Calculus · Integrals with a Parameter
An improper integral can converge for every value of a parameter and still produce a badly behaved function. The missing ingredient is uniform convergence. This post explains the idea with pictures in words, gives the two standard tests (Cauchy and Weierstrass), and shows them at work on the Gamma function.
In the previous post we saw that a proper integral \( \int_a^b f(x, m)\,dx \) of a continuous function is automatically a continuous function of \(m\). For improper integrals this fails. Consider
\[ I(m) = \int_0^{\infty} m\,e^{-mx}\,dx \qquad (m \ge 0). \]For \(m > 0\), \( I(m) = \big[-e^{-mx}\big]_0^\infty = 1 \). For \(m = 0\), the integrand is \(0\), so \(I(0) = 0\). The integrand is continuous everywhere, the integral converges for every \(m \ge 0\), and yet
\[ I(m) = \begin{cases} 1, & m > 0,\\ 0, & m = 0 \end{cases} \]jumps at \(m = 0\). What went wrong?
Look at the tail of the integral, the part beyond some point \(b\):
\[ \int_b^{\infty} m\,e^{-mx}\,dx = e^{-mb}. \]For each fixed \(m > 0\), this goes to zero as \(b \to \infty\), which is just convergence. But how far out you have to go depends on \(m\): to make \(e^{-mb} < 0.01\) you need \(b > \frac{\ln 100}{m}\), and as \(m \to 0\) that point runs away to infinity. No single \(b\) works for all \(m\) at once. The mass of the integrand slides off to infinity as \(m \to 0\), and the integral loses track of it.
Uniform convergence is exactly the requirement that one cut-off point works for all parameter values simultaneously.
Compare with ordinary (pointwise) convergence: there, \(M\) may depend on both \(\varepsilon\) and \(m\). The whole difference is the order of the quantifiers: "for every \(m\) there is an \(M\)" versus "there is an \(M\) for every \(m\)".
A practical reformulation: the integral converges uniformly exactly when
\[ \sup_{m_1 < m < m_2}\left|\int_b^{\infty} f(x, m)\,dx\right| \longrightarrow 0 \quad \text{as } b \to \infty. \]So to check uniform convergence, compute (or estimate) the tail, then take the worst case over \(m\).
Chop the half-line \([a, \infty)\) into pieces with points \( a = a_0 < a_1 < a_2 < \cdots \), \( a_n \to \infty \). Then the integral is a sum of proper integrals:
\[ \int_a^\infty f(x, m)\,dx = \sum_{n=0}^{\infty} \int_{a_n}^{a_{n+1}} f(x, m)\,dx. \]This is more than a curiosity: it means every theorem you know about uniformly convergent series of functions has a twin for integrals. The tests below are those twins. (Note the word every: a single subdivision is not enough, because a sum over a coarse subdivision can hide oscillations inside each piece.)
Why it works (the "\(\varepsilon/3\)" argument): split \(I(m) - I(m_0)\) into three parts,
\[ I(m) - I(m_0) = \int_a^b \big[f(x, m) - f(x, m_0)\big]dx + \int_b^\infty f(x, m)\,dx - \int_b^\infty f(x, m_0)\,dx. \]Uniformity is used precisely to fix \(b\) before letting \(m\) approach \(m_0\). In Example 1 on \((0,\infty)\) this is impossible, and that is why \(I\) jumped.
Continuity means we can pass to the limit under the integral:
\[ \lim_{m\to m_0}\int_a^\infty f(x, m)\,dx = \int_a^\infty f(x, m_0)\,dx. \]The definition requires knowing the tail \( \int_b^\infty f\,dx \), which presupposes convergence. The Cauchy criterion avoids this by looking only at finite pieces far out.
In words: far enough out, every piece of the integral is small, uniformly in \(m\). This is the exact analogue of the Cauchy criterion for series from the first post in this series, and it is the main tool for proving other tests. It is also useful for proving that convergence is not uniform: find, for some \(\varepsilon\), pieces \([b', b'']\) arbitrarily far out and parameter values \(m\) for which the piece is at least \(\varepsilon\).
This is the test you'll use most in practice. The idea: if every integrand in the family is trapped under one fixed function whose integral converges, then all the tails are controlled by that one function's tail.
The proof is one line: \( \left|\int_b^\infty f(x, m)\,dx\right| \le \int_b^\infty F(x)\,dx \), and the right side doesn't depend on \(m\) and tends to 0.
Everything above carries over to the other kind of improper integral: a finite interval where the integrand blows up at an endpoint. If \(f(x, m)\) is continuous for \( a < x \le b \) but may be unbounded near \(x = a\), then \( \int_a^b f(x, m)\,dx \) converges uniformly on \((m_1, m_2)\) if
\[ \sup_{m_1 < m < m_2}\left|\int_a^{a+\delta} f(x, m)\,dx\right| \longrightarrow 0 \quad \text{as } \delta \to 0^+. \]The "tail" is now the small piece next to the singular point instead of the far-away piece. The continuity theorem, the Cauchy criterion and the Weierstrass test all hold in the same form.
