\), \(f_a(x)\to x^2e^{-2x}\). The envelope test applies, so the integrals converge to \(\int_0^\infty x^2e^{-2x}\,dx\). Substituting \(u=2x\), with \(dx=du/2\), evaluates this limit as

\int_0^\infty x^2e^{-2x}\,dx =\frac18\int_0^\infty u^2e^{-u}\,du =\frac18\cdot2=\frac14.

Indeed, for every \(a>0\), the same substitution gives \(\int_0^\infty x^2e^{-ax}\,dx=2/a^3\), so at \(a=2\) the integral is \(2/8=1/4\), agreeing with the dominated limit.

The envelope can also come from controlling a power near an endpoint. One first checks the exponent at the limiting parameter: on \((0,1)\), \(x^\gamma\) is integrable precisely when \(\gamma>-1\). A nearby family of powers can then be bounded by the least favorable exponent on that parameter interval.

Worked Example: Powers Near a Singular Endpoint

On \((0,1)\) with Lebesgue measure, let \(f_a(x)=x^a\) and take \(a\to0\) with \(a\in[-1/2,1/2]\). Since \(0<x<1\), the function \(x^a\) decreases as \(a\) increases, so \(0\le x^a\le x^{-1/2}\) throughout this parameter interval. The proposed envelope is integrable, since

\int_0^1x^{-1/2}\,dx=\left[2\sqrt{x}\right]_0^1=2.

At every \(x\in(0,1)\), \(x^a\to1\) as \(a\to0\). The envelope test therefore gives \(\lim_{a\to0}\int_0^1x^a\,dx=\int_0^1 1\,dx=1\). This agrees with the direct evaluation \(\int_0^1x^a\,dx=1/(a+1)\), valid here because \(a>-1\); substituting \(a=0\) gives \(1\). The restriction to \(a\ge-1/2\) is useful: it supplies a single integrable bound all the way to the endpoint \(x=0\).

Rational Families and Uniform Tail Bounds

For rational functions on an unbounded domain, it is often effective to split the domain into a bounded region and a tail. On the bounded region, a simple constant bound may work. On the tail, use the size of the denominator to obtain decay, taking care that the estimate holds for every parameter under consideration.

Worked Example: A Shifted Rational Function

On \(\mathbb{R}\) with Lebesgue measure, let \(f_t(x)=1/(1+(x-t)^2)\), with \(t\to0\) and \(|t|\le1\). If \(|x|\ge2\), then \(|x-t|\ge|x|-|t|\ge|x|-1\ge|x|/2\). Thus \(f_t(x)\le4/x^2\) on this tail. On \(|x|<2\), \(f_t(x)\le1\). A single envelope is therefore

G(x)= \begin{cases} 1,&|x|<2,\\ 4/x^2,&|x|\ge2. \end{cases}

It is integrable, because the bounded part contributes \(4\), and the two tails contribute \(4\int_2^\infty x^{-2}\,dx+4\int_{-\infty}^{-2}x^{-2}\,dx=4(1/2)+4(1/2)=4\). Hence \(\int_{\mathbb{R}}G\,dm=8\). For every \(x\), \(f_t(x)\to f_0(x)\), so the envelope test applies. Translation \(y=x-t\) and the antiderivative \(\arctan y\) give, for each \(|t|\le1\),

\int_{\mathbb{R}}\frac{1}{1+(x-t)^2}\,dx =\int_{\mathbb{R}}\frac{1}{1+y^2}\,dy =\left[\arctan y\right]_{-\infty}^{\infty} =\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi.

Thus both the limit of the integrals and the integral of the pointwise limit equal \(\pi\), as the envelope test predicts.

A moving denominator can also be useful when its limiting singularity is integrable. The comparison should be made with the limiting power, not by bounding the denominator away from zero uniformly when that is impossible.

Worked Example: A Rational Family with a Vanishing Parameter

On \((0,1)\) with Lebesgue measure, set \(f_t(x)=x^2/(x+t)\) for \(0<t\le1\), and consider \(t\to0\). Since \(x+t\ge x>0\),

0\le f_t(x)=\frac{x^2}{x+t}\le x.

The envelope \(G(x)=x\) is integrable, with \(\int_0^1x\,dx=1/2\). For every \(x\in(0,1)\), \(f_t(x)\to x\), so the limit of the integrals is \(\int_0^1x\,dx=1/2\). To check this independently, divide \(x^2\) by \(x+t\): \(x^2/(x+t)=x-t+t^2/(x+t)\). Therefore, for \(t>0\),

\int_0^1\frac{x^2}{x+t}\,dx =\frac12-t+t^2\ln\!\left(\frac{1+t}{t}\right).

The last term tends to zero. For \(0<t\le1\), \(\ln((1+t)/t)\le\ln2+\ln(1/t)\), and \(\ln(1/t)\le1/t\); hence \(0\le t^2\ln((1+t)/t)\le t^2\ln2+t\to0\). The direct formula thus also tends to \(1/2\).

Using Derivative Bounds to Find an Envelope

For differentiation under the integral, a useful envelope is often obtained from a bound on the parameter derivative. The following statement specifies the conditions. The functions and parameter derivatives are real-valued; measurability of the difference quotients and of the integrands is assumed.

