In this recitation, we covered improper integrals.

Improper Integrals

Essentially, an improper integral occurs when we have a vertical asymptote (the function is unbounded on the interval) or when one of the bounds of integration is \(\pm \infty\). To deal with this, we need to break up the problem into smaller chunks and take limits. Suppose we are looking at

\[\int_a^b f(x)dx.\]

The rule of thumb is the following:

• If \(f(x)\) is unbounded at \(x=c\), where \(a \leq c \leq b\), then we write

\[\lim_{\alpha \rightarrow c^-} \int_a^\alpha f(x)dx + \lim_{\beta \rightarrow c^+} \int_\beta^b f(x)dx.\]

• If \(f(x)\) is unbounded at \(x=a\), then we write

\[\int_a^b f(x)dx = \lim_{\alpha \rightarrow a^+} \int_\alpha^b f(x)dx.\]

If \(f(x)\) is unbounded at \(x=b\), then we write

\[\int_a^b f(x)dx = \lim_{\alpha \rightarrow b^-} \int_a^\alpha f(x)dx.\]

This also applies for when \(a = -\infty\) or \(b = + \infty\) (in which case we don’t need to write the \(\pm\) superscript).

• We repeat this procedure until all of the integrals under the limits are proper (i.e. the function is bounded on the interval and the bounds do not include \(\pm \infty\).) Notice that we sometimes have to choose an arbitrary number to ensure we don’t have a double limit on the integral; this may happen if the function is unbounded at \(x=a\) and \(x=b\). As long as the function is bounded for all \(a < x < b\), we can choose any \(c\) in this interval and write

\[\int_a^b f(x)dx = \lim_{\alpha \rightarrow a^+} \int_a^c f(x)dx + \lim_{\beta \rightarrow b^-} \int_c^b f(x)dx.\]

As long as the limits are well-defined (meaning they converge to some number), then the choice of \(c\) doesn’t matter by the fundamental theorem of calculus: if \(F'(x) = f(x)\), then

\[\int_a^b f(x)dx = F(c) - \lim_{\alpha \rightarrow a^+} F(\alpha) + \lim_{\beta \rightarrow b^-} F(\beta) - F(c) = \lim_{\beta \rightarrow b^-} F(\beta) - \lim_{\alpha \rightarrow a^+} F(\alpha).\]

If the limits do not exist, then this also happens independent of the choice of \(c\), so we’re allowed to pick our favorite.

If one of the limits doesn’t exist in the above procedure or is equal to \(\pm \infty\), then we say that the limit diverges. Otherwise, the limit converges.

Note this can sometimes lead to weird answers, but these integrals are important for physics/probability. For example, improper integrals are important for calculating probabilities involving the Gaussian distribution.

Important things to remember from this section are:

• identifying whether or not an integral is improper,
• writing out the integral as a sum of limits, and
• evaluating the integrals/limits you get at the end.

Example: Let’s look at

\[\int_{1}^\infty \frac{dx}{x-1}.\]

The function on the inside of the integral has a vertical asymptote at \(x=1\), so we need to break up the integral based on that. There’s also an \(\infty\) in the bound, so we need to write this as a limit. Finally, there are no vertical asymptotes after \(x=1\), so we can pick our favorite number between \(1\) and \(\infty\) to break up the integral. Let’s choose \(x=2\) as our break up point. We get

\[\int_1^\infty \frac{dx}{x-1} = \lim_{a \rightarrow 1^+} \int_a^2 \frac{dx}{x-1} + \lim_{b \rightarrow \infty} \int_2^b \frac{dx}{x-1}.\]

Now we use the fundamental theorem of calculus to get

\[\int_1^\infty \frac{dx}{x-1} = \ln(1) - \lim_{a \rightarrow 1^+}\ln(a-1) + \lim_{b \rightarrow \infty} \ln(b - 1) - \ln(1).\]

Simplifying:

\[\int_1^\infty \frac{dx}{x-1} = + \lim_{b \rightarrow \infty} \ln(b - 1) - \lim_{a \rightarrow 1^+}\ln(a-1).\]

Now we see that \(\lim_{a \rightarrow 1^+} \ln(a-1) = \lim_{a \rightarrow 0^+} \ln(a) = -\infty\) and \(\lim_{b \rightarrow \infty} \ln(b-1) = \lim_{b \rightarrow \infty} \ln(b) = \infty.\) Since the limits are not finite, the integral diverges.

Remark: We could have done just one of the limits and noticed that it doesn’t converge to a number and stopped.