Here is where all of this pays off. Euler's Gamma function, the famous extension of the factorial, is defined by an improper integral with a parameter:
\[ \Gamma(m) = \int_0^\infty x^{m-1}e^{-x}\,dx \qquad (m > 0). \]It is improper in two ways: at \(\infty\), and at \(0\) when \(m < 1\). Is \(\Gamma\) continuous? Fix any \(0 < \alpha < \beta\) and split the integral at \(x = 1\):
So the integral converges uniformly on every \([\alpha, \beta] \subset (0, \infty)\), and \(\Gamma\) is continuous on \((0, \infty)\). Since every point \(m > 0\) lies inside such an interval, that's all we need. (On the whole of \((0, \infty)\) the convergence is not uniform: both ends \(m \to 0\) and \(m \to \infty\) cause trouble. This "uniform on every closed subinterval" situation is very typical.)
Integrating by parts gives \( \Gamma(m + 1) = m\,\Gamma(m) \), and with \(\Gamma(1) = 1\) this yields \( \Gamma(n + 1) = n! \). Uniform convergence is what makes this smooth interpolation of the factorials a legitimate, continuous function.
| Concept | Statement | Remark |
|---|---|---|
| Uniform convergence | \( \sup_m \left|\int_b^\infty f\,dx\right| \to 0 \) as \(b \to \infty\) | one cut-off \(M(\varepsilon)\) for all \(m\) |
| Series equivalence | uniform \(\iff\) every series \( \sum \int_{a_n}^{a_{n+1}} f\,dx \) is uniform | transfers series theorems to integrals |
| Continuity | \(f\) continuous + uniform convergence \(\Rightarrow I(m)\) continuous | limit can go under the integral |
| Cauchy criterion | \( \left|\int_{b'}^{b''} f\,dx\right| < \varepsilon \) for \(b', b'' > M\), all \(m\) | necessary and sufficient |
| Weierstrass test | \( |f(x, m)| \le F(x) \), \( \int_a^\infty F < \infty \) | sufficient only; gives absolute convergence |
Try these before opening the solution. More solved problems are in Definite Integrals and Series.
Majorant: \( |e^{-mx}\sin x| \le e^{-m_0x} \), and \( \int_0^\infty e^{-m_0x}\,dx = \frac{1}{m_0} \). Uniform by Weierstrass.
\( |e^{-x^2}\cos(mx)| \le e^{-x^2} \), and \( \int_0^\infty e^{-x^2}\,dx = \frac{\sqrt\pi}{2} \). Uniform by Weierstrass, so the integral is a continuous function of \(m\).
On \([m_0, \infty)\): yes, by Weierstrass with the majorant \( \frac{1}{x^2 + m_0^2} \).
On \((0, \infty)\): also yes, but here a single majorant on all of \([0, \infty)\) doesn't exist, since \( \sup_{m>0}\frac{1}{x^2+m^2} = \frac{1}{x^2} \) is not integrable near \(x = 0\). Go back to the definition instead: for every \(m > 0\), \[ \int_b^\infty \frac{dx}{x^2 + m^2} \le \int_b^\infty \frac{dx}{x^2} = \frac1b \longrightarrow 0, \] independently of \(m\). So the convergence is uniform. (Lesson: uniform convergence is about the tail only; the integral's value \( \frac{\pi}{2m} \) may still be unbounded as \(m \to 0\), and it is, but it is continuous on \((0,\infty)\), as the theorem promises.)
The tail is \( \int_b^\infty \frac{m\,dx}{1 + m^2x^2} = \frac\pi2 - \arctan(mb) \), which tends to \( \frac\pi2 \) as \(m \to 0^+\) for every fixed \(b\). So the supremum doesn't go to zero: not uniform. Consistently, the integral equals \( \frac\pi2 \) for \(m > 0\) but \(0\) at \(m = 0\), a jump.
For \(0 < x \le 1\) and \(m \ge m_0\): \( |x^{m-1}\ln x| \le x^{m_0-1}|\ln x| \). This majorant is integrable on \((0,1]\), since \( \int_0^1 x^{m_0-1}|\ln x|\,dx = \frac{1}{m_0^2} \) (differentiate \( \int_0^1 x^{m-1}dx = \frac1m \) with respect to \(m\)). Uniform by Weierstrass.
Uniform convergence is the price of admission for treating an improper integral as a nice function. The intuition to keep: the tail must become small at the same rate for all parameter values, so nothing escapes to infinity (or into a singularity) as the parameter moves. In practice, look for a single majorant; if the trouble sits at an endpoint of the parameter interval, move a fixed distance away from it, as with the Gamma function.
Next in the series: differentiating and integrating improper integrals with respect to a parameter, and the Euler integrals \(\Gamma\) and \(\mathrm{B}\).