Integrating Infinities: Zero or Infinity?

[bmbuncher]

Hi! I have a question about integrating a function with an infinite value. If you integrate a function with a place where the integrand diverges to infinity, I understand that the value of the integral should diverge to infinity. However, what happens when you set both bounds to be the value where the integral diverges? For example:

\int_{π/2}^{π/2} tan(x)\,dx

In this example, tan(x), which diverges at π/2, is integrated from π/2 to π/2. I understand that normally, when you integrate with both bounds being the same, the result is zero because there is no length covered in the x direction. Is it the same in this case, or does integrating where the value of the integrand approaches infinity change this? It seems to me to resemble multiplying zero by infinity (where zero is the length and infinity is the height), which is undefined, but I don't know whether that is an adequate description of what is occurring for that to be the answer. Please let me know what you think. Thank you!

 

[micromass]

Hi! I have a question about integrating a function with an infinite value. If you integrate a function with a place where the integrand diverges to infinity, I understand that the value of the integral should diverge to infinity.

Not necessarily. Take for example

\int_0^1 \frac{1}{\sqrt{x}}dx = 2

So this integral is finite, while the function does diverge to infinity in $0$. So it is very much possible for a function to have a finite area, while the the "perimeter" is infinite. This confused the ancient mathematicians very much. See for example Gabriel's horn: http://en.wikipedia.org/wiki/Gabriel's_Horn

However, what happens when you set both bounds to be the value where the integral diverges? For example:

\int_{π/2}^{π/2} tan(x)\,dx

In this example, tan(x), which diverges at π/2, is integrated from π/2 to π/2. I understand that normally, when you integrate with both bounds being the same, the result is zero because there is no length covered in the x direction. Is it the same in this case, or does integrating where the value of the integrand approaches infinity change this? It seems to me to resemble multiplying zero by infinity (where zero is the length and infinity is the height), which is undefined, but I don't know whether that is an adequate description of what is occurring for that to be the answer. Please let me know what you think. Thank you!

The general case is a function $f:(a,b)\rightarrow \mathbb{R}$, so $f$ is only defined on an open interval. The limits \lim_{x\rightarrow a} f(x) might be infinite, the same for $b$. In this case, we have $(a,b) = (-\pi/2,\pi/2)$.

We want to make sense of the integral

\int_a^b f(x)dx

By definition, this integral is defined if there exists some $c\in (a,b)$ so that the two integrals

\int_a^c f(x)dx~\text{and}~\int_c^b f(x)dx

exist. This is equivalent to demanding that the above two integrals exist for any $c\in (a,b)$.

So, in your case, your integral makes sense if both integrals

\int_0^{\pi/2} \tan(x)dx~\text{and}~\int_{-\pi/2}^0\tan(x)dx

exist. This is not the case, so we say the integral does not exist.

However, you have the idea of using the following:

\lim_{x\rightarrow \pi/2} \int_{-x}^x \tan(x)dx

This limit does exist and actually equals zero. However, we do not call this the integral of $\tan(x)$. Rather, we call this the Cauchy principal value of the integral.

 

[bmbuncher]

@Micromass: I see, that makes some sense. However, what happens when both bounds of the integrand are the same (namely, they are both positive π/2)? Would there not be a valid value c ∈ (a,b) because a and b are the same?

 

[bmbuncher]

Oh, I see, because (a,b) is an open interval, when a=b, it is an empty set, so there is no c such that c ∈ (a,b) (I think, correct me if I'm wrong). Is it therefore not a valid integral to integrate from a to a?

 

[micromass]

@Micromass: I see, that makes some sense. However, what happens when both bounds of the integrand are the same (namely, they are both positive π/2)? Would there not be a valid value c ∈ (a,b) because a and b are the same?

Oh, I misread your OP.

Well, the usual Riemann integration theory usually only deals with functions $f:(a,b)\rightarrow \mathbb{R}$ where $a<b$. So it doesn't exactly answer your question. It is of course easy to extend the theory to $a=b$, but this isn't very interesting.

However, when doing Lebesgue integration, then our integration theory is generalized a lot. In particular, we now consider functions $f:\mathbb{R}\rightarrow \mathbb{R}\cup\{-\infty,+\infty\}$. So function values like $f(0) = +\infty$ are allowed then. In particular, the $\tan(x)$ function can be defined on entire $\mathbb{R}$.
That's not the only thing, we can also integrate over much more general domains, such as $\{\pi/2\}$. So things like

\int_{\pi/2}^{\pi/2} \tan(x)dx

are perfectly legitimate questions in measure theory and Lebesgue integration. Like you noticed, the answer should be equal to $0\cdot (+\infty)$, whatever that means. Well, in the theory of Lebesgue integration, that is defined as zero. This is simply a definition, other definitions are possible. But it makes sense because if we consider an infinite straight line, then it's intuitively clear that its area should be $0$. Anyway, with this convention, we have

\int_{\pi/2}^{\pi/2}\tan(x) dx = 0

 

[bmbuncher]

I see, thank you! I'll do some more research on the different types of integration, then, this is very interesting!