Improper integrals review

Improper integrals are definite integrals that cover an unbounded area.

One type of improper integrals are integrals where at least one of the endpoints is extended to infinity. For example, ${\displaystyle {\int }_1^{\mathrm{\infty }}{\displaystyle \frac{1}{x^2}}dx}$‍ is an improper integral. It can be viewed as the limit ${\displaystyle \underset{b\to \mathrm{\infty }}{lim}{\int }_1^b{\displaystyle \frac{1}{x^2}}dx}$‍.

Another type of improper integrals are integrals whose endpoints are finite, but the integrated function is unbounded at one (or two) of the endpoints. For example, ${\displaystyle {\int }_0^1{\displaystyle \frac{1}{\sqrt{x}}}dx}$‍ is an improper integral. It can be viewed as the limit ${\displaystyle \underset{a\to 0^{+}}{lim}{\int }_a^1{\displaystyle \frac{1}{\sqrt{x}}}dx}$‍.

An unbounded area that isn't infinite?! Is that for real?! Well, yeah! Not all improper integrals have a finite value, but some of them definitely do. When the limit exists we say the integral is convergent, and when it doesn't we say it's divergent.

Let's evaluate, for example, the improper integral ${\displaystyle {\int }_1^{\mathrm{\infty }}{\displaystyle \frac{1}{x^2}}dx}$‍. As mentioned above, it's useful to view this integral as the limit ${\displaystyle \underset{b\to \mathrm{\infty }}{lim}{\int }_1^b{\displaystyle \frac{1}{x^2}}dx}$‍. We can use the fundamental theorem of calculus to find an expression for the integral:

Now we got rid of the integral and we have a limit to find:

Let's evaluate, for example, the improper integral ${\displaystyle {\int }_0^1{\displaystyle \frac{1}{\sqrt{x}}}dx}$‍. As mentioned above, it's useful to view this integral as the limit ${\displaystyle \underset{a\to 0}{lim}{\int }_a^1{\displaystyle \frac{1}{\sqrt{x}}}dx}$‍. Again, we use the fundamental theorem of calculus to find an expression for the integral:

Now we got rid of the integral and we have a limit to find: