Bound the integrand by something simpler

Quick description

This is a generalisation of the base times height bound: instead of bounding the magnitude $|f(x)|$ of the integrand by some fixed constant $M$ , instead bound it by some quantity $M(x)$ which still depends on $x$ , but in a "simpler" fashion than the original integrand $f$ , where "simpler" means "easier to integrate". Then one clearly has $|\int_E f(x)\ dx| \leq \int_E M(x)\ dx$ . Note that in many cases, one is prepared to lose a multiplicative constant C in the final bound on the integral, in which case one is also free to lose a constant on the integrand, i.e. one will settle for $|f(x)| \leq C M(x)$ rather than $|f(x)| \leq M(x)$ .

Prerequisites

Undergraduate calculus

Example 1

Obtain an upper bound for $\int_0^\infty \frac{1}{a^7 + x^7}\ dx$ , where $a >0$ is a parameter.

The integrand $\frac{1}{a^7+x^7}$ looks somewhat unpleasant to deal with by exact integration methods (substitution, integration by parts, contour integration, etc.). However, we only seek an upper bound. Note that up to a factor of two, a sum $X+Y$ of two positive quantities is comparable to the maximum $\max(X,Y)$ of these quantities. Taking reciprocals, we see that $\frac{1}{a^7+x^7}$ is comparable to $\min( \frac{1}{a^7}, \frac{1}{x^7} )$ (in fact, the former is less than or equal to the latter).

Now, the quantity $\min( \frac{1}{a^7}, \frac{1}{x^7} )$ is simpler to integrate than $\frac{1}{a^7+x^7}$ . Why? Well, freshman calculus already tells us how to integrate $\frac{1}{a^7}$ and $\frac{1}{x^7}$ separately. And dealing with a minimum (or maximum) in an integrand is not too scary; one simply needs to divide the region of integration into two sets, one where the first term attains the minimum, and one where the second term does, and divide and conquer.

In this case, $\frac{1}{a^7}$ attains the minimum when $x \leq a$ , and $\frac{1}{x^7}$ does otherwise. Thus we obtain an upper bound of

$\int_0^a \frac{1}{a^7}\ dx + \int_a^\infty \frac{1}{x^7}\ dx$

which can now be easily computed via freshman calculus as

$\frac{a}{a^7} + \frac{1}{6} a^{-6} = \frac{7}{6} a^{-6}.$

In the converse direction, to get a lower bound, observe on the first region $x \leq a$ , $a^7 + x^7$ never exceeds $2a^7$ , and on the second region it never exceeds $2x^7$ , so we obtain a lower bound of

$\int_0^a \frac{1}{2a^7}\ dx + \int_a^\infty \frac{1}{2x^7}\ dx = \frac{7}{12} a^{-6}.$

Thus our upper and lower bounds only differ by a factor of two (this is the same factor of two that appeared earlier in the discussion).

Note that one can also guess the bound by the base times height heuristic. If one sketches a graph of the function $\frac{1}{a^7+x^7}$ , one sees that it has "height" about $\frac{1}{a^7}$ and is concentrated on a "base" of width about $a$ , so it is reasonable to expect the final integral to be of size about $a^{-6}$ , which is indeed what happens.

General discussion

Of course, for this strategy to be effective, the upper bound $M(x)$ should be easier to integrate than the original function $f(x)$ . See "Which integrals are simpler to integrate?" for more discussion of this issue.

The bound is only efficient if one believes that $f(x)$ is generally not much smaller than $M(x)$ , and also if one believes that $f(x)$ does not oscillate enough to provide additional cancellation.

Remark The strategy can be thought of as the analogue for integrals of the comparison test, which is the basic strategy for bounding infinite sums.