Example: Let’s look at

\[\int_{-\infty}^\infty \sin(x)dx.\]

It’s tempting to say that this integral is \(0\) because for any \(a > 0\) we have

\[\int_{-a}^a \sin(x)dx = 0.\]

However, according to the definition of improper integrals, we have to do the limits separately. Following the definition, we choose our favorite value on the real line and break the integral up into two integrals. My favorite number is \(x=0\), so

\[\int_{-\infty}^\infty \sin(x)dx = \lim_{a \rightarrow -\infty} \int_a^0 \sin(x)dx + \lim_{b \rightarrow \infty} \int_0^b \sin(x)dx.\]

Now unlike last time, let’s just evaluate one of the integrals. Notice that

\[\lim_{a \rightarrow -\infty} \int_a^0 \sin(x)dx = \lim_{a \rightarrow -\infty}( -\cos(0) + \cos(a)) = \lim_{a \rightarrow -\infty} (\cos(a) - 1).\]

Does this converge? Let’s take two different paths to \(-\infty\) and see what happens. Notice that \(\lim_{n \rightarrow \infty} -2\pi n = -\infty\) (here \(n \geq 1\) a whole number), and so if this limit converged we would have

\[\lim_{a \rightarrow -\infty} \cos(a) = \lim_{n \rightarrow \infty} \cos(-2\pi n) = 1.\]

On the other hand, we also have \(\lim_{n \rightarrow \infty} -(n \pi + \pi/2) = -\infty,\) and \(\cos(- (n\pi + \pi/2)) = 0\) for each \(n \geq 1\) a whole number. So if this limit converged, we would have

\[\lim_{a \rightarrow -\infty} \cos(a) = \lim_{n \rightarrow \infty} \cos(-(n\pi + \pi/2)) = 0.\]

Uh oh! If the limit did converge, I would have \(1 = 0\). We see the limit doesn’t converge (note this is also easy to see from a graph of cosine and you don’t have to do as much work). Since one of the limits doesn’t converge, we have that the integral diverges.

Example: Evaluate

\[\int_{0}^{\pi/2} \tan(2x) \sec^4(2x)dx.\]

If we don’t think about it, it looks like there’s no vertical asymptotes because there is no denominator. However, remember that these trig functions are hiding something in the denominator. We have

\[\tan(2x) = \frac{\sin(2x)}{\cos(2x)} \text{ and } \sec^4(2x) = \frac{1}{\cos^4(2x)}.\]

Now it’s a little annoying to deal with this \(2x\) on the inside, so let’s get rid of that by doing a \(u\)-substitution. Let \(u = 2x\), \(du = 2 dx\). Then

\[\int_{0}^{\pi/2} \tan(2x) \sec^4(2x)dx = \frac{1}{2}\int_0^{\pi} \tan(u) \sec^4(u)du = \frac{1}{2}\int_0^{\pi} \frac{\sin(u)}{\cos^5(u)}du.\]

Now on the interval \([0,\pi],\) we have that \(\cos(x) = 0\) when \(x = \pi/2\) (I recommend drawing a graph). So we need to break up our integral at this point. We write

\[\int_0^{\pi/2} \tan(2x) \sec^4(2x)dx = \frac{1}{2} \left\{ \lim_{a \rightarrow (\pi/2)^-}\int_0^{a} \frac{\sin(u)}{\cos^5(u)}du + \lim_{b \rightarrow (\pi/2)^+}\int_b^{\pi} \frac{\sin(u)}{\cos^5(u)}du \right\}.\]

Let’s look at just one of these limits. To evaluate the integral, we can do a \(u\)-substitution. Sadly I already used \(u\), but let’s use \(w\) instead. Let \(w = \cos(u)\), so \(dw = -\sin(w)dw\). Then

\[\int_0^a \frac{\sin(u)}{\cos^5(u)}du = \int_1^{\cos(a)} w^{-5}dw = -\frac{1}{4} w^{-4} \Big|_{w=1}^{\cos(a)}.\]

Notice it’s okay to let the bounds depend on \(a\) since this is fixed until the end. Plugging things in:

\[\int_0^a \frac{\sin(u)}{\cos^5(u)}du = -\frac{1}{4} \cos(a)^{-4} + \frac{1}{4}.\]

Now we take a limit as \(a \rightarrow (\pi/2)^-\). But cosine is in the denominator, and as \(a \rightarrow (\pi/2)^-\) we see that this is blowing up to \(\infty\). Since one of the limits blows up, the integral must be divergent, and we can stop.

Example: We’ve done a lot of divergent improper integrals, is there any chance that these integrals converge? Let’s explore a family of examples. Let \(p > 0\) (not necessarily a whole number) and look at

\[\int_1^\infty \frac{dx}{x^p}.\]

This only problem can happen at \(x=0\), which we are excluding from our domain of integration. It makes sense to write

\[\int_1^\infty \frac{dx}{x^p} = \lim_{a \rightarrow \infty} \int_1^a \frac{dx}{x^p}.\]

Now we can use fundamental theorem of calculus to get two different answers depending on \(p\): If \(p \neq 1\), then we can use the power rule for integration to get

\[\int_1^\infty \frac{dx}{x^p} = \lim_{a \rightarrow \infty} \left[ \frac{x^{1-p}}{1-p} \Big|_{x=1}^a \right] = \lim_{a \rightarrow \infty} \frac{a^{1-p}}{1-p} - \frac{1}{1-p}.\]