Derivative test: Let \(f(t,x)\) be defined for \(t\) in an open interval about \(t_0\) and for \(\mu\)-almost every \(x\in X\). Suppose \(f(t_0,\cdot)\) is integrable, \(f(\cdot,x)\) is differentiable on that interval for almost every \(x\), and there is \(G\in L^1(\mu)\) such that \(|\partial_t f(t,x)|\le G(x)\) for all \(t\) in a neighborhood of \(t_0\) and almost every \(x\). Then \(F(t)=\int f(t,x)\,d\mu(x)\) is defined near \(t_0\) and \(F'(t_0)=\int\partial_t f(t_0,x)\,d\mu(x)\).

Proof. The mean value theorem gives \(|f(t,x)-f(t_0,x)|\le |t-t_0|G(x)\) for almost every \(x\). Thus \(f(t,\cdot)\) is integrable: its absolute value is bounded by \(|f(t_0,\cdot)|+|t-t_0|G\), an integrable function. For \(t\ne t_0\), define \(q_t(x)=(f(t,x)-f(t_0,x))/(t-t_0)\). The same mean value theorem gives \(|q_t(x)|\le G(x)\) almost everywhere, uniformly for nearby \(t\). At almost every \(x\), differentiability gives \(q_t(x)\to\partial_t f(t_0,x)\). Apply the Dominated Convergence Theorem to any sequence \(t\to t_0\), \(t\ne t_0\). It yields \(\int q_t\,d\mu\to\int\partial_t f(t_0,\cdot)\,d\mu\). Since \(\int q_t\,d\mu=(F(t)-F(t_0))/(t-t_0)\), this is the claimed derivative formula. \(\square\)

Worked Example: Differentiating a Power-Exponential Integral

On \([0,\infty)\) with Lebesgue measure, let \(f(t,x)=e^{-(1+t)x}\) for \(|t|<1/2\). Its parameter derivative is \(\partial_t f(t,x)=-xe^{-(1+t)x}\). Because \(1+t>1/2\),

|f(t,x)|\le e^{-x/2}, \qquad |\partial_t f(t,x)|\le xe^{-x/2}.

Both bounds are integrable: \(\int_0^\infty e^{-x/2}\,dx=2\), and integration by parts gives \(\int_0^\infty xe^{-x/2}\,dx=4\). Thus the derivative test applies at \(t_0=0\), giving \(F'(0)=\int_0^\infty -xe^{-x}\,dx=-1\); the last integral equals \(-1\) by integration by parts. Independently, direct integration gives \(F(t)=1/(1+t)\) for \(|t|<1/2\), so \(F'(0)=-1/(1+0)^2=-1\). The formula and the derivative bound agree.

When a Proposed Bound Fails

A frequent mistake is to check that each \(f_t\) is integrable and infer that the family has an integrable envelope. These are different claims. For instance, on \([0,\infty)\) with Lebesgue measure, \(f_t(x)=e^{-tx}\) is integrable for each \(t>0\), with integral \(1/t\). But as \(t\to0\), \(f_t(x)\to1\) at every \(x\), and the limit function is not integrable on this infinite-measure space. There cannot be an integrable \(G\) dominating \(f_t\) for all \(t\) in any interval \((0,\delta)\): if there were, Fatou's lemma applied to any sequence \(t_n\downarrow0\) would give \(\int_0^\infty 1\,dx\le\int_0^\infty G\,dx<\infty\), a contradiction.

A specific incorrect bound illustrates the issue. One might claim \(e^{-tx}\le e^{-x}\) for \(0<t<1\); at \(x=1\) and \(t=1/2\), this would say \(e^{-1/2}\le e^{-1}\), which is false. The correction is to check the inequality at the actual parameter range and at the same \(x\). In this case, the pointwise limit itself rules out an integrable envelope. In other problems, an envelope may exist, but only after restricting the parameter range or splitting the domain to obtain a valid tail estimate.

Before exchanging a limit or derivative with an integral, check the relevant conditions explicitly:

Is each integrand measurable, and does it converge almost everywhere to the claimed limit?
Is there one integrable bound valid for all parameters sufficiently near the target?
For differentiation, is the bound on the parameter derivative valid throughout an interval, so it also bounds the difference quotients by the mean value theorem?

Check Your Understanding

For each question, identify the parameter range and domain before proposing an envelope.

On \([0,\infty)\), find an integrable envelope for \(x^3e^{-ax}\) when \(a\to2\) and \(a\in[1,3]\). What is the pointwise limiting function?
On \((0,1)\), for which values of \(\alpha\) is \(x^\alpha\) integrable? Use this to choose a local envelope for \(x^a\) as \(a\to1/2\).
For \(1/(1+(x-t)^2)\) on \(\mathbb{R}\), explain why the bound on the tail must account for every \(|t|\le1\), not just \(t=0\).
For \(x^2/(x+t)\) on \((0,1)\), verify the pointwise limit as \(t\downarrow0\) and state an integrable envelope.
For \(e^{-(1+t)x}\) on \([0,\infty)\), find an integrable bound for the parameter derivative when \(|t|<1/2\), and explain how it controls the difference quotients.