Introduction

Welcome to this guide on Evaluating Improper Integral of the FiveHive Calculus BC course. This article will guide you through the following:

Defining Improper Integrals

We define improper integral as:

Calculating Improper Integrals

Infinite Discontinuity in Integration Bound

Consider this integral:

${\int }_0^1\frac{1}{x}dx$

Since we know that $f(x)=\frac{1}{x}$ is not defined at $x=0$, we need to use some clever trick to calculate this integral.

Let's rewrite the lower bound as $a$:

${\int }_a^1\frac{1}{x}dx$

where $a>0$.

It is obvious that this integral is not equal to the original integral we want to find, since the lower bound is not equal, but if we take the limit as $a\to 0$, this integral will approach the integral we want to find, then we can apply the Fundamental Theorem of Calculus and find the result:

$\underset{a\to 0^{-}}{\lim }{\int }_a^1\frac{1}{x}dx=\underset{a\to 0^{-}}{\lim }\ln a-\ln 1$

$=\infty -0$

$=\infty$

Hence we can see that this integral diverges (the limit goes to infinity).

Let's take a look at where the integral converges:

${\int }_0^1\frac{1}{\sqrt{x}}dx$

We can apply the same trick we used:

$\underset{a\to 0^{+}}{\lim }{\int }_a^1\frac{1}{\sqrt{x}}dx=\underset{a\to 0^{+}}{\lim }2\sqrt{a}+2\sqrt{1}$

$=0+2$

$=2$  

Which means this integral converges (the limit has a finite value).

Defining Convergence and Divergence of Improper Integral

Through the examples above, we can see that we could always rewrite an improper integral into a limit. If the limit has a finite value, then the integral converges. If the limit does not exist (has an infinite value), then the integral diverges.

Infinite Integration Bound

We will use some example to illustrate the idea of integrating on a infinite bound:

${\int }_1^{\infty }\frac{1}{x}dx$

Let's first replace the upper bound to $b$, and note when $b$ become very large, the integral will get very close to the original integral:

${\int }_1^{\infty }\frac{1}{x}dx=\underset{b\to \infty }{\lim }{\int }_1^b\frac{1}{x}dx$

$=\underset{b\to \infty }{\lim }\ln b-\ln 1$

$=\infty -0$

$=\infty$

Which means this integral diverges, we can see that there is no difference between this improper integral and the one listed above, it is both taking a limit. The same also applies for negative infinity, the lower bound will approach negative infinity.

Let's see a integral that converges:

${\int }_1^{\infty }\frac{1}{x^2}dx$

${\int }_1^{\infty }\frac{1}{x^2}dx=\underset{b\to \infty }{\lim }{\int }_1^b\frac{1}{x^2}dx$

$=\underset{b\to \infty }{\lim }-\frac{1}{b}+1$

$=0+1$

$=1$

Both Infinite Discontinuity and Infinite Integration Bound

There are other improper integrals that can be seen as a combination of both case 1 and case 2. To determine their convergence, we need to consider things separately.

Let ${\int }_a^{b}f(x)dx$ be an improper integral, the integral converges only if ${\int }_a^{c}f(x)dx\text{ and }{\int }_c^{b}f(x)dx$ both converges, given that $b<c<a$.

This gives a method to calculate some other improper integrals:

${\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx$

We can rewrite the integral as:

${\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx={\int }_{-\infty }^0\frac{1}{1+x^2}dx+{\int }_0^{\infty }\frac{1}{1+x^2}dx$

$=\underset{b\to -\infty }{\lim }{\int }_b^0\frac{1}{1+x^2}dx+\underset{a\to \infty }{\lim }{\int }_0^a\frac{1}{1+x^2}dx$

$=\arctan 0-\underset{b\to -\infty }{\lim }\arctan b+\arctan 0-\underset{a\to \infty }{\lim }\arctan a$

$=-\frac{-\pi }{2}+\frac{\pi }{2}$

$=\pi$

Which means this integral converges to $\pi$.

Here is another example:

${\int }_0^{\infty }\frac{1}{x^2}\mathrm{d}x$

It is convenient to split this integral into two parts and analyze them separately:

${\int }_0^{\infty }\frac{1}{x^2}dx={\int }_0^1\frac{1}{x^2}dx+{\int }_1^{\infty }\frac{1}{x^2}dx$

$=\underset{b\to 0^{+}}{\lim }{\int }_b^1\frac{1}{x^2}dx+\underset{a\to \infty }{\lim }{\int }_1^a\frac{1}{x^2}dx$

$=-\frac{1}{1}+\underset{b\to 0^{+}}{\lim }\frac{1}{b}+1$

$=\infty$

There is a quicker way to do this:

${\int }_0^1\frac{1}{x^2}dx=-\frac{1}{1}+\underset{b\to 0^{+}}{\lim }\frac{1}{b}$

$=-1+\infty$

$=\infty$

This integral diverges, which by the theorem introduced earlier, the entire integral diverges.

It is extremely important to check the infinite discontinuity within the integration bound, consider the following example:

${\int }_{-1}^1\frac{1}{x^2}dx$

If one ignores that at $x=0$, the function has an infinite discontinuity, and directly applies the Fundamental Theorem of Calculus, one will get the incorrect result of $-2$:

${\int }_{-1}^1\frac{1}{x^2}dx=-\frac{1}{x}{|}_{-1}^1$

$=-1-1$

$=-2$

The correct approach is to recognize there is an infinite discontinuity at $x=0$, thus

${\int }_{-1}^1\frac{1}{x^2}dx={\int }_{-1}^0\frac{1}{x^2}dx+{\int }_0^1\frac{1}{x^2}dx$

$=\underset{a\to 0^{-}}{\lim }{\int }_{-1}^a\frac{1}{x^2}dx+\underset{b\to 0^{+}}{\lim }{\int }_b^1\frac{1}{x^2}dx$

$=\underset{a\to 0^{-}}{\lim }-\frac{1}{a}-1+(-1)-\underset{b\to 0^{+}}{\lim }-\frac{1}{b}$

$=\infty$

Do not assume that 2 infinities can cancel each other, in other words $\infty -\infty \ne 0$, this expression is undefined.

Practice