If \(p = 1\), then we have the one exception to the power rule for integration

\[\int_1^\infty \frac{dx}{x} = \lim_{a \rightarrow \infty} \left[ \ln(x) \Big|_{x=1}^a \right] = \lim_{a \rightarrow \infty} \ln(a) - \ln(1).\]

Notice we can already rule out \(p =1\) as a possibility, since this limit is \(\infty\). Thus, as long as \(p \neq 1\), the integral is convergent if

\[\lim_{a \rightarrow \infty} \frac{a^{1-p}}{1-p}\]

exists and is finite. We have two cases to think about. First, if \(p < 1\), then \(1-p > 0\), so there is some power of \(a\) in the numerator. Taking the limit will have this converge to \(\infty\), so we have to ignore \(0 < p < 1\). Next, if \(p > 1\), then \(1-p < 0\), so \(\lim_{a \rightarrow \infty} a^{1-p} = 0\). This tells us that the integral is convergent if and only if \(p > 1\), and furthermore the answer is

\[\int_1^\infty \frac{dx}{x^p} = - \frac{1}{1-p} = \frac{1}{p-1}.\]

For example, if \(p =2\) then we get that the integral is \(1\) (and we finally have an example of a convergent improper integral).

Example: In the opposite direction, let’s look at \(p > 0\) a real number and the integral

\[\int_0^1 \frac{dx}{x^p}.\]

Again, because of the power rule we have that the case \(p=1\) is special, and so we should look at it separately. For \(p=1\) we get

\[\int_0^1 \frac{dx}{x} = \lim_{a \rightarrow 0^+} \int_a^1 \frac{dx}{x} = \lim_{a \rightarrow 0^+} \ln(1) - \ln(a) = \lim_{a \rightarrow 0^+} - \ln(a) = \infty.\]

So it is divergent. If \(p \neq 1\), then the power rule tells us

\[\int_0^1 \frac{dx}{x} = \lim_{a \rightarrow 0^+} \int_a^1 \frac{dx}{x} = \lim_{a \rightarrow 0^+} \left(\frac{1}{1-p} - \frac{a^{1-p}}{1-p} \right).\]

Now if \(p < 1\), then \(1-p > 0\), so \(\lim_{a \rightarrow 0^+} a^{1-p} = 0\). We conclude that if \(p < 1\), then the integral converges to \(1/(1-p)\). If \(p > 1\), then \(1-p < 0\), and we get \(\lim_{a \rightarrow 0^+} a^{1-p} = + \infty\), so the integral diverges. This gives us another family of convergent improper integrals.

Remark: These kinds of integrals are important for something called the \(p\)-series test, which we will come back to later. We’re starting to get into deep math!

Warning: There is something called Cauchy’s principal value which is a way of assigning values to certain divergent improper integrals (the idea is to bring the limit outside and use the same variable in both integrals). We are not learning this in class, however, integration calculators sometimes like to use this value when calculating improper integrals. For example, according to our definition of an improper integral, we have that

\[\int_{-1}^1 \frac{dx}{x}\]

diverges, since we can write this as

\[\int_{-1}^0 \frac{dx}{x} + \int_0^1 \frac{dx}{x},\]

and we know from our prior example that the integral on the right diverges (so everything diverges). However, the Cauchy principal value of this integral is \(0\) (in fact, the Cauchy principal value of an odd function over a symmetric interval is always zero). Calculators like Desmos will tell you this integral is \(0\) because it is using this alternative definition (for some reason). You can see here that Wolfram Alpha avoids this issue. Interestingly, Desmos seems inconsistent about this. If you plug in

\[\int_{-\infty}^\infty \frac{2x}{x^2+1}dx,\]

it tells you (correctly) that the integral is undefined, but the Cauchy principal value of this integral is \(0\).

Looking at some threads, I’ve found some other bugs when it comes to Desmos. If you ask

\[\int_3^\infty \frac{(2 + \cos(x))}{x}dx,\]

then Desmos will tell you this is zero. Notice this doesn’t make any sense. Notice that \(-1 \leq \cos(x) \leq 1\), so \(1 \leq 2 + \cos(x) \leq 3.\) Hence for any \(a > 3\), we have

\[\int_3^a \frac{dx}{x} \leq \int_3^a \frac{2 + \cos(x)}{x} \leq \int_3^a \frac{3}{x}dx = 3 \int_3^a \frac{dx}{x}.\]

Notice that

\[\int_3^a \frac{dx}{x} = \ln(a) - \ln(3).\]

If we take \(a \rightarrow \infty\), this converges to \(+\infty\), so by the squeeze theorem we should have

\[\lim_{a \rightarrow \infty} \int_3^a \frac{2 + \cos(x)}{x} dx = \infty,\]

or, in other words, the integral doesn’t converge. The moral of the story is that Desmos lies and we should not trust it for improper integrals.