Last time we looked at integrals which weren’t proper integrals because the limits of integration were infinite (either on one side, or both). This was one of the constraints we had on a well defined Riemann sum. The other constraint we had was that there were no infinite discontinuities in the integrand. Here we will show that sometimes we can indeed define an improper integral which does include such a discontinuity within the limits of integration.

Improper Integrals of the second kind: Infinite discontinuities in the integrand

We have seen what happens when we integrate $x$ from $\infty$ or to $-\infty$. Sometimes it gives a convergent value and we can find the integral, sometimes the improper integral does not converge and it gives us an answer of $\pm\infty$. Indeed sometimes it doesn’t blow up, but it just doesn’t converge – as in the integral of $\cos x$ over an infinite range.

Now we are going to look at what happens if the function itself has an infinite discontinuity in it and we want to integrate over this region, or include it at one of the limits of integration. The formal definition of this can be found in Stewart page 418. Here we will look at the examples and examine what is happening.

The function $f(x)=\frac{1}{\sqrt{x-4}}$ is continuous on the interval $(4,\infty)$ and looks like: If we are trying to integrate this from $x=4$ then clearly we will be including the infinite discontinuity at $x=4$.  We can define this integral however as a limit. Let’s see what happens when we do this: $\int_4^{10}\frac{1}{\sqrt{x-4}}dx=\lim_{t\rightarrow 4^+}\int_t^{10}\frac{1}{\sqrt{x-4}}dx=\lim_{t\rightarrow 4^+}\left.2\sqrt{x-4}\right|_t^{10}$

but, just as in the cases of some of the infinite integrating regions that we saw last time, we can take this limit and it will give us a convergent quantity: $\lim_{t\rightarrow 4^+}\left.2\sqrt{x-4}\right|_t^{10}=2\left(\sqrt{10-4}-\sqrt{4-4}\right)=2\sqrt{6}$

A logarithmic divergence $\int_0^1\ln x dx$

This diverges at $x=0$ and so we can define the improper integral as: $\int_0^1\ln x dx=\lim_{t\rightarrow 0^{+}}\int_t^1\ln x dx=\lim_{t\rightarrow 0^{+}} -t\ln t-1+t$

Clearly the last terms converge, but how about $t\ln t$. We know that $t$ goes to zero and $\ln t$ goes to $-\infty$ so we have to be careful about this. We can invoke L’Hospital’s rule again: $\lim_{t\rightarrow 0^{+}} t\ln t=\lim_{t\rightarrow 0^{+}}\frac{\ln t}{t^{-1}}=\lim_{t\rightarrow 0^{+}}\frac{\frac{1}{t}}{\frac{-1}{t^2}}=\lim_{t\rightarrow 0^{+}}(-t)=0$ $\int_0^1\ln x dx=-1$

A non-convergent example $\int_0^1\frac{1}{x^2} dx=\lim_{t\rightarrow 0^+}\int_t^1\frac{1}{x^2}=\lim_{t\rightarrow 0^+}\left.\frac{-1}{x}\right|_t^1=\infty$

The limit does not converge.

When a part of an integral is divergent, the whole thing is divergent

For the time being, if we have an integral where any part of it is divergent, then it is said to be divergent as a whole. For instance: $\int_{-1}^1\frac{1}{x^2}dx$

can be split up as: $\int_{-1}^1\frac{1}{x^2}dx=\lim_{t\rightarrow 0^-}\int_{-1}^t\frac{1}{x^2}dx+\lim_{t\rightarrow 0^+}\int_{t}^1\frac{1}{x^2}dx$

As the first term in this expression diverges, so the whole thing is said to be divergent.

Watch out for examples when neither of the limits have a problem, but the problem is somewhere in the range $[a,b]$

The Comparison Theorem

For two continuous functions, $f(x)$ and $g(x)$, if $f(x)\ge g(x)\ge 0$ for $x\ge a$ then:

1. If $\int_a^\infty f(x) dx$ is convergent, then so it $\int_a^\infty g(x) dx$.
2. If $\int_a^\infty g(x) dx$ is divergent, then so is $\int_a^\infty f(x) dx$.

Let’s look at an example:

If we have the integral: $\int_0^\infty e^{-x^2}dx$

We may not know how to do this integral. However, we can still say something about whether it converges or diverges. We use a trick for this and split the integral into two regions: $\int_0^\infty e^{-x^2}dx=\int_0^1 e^{-x^2}dx+\int_1^\infty e^{-x^2}dx$

Certainly the first part is finite because it is continuous and the limits are finite. The question is whether the second part is finite. The reason that we split the integral up like this is because there is a simple function which is always greater than $e^{-x^2}$ between 1 and $\infty$ (but not between 0 and 1 which is why we made the split).

Here, the graphs of $e^{-x}$ in red and $e^{-x^2}$ in blue: We can show that: $e^{-x}\ge e^{-x^2}$

for $x\ge 1$. Thus, if the area under the red curve is finite, then the area under the blue curve is finite. However, if the area under the red curve is infinite, we can’t say anything about the area under the blue curve. We can now ask whether the integral: $\int_1^\infty e^{-x}dx$

converges, and indeed we can show with the methods that we’ve learnt above, that it does. Thus, we can use the comparison theorem and say that because $e^{-x}$ is always greater than or equal to $e^{-x^2}$ in the region between 1 and $\infty$ and that the integral over the former converges, that the integral over $e^{-x^2}$ must also converge.

Let’s look at some more examples:

Let’s look at the integral: $\int_2^\infty\frac{\cos^2 x}{x^2}dx$

This integral is actually not possible to write down in terms of elementary functions, but we can still say whether or not it converges. Because $\cos^2x$ is bounded between 0 and 1, we know that: $\frac{1}{x^2}\ge\frac{\cos^2x}{x^2}$

So if we can show that the integral over the larger function converges, then the smaller function must also converge. Indeed we know that $\int_2^\infty\frac{1}{x^2}$ converges, and so the original integral also converges. $\int_3^\infty\frac{1}{x-e^{-x}}dx$

The trick is to look at the behaviour of the function in the large $x$ limit. On the bottom we have $x$ and $e^{-x}$. As $x\rightarrow \infty$, $x$ goes large and $e^{-x}$ goes to zero so we can see that in the large $x$ limit: $\lim_{x\rightarrow\infty}\frac{1}{x-e^{-x}}\sim \frac{1}{x}$

Let’s use this as the comparison. Indeed between 3 and $\infty$: $\frac{1}{x-e^{-x}}>\frac{1}{x}$

and the integral: $\int_3^\infty \frac{1}{x}dx$

diverges. Because this integral is less than the original integral, we can say that, by the comparison theorem, the original integral also diverges.

Let’s look at a similar looking example: $\int_3^\infty\frac{1}{x+e^x}dx$

However, in this case, as $x\rightarrow\infty$ the part in the denominator is dominated by $e^x$ and so: $\lim_{x\rightarrow\infty}\frac{1}{x+e^{x}}\sim \frac{1}{e^x}$

Let’s use this as the comparison. We know that, in the region of interest: $\frac{1}{x+e^{x}}<\frac{1}{e^x}$

and $\int_3^\infty \frac{1}{e^x}dx=e^3$

so this converges and because this integral is certainly greater than the original integral, by the comparison theorem, the original integral also converges.

 How clear is this post?