Improper Integrals - Mathematics.lu

Given the graph of an integrable function $f(x)$ on some interval $x=a$ to $x=b$ , the integral $\int_a^b f(x)dx$ represents the net area between the graph of the function and the $x$ -axis.

An improper integral is a special type of integral which we will discuss in this section. The technique to find such an integral is not difficult, and we will see that even though some areas may at first seem infinite, they will in fact be finite. We begin with a definition.

Definition: An improper integral is a definite integral meeting at least one of the following two conditions:

Infinite Intervals: One or both of the limits of integration are infinite, meaning the interval over which you integrate is unbounded (for example, $\int_1^\infty f(x) \, dx$ or $\int_{-\infty}^{\infty} f(x) \, dx$ ).
Finite Interval with Discontinuity: The integrand function $f(x)$ approaches infinity at some point within or at the boundaries of the interval of integration $[a, b]$ , creating a vertical asymptote.

We evaluate an improper integral as follows:

An integral of the form $\int_a^\infty f(x)dx$ becomes $\lim_{b\to\infty}\int_a^b f(x)dx$ . We evaluate this integral and then take $b\to\infty$ .
An integral of the form $\int_{-\infty}^b f(x)dx$ is evaluated analogously. I.e.: Taking $\lim_{a\to-\infty}\int_a^bf(x)dx$ .
An integral of the form $\int_{-\infty}^\infty f(x)dx$ can be divided into two integrals as above and calculated separately. If either part is infinity, the total part is infinity.
An integral of the form $\int_a^bf(x)dx$ where at some point $a\leq c\leq b$ the function has a vertical asymptote can be decomposed into two integrals (before and after said point). E.g.:
$\int_{-1}^1 \frac{1}{x}dx=\int_{-1}^0 f(x)dx+\int_0^1 f(x)dx$

Then each integral is calculated respectively.

If the limit yields a finite value, the integral converges. If the limit does not exist or equals infinity, the integral diverges.

Table of Contents

Improper Integrals Over Infinite Intervals

Taking into consideration that the integral represents the area, you could think that every integral with an infinite integral will be infinite. However, when your function has a horizontal asymptote at $y=0$ , your integral might end up being finite. This is very closely related to an optional topic in advanced maths called Infinite Series and in the sequences part of the programme in 5 periods you discuss infinite sums (e.g.: with geometric sequences). Since an integral is by definition an infinite sum, it makes sense these are related.

Example: Evaluate the following improper integral $\int_1^\infty \frac{1}{x} \, dx$ to determine whether it converges or diverges.

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Example: Evaluate the following improper integral $\int_1^\infty \frac{1}{x^2} \, dx$ to determine whether it converges or diverges.

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The next example will show an integral where both endpoints of the interval are infinity. This integral does not form part of the 5 periods programme (this is seen in advanced maths), but we can simply accept the fact that $\int\frac{1}{x^2+1}dx=\arctan(x)$ and still take advantage of an example that covers one of our special cases.

Example: Evaluate the following improper integral $\int_{-\infty}^{\infty} \frac{1}{x^2+1} \, dx$ to determine whether it converges or diverges.

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Common Mistake with Improper Integrals

When taking an improper integral of the form $\int_{-\infty}^\infty f(x)dx$ , a common mistake is to try and evaluate the integral “simultaneously”. Our method tells us we need to divide this into two integrals:

$\int_{-\infty}^0 f(x)dx+\int_0^\infty f(x)dx$

However, often students will forget this and calculate everything as one limit so the $F(\infty)-F(-\infty)$ seems to cancel out (depending on the function) due to the minus between them. Let us see this in an example.

Example: Evaluate the improper integral $\int_{-\infty}^{\infty} \frac{x}{x^2+1} \, dx$ to determine whether it converges or diverges.

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Improper Integral over a Finite Interval with Discontinuity

We cover our final case, where the function has a point of discontinuity within the interval of integration. Just because your function has a vertical asymptote it does not imply that the integral will be infinite.

Example: Evaluate the improper integral $\int_0^1 \frac{1}{\sqrt{x}} \, dx$ to determine whether it converges or diverges.

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Exercises

Determine whether each of the following improper integrals converges or diverges. If it converges, find its exact value.

1. Evaluate $\int_{0}^{\infty} e^{-2x} \, dx$

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2. Evaluate $\int_{2}^{\infty} \frac{1}{x^3} \, dx$

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3. Evaluate $\int_{e}^{\infty} \frac{1}{x \ln(x)} \, dx$

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4. Evaluate $\int_{-\infty}^{0} e^{x} \, dx$

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5. Evaluate $\int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx$

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Determine whether each of the following Type II improper integrals converges or diverges. If it converges, find its exact value.

1. Evaluate $\int_{0}^{1} \frac{1}{x} \, dx$

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2. Evaluate $\int_{1}^{2} \frac{1}{(x-1)^2} \, dx$

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3. Evaluate $\int_{0}^{e} \ln(x) \, dx$

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Determine whether the following improper integral converges or diverges. If it converges, find its exact value. Disclaimer: This type of integral should no longer be in the programme, but the hint below tells you how to approach it.

$\int_{-\infty}^{\infty} e^{-|x|} \, dx$

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Hint: To integrate a function with absolute value, split the integral at $x = 0$ into two separate halves with independent limits:

$\int_{-\infty}^{\infty} e^{-|x|} \, dx = \int_{-\infty}^{0} e^{x} \, dx + \int_{0}^{\infty} e^{-x} \, dx$

This is because in each interval the sign of $x$ does not change so you can understand what the absolute value does to the variable.

Analyse the behaviour of the following improper integral based on the value of the constant exponent $n$ :

$\int_{1}^{\infty} x^n \, dx$

Find a condition on $n$ for which this improper integral converges, and the condition for which it diverges.

Hint: Make sure to test the specific case where $n = -1$ separately, as it uses a different integration rule.