Integrating a function that blows up

Available under Creative Commons-ShareAlike 4.0 International License.

When we integrate a function that blows up to infinity at some point in the interval we’re integrating, the result may be either finite or infinite.

Example 75

Integrate the function $y=1\sqrt{x}$ from x = 0 to x = 1.
The function blows up to infinity at one end of the region of integration, but let’s just try evaluating it, and see what happens.
$\begin{align*} \int_{0}^{1}x^{-1/2}dx &=2x^{1/2}\mid _{0}^{1} \\&=2 \end{align*}$

The result turns out to be finite. Intuitively, the reason for this is that the spike at x= 0 is very skinny, and gets skinny fast as we go higher and higher up.

Figure 6.1

a The integral $\int_{0}^{1}dx/\sqrt{x}$ is finite.

Figure 6.2

Example 76

Integrate the function $y=1/x^{2}$ from x= 0 to x= 1.
$\begin{align*} \int_{0}^{1}x^{-2}dx&=x^{-1}\mid _{0}^{1}\\&=-1+\frac{1}{0} \end{align*}$

Division by zero is undefined, so the result is undefined.
Another way of putting it, using the hyperreal number system, is that if we were to integrate from $\varepsilon$ to 1, where was an infinitesimal number, then the result would be - 1 + 1= $\varepsilon$ , which is infinite. The smaller we make , the bigger the infinite result we get out.
Intuitively, the reason that this integral comes out infinite is that the spike at x= 0 is fat, and doesn’t get skinny fast enough.

Figure 6.3

b / The integral $\int_{0}^{1}dx/x^{2}$ is infinite.

These two examples were examples of improper integrals.

2267 reads

You are here

Integrating a function that blows up