Vertical Asymptotes

Posted: 02/03/2025

I will be referring to parts of this desmos graph throughout this page. I can not remember if we learned this method in MAT187 last year to solve these kinds of integrals, but I remember using my method for other purposes and it feels the most logical to me.

I have heard some first-year engineering students believe that any integral with a vertical asymptote must be divergent. Let's try one out:
\[\int_{0}^{4} \frac{2}{\sqrt{x}} \, dx \] In the graph, I have divided this area into the blue and green regions. The blue region can be solved by calculating the area of the rectangle. We are much more interested in the green region, which contains the asymptote. To make this finding the green region easier for ourselves, we can integrate with respect to \(y\) rather than to \(x\). To do this, we must rearrange the above equation.
\[y = \frac{2}{\sqrt{x}}\] \[\sqrt{x} = \frac{2}{y}\] \[x = \frac{4}{y^2}\] With \(y = 1\) as the top of our rectangle, that will be where the integral for finding the green region starts. Thus, we will now need to solve the below integral:
\[\int_{1}^{\infty} \frac{4}{y^2} \, dy \] When solving improper integrals like this one, we want to find the limit as a variable approaches infinity (i.e. \(\lim_{a\to\infty}\int_{1}^{a} \frac{1}{y^2}\, dy\)). Using power rule on this integral, we get the below equation:
\[\lim_{a\to\infty}(-\frac{4}{a}) - (-\frac{4}{1})\] We know the limit as \(a\) approaches infinity for \(-\frac{4}{a}\) is \(0\), so we can remove it from the equation. That leaves us with \(\frac{4}{1} = 4\). We can then add the area of the rectangle to find the answer to \(\int_{0}^{4} \frac{2}{\sqrt{x}} \, dx\).
\[4 + 4 = \fbox{8}\] This is not to say every integral on a vertical asymptote is convergent. If you try the same method on \(\int_0^1\frac{1}{x^2}\, dx\), you will find the integral diverges. In fact, this is true of any integral \(\int_0^c\frac{1}{x^b}\, dx\), where \(b \ge 1\) and \(c\) is any real number. I encourage you to try this yourself if you have time.

A previous version of this page suggested that the integral \(\int_0^c\frac{1}{x^b}\, dx\) would diverge if \(b \leq 1\). This has been corrected.