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7. Week 6: Infinite Finite Integrals and Finite Infinite Integrals and Infinite Infinite Integrals

≪ 6. Week 5: Trigonometric Substitutions | Table of Contents | 8. Week 7: Convergence of Series ≫

To motivate our discussion today, let us consider the following:

It’s very tempting to say that the first two integrals are both

\[ \int _{-1}^{1} \frac{1}{x^{1/3}}dx = \int _{-1}^{1} \frac{1}{x} dx = 0, \]

since both integrands have symmetry and the positive/negative portions of the area these integrals represent should cancel each other out. And indeed, one does have that \(\int _{-1}^{1}\frac{1}{x^{1/3}}dx = 0\). However, the integral \(\int _{-1}^{1} \frac{1}{x} dx\) does not exist.

Similarly, one may argue that the positive and negative “lobes” of \(\sin x\) pair up perfectly and ought to cancel out, but alas \(\int _{0}^{\infty} \sin x\ dx\) also does not exist.

The issue is one of infinites: the positive and negative areas in the first two integrals are infinitely tall and in particular are potentially infinite. If these areas are finite, it’s perfectly reasonable to say that they sum to \(0\), but if the two areas are infinite, it makes no sense to say \(-\infty + \infty = 0\). Likewise, in \(\int _{0}^{\infty} \sin x\ dx\), we cannot say that infinitely many positive areas cancel out with infinitely many negative areas!

All of the above are called improper integrals, and they are incompatible with our definition of the integral via Riemann sums. In the first two integrals, Riemann sums with an even number of rectangles will necessarily contain an infinitely tall rectangle… unacceptable. In the third integral, one can’t even write down a finite Riemann sum to begin with!

We handle these improper integrals by rewriting them as limits of well-defined finite and proper integrals. For instance,

\[\begin{align*} \int _{-1}^{1} \frac{1}{x^{1/3}} dx &= \lim _{\epsilon \to 0^-} \int _{-1}^{\epsilon } \frac{1}{x^{1/3}} dx + \lim _{\epsilon \to 0^+} \int _{\epsilon }^{1} \frac{1}{x^{1/3}} dx \\ &= \lim _{\epsilon \to 0^-} \left. \frac{3}{2} x ^{2/3}\right\rvert_{-1} ^{\epsilon } + \lim _{\epsilon \to 0^+} \left. \frac{3}{2} x ^{2/3}\right\rvert_{\epsilon }^1 \\ &= -\frac{3}{2} + \frac{3}{2} \\ &= 0. \end{align*}\]

On the other hand,

\[\begin{align*} \int _{-1}^{1} \frac{1}{x} dx &= \lim _{\epsilon \to 0^-} \int _{-1}^{\epsilon }\frac{1}{x} dx + \lim _{\epsilon \to 0^+} \int _{\epsilon }^{1}\frac{1}{x}dx \\ &= \lim _{\epsilon \to 0^-} \left. \ln \left\lvert x \right\rvert \right\rvert_{-1} ^{\epsilon} + \lim _{\epsilon \to 0^+} \left. \ln \left\lvert x \right\rvert \right\rvert_{\epsilon}^1. \end{align*}\]

Neither of these limits exist!

Finally,

\[ \int _{0}^{\infty} \sin x\ dx = \lim _{M \to \infty} \int _{0}^{M} \sin x\ dx = \lim _{M\to\infty} \left. - \cos x \right\rvert_{0} ^{M}, \]

and this limit does not exist either.

Writing limits for integrals is a surefire way to not only see if an integral converges or not but also to determine its value if it does. There are some situations, however, where one needs only determine whether or not an improper integral converges. Consider, for instance, the following problems:

In the first integral, there is a very long “tail” that gets infinitesimally thin; in the second, there is a huge spike near \(x=0\). The heuristic is, if the “tail” is thin enough, then the integral converges; likewise, if the “spike” is narrow enough, the integral converges. There is a pair of tests that usually helps with this: the comparison test and the \(p\)-integral test.

The comparison test says, if your improper integral can be bounded by a finite area, then your improper integral converges; if your improper integral bounds an infinite area, then it diverges.

The \(p\)-test gives us a set of “reference integrals” that have spikes and tails of varying thicknesses. Specifically, it says:

• If \(p > -1\), then the “spike” for \(x^p\) is integrable near \(x=0\) and the “tail” as \(x\to \pm \infty\) is too fat to integrate.
• If \(p < -1\), then the “spike” for \(x^p\) is too fat near \(x=0\) and the “tail” as \(x\to\pm\infty\) is thin enough to integrate.
• If \(p = -1\), both the “spike” and the “tail” for \(x ^{-1}\) are too fat to integrate.

Let us return to the question of if \(\int _{0}^{4} \frac{1}{x^2 + x\cos x}dx\) converges. We need to determine if the “spike” at \(x=0\) is thin enough to integrate. When \(x\) is very, very small, \(x^2\) will be way smaller than \(x\cos x\). In fact, \(\cos x \approx 1\) when \(x \approx 0\), so \(x^2 + x\cos x \approx x\). We therefore expect a spike of size comparable to \(\frac{1}{x}\) near \(x = 0\), and the \(p\)-test says this spike is way too fat.

To make this comparison rigorous, though, we need to use the comparison test: When \(0 < x \leq 1\), \(x^2 \leq x\) and \(x\cos x \leq x\), so \(x^2 + x \cos x \leq 2x\). Therefore, \(\frac{1}{x^2 + x\cos x} \geq \frac{1}{2x}\), and we have

\[ \int _{0}^{4} \frac{1}{x^2 + x\cos x} dx \geq \int _{0}^{1} \frac{1}{2x} dx + \int _{1}^{4} \frac{1}{x^2+x\cos x} dx. \]

By the \(p\)-test, the latter improper integral diverges; by the comparison test, \(\int _{0}^{4} \frac{1}{x^2 + x\cos x} dx\) diverges too.

Let’s now consider \(\int _{2}^{\infty}\frac{1}{\sin^2x + x^2} dx.\) Similar story, but this time we have a long tail as \(x\to\infty\). When \(x\) is huge, \(\sin ^2 x \) is very small compared to \(x^2\), and therefore \(\sin^2x + x^2 \approx x^2\). Thus we’re looking at a tail of size comparable to \(\frac{1}{x^2}\), which we know converges by the \(p\)-test.

To make this comparison rigorous, we should bound \(\sin^2x + x^2 \geq x^2\) always, hence \(\frac{1}{\sin^2x + x^2} \leq \frac{1}{x^2}\) for all \(x\). Thus,

\[ \int _{2}^{\infty} \frac{1}{\sin^2x + x^2} dx \leq \int _{2}^{\infty} \frac{1}{x^2} dx. \]

By the \(p\)-test, the latter integral converges; by the comparison test, the former integral converges too.