## Section 2.9 | Improper Integrals

What are improper integrals? |

It should seem somewhat odd that we have gone through so many integration techniques and yet have not seen a single integral of the form, say, \[\int_2^{\infty}{e^{-2x} dx}.\]I mean, we are mired in calculus, up to our necks in limits, derivatives, infinity, and the infinitesimal, yet we never have asked the fundamental question containing each of these elements. What does it mean to evaluate a definite integral that has an infinite upper or lower limit? Wait... what if both limits are infinite? (cue the creepy music)

Worse yet, we have completely ignored (unbeknownst to many of us) integrals like \[\int_1^7{\frac{1}{x - 5} dx}.\]In fact, many may not even know what the problem is with that integral. I'm sure you're looking at it right now and saying, "What's so hard about integrating that? The antiderivative of \(\frac{1}{x - 5}\) is \(\ln{|x - 5|}\). That's easy!"

Hold on there, wise guy.

Let me give you a hint. Between 1 and 7 is the pesky, troublesome number 5. What do we do here? How do we overcome, or at least try to overcome, this domain issue?

These types of integrals, those with infinite limits or domain issues across the limits, deserve a special name - something that describes an intolerable type of integral. How about improper?

Hold on there, wise guy.

Let me give you a hint. Between 1 and 7 is the pesky, troublesome number 5. What do we do here? How do we overcome, or at least try to overcome, this domain issue?

These types of integrals, those with infinite limits or domain issues across the limits, deserve a special name - something that describes an intolerable type of integral. How about improper?

**DEFINITION | IMPROPER INTEGRAL**

An

**improper integral**is an integral with any of the following properties.

- The upper limit is infinite.
- The lower limit is infinite.
- The domain of the integrand is undefined on the interval of integration.
- The integrand is discontinuous at some point on the interval of integration.

Technically speaking, condition 3 is a subset of condition 4, so there is truly no real need to write it; however, I find that it is important enough (and common enough) to distinguish condition 3 as a condition to be an improper integral.

To be absolutely clear, it is important that we define the notation used for improper integrals and how they are defined as limits.

To be absolutely clear, it is important that we define the notation used for improper integrals and how they are defined as limits.

DEFINITION | IMPROPER INTEGRAL NOTATION- If the integral exists \(\forall t \geq a\), then we define \[\int_a^{\infty}{f(x) dx} = \lim_{t \to \infty}{\int_a^t{f(x) dx}},\] provided this limit exists and is finite.
- If the integral exists \(\forall t \leq a\), then we define \[\int_{-\infty}^a{f(x) dx} = \lim_{t \to -\infty}{\int_t^a{f(x) dx}},\] provided this limit exists and is finite.
- If both \(\int_a^{\infty}{f(x) dx}\) and \(\int_{-\infty}^a{f(x) dx}\) exist, then we define \[\int_{-\infty}^{\infty}{f(x) dx} = \int_{-\infty}^a{f(x) dx} + \int_a^{\infty}{f(x) dx}\]
- If the integrand \(f\) is continuous on \([a,c)\) and is discontinuous at \(c\), then we define \[\int_a^c{f(x) dx} = \lim_{t \to c^-}{\int_a^t{f(x) dx}}\] provided this limit exists and is finite.
- If the integrand \(f\) is continuous on \((c,b]\) and is discontinuous at \(c\), then we define \[\int_c^b{f(x) dx} = \lim_{t \to c^+}{\int_t^b{f(x) dx}}\] provided this limit exists and is finite.
- If \(f\) has a discontinuity at \(c\) and both 4. and 5. exist, then we define \[\int_a^b{f(x) dx} = \int_a^c{f(x) dx} + \int_c^b{f(x) dx}\]
improper integral notation. |
HELPFUL HINTSome people call the improper integrals having infinite limits, "Type 1" improper integrals. !WARNING!Symmetry cannot be used to simplify the process of evaluating an improper integral. HELPFUL HINTAgain, some people call the improper integrals having discontinuous integrands, "Type 2" improper integrals. |

That list of definitions may

What happens when the limit associated with an improper integral does not exist or is not finite? We've got a definition for that!

*seem*daunting, but you will soon find yourself intimately familiar with the forms. The important thing to remember is that statement, "provided this limit exists and is finite." If it has been a while since you have played around with limits, you might want to review them now as evaluating an improper integral requires a strong knowledge of l'Hospital's Rule and the limit laws.What happens when the limit associated with an improper integral does not exist or is not finite? We've got a definition for that!

**DEFINITION | CONVERGENT and DIVERGENT INTEGRAL**

An improper integral is called

**convergent**if the corresponding limit exists (and is finite) and

**divergent**if this limit does not exist (or is infinite).

Convergence or divergence of a given improper integral is often the main desired piece of information. Moreover, once we know that an improper integral is convergent, we can

Why "often" and not "always"? That conversation is going to be held for next section. For now, we will only be considering improper integrals that we have a pretty good chance of evaluating (or determining they diverge).

Before trying our hand at a few improper integrals, let's establish a theorem that will be incredibly useful for us. I should note that the following theorem has significant implications in the analysis of infinite series as well.

*often*establish its value by evaluating the corresponding limit.Why "often" and not "always"? That conversation is going to be held for next section. For now, we will only be considering improper integrals that we have a pretty good chance of evaluating (or determining they diverge).

Before trying our hand at a few improper integrals, let's establish a theorem that will be incredibly useful for us. I should note that the following theorem has significant implications in the analysis of infinite series as well.

**THEOREM**

If \(a > 0\), then \[\int_a^{\infty}{\frac{1}{x^p} dx}\] is convergent if \(p > 1\) and divergent if \(p \leq 1\).

**EXAMPLE 1 | Proving a theorem**

Prove the previous theorem.

**EXAMPLE 2 | Subbing to infinity**

Determine if the integral converges or diverges. If it converges, determine its value. \[\int_{-\infty}^0{\frac{1}{\sqrt{3 - x}} dx}\]

EXAMPLE 3 | Cubing funDetermine if the integral converges or diverges. If it converges, determine its value. \[\int_{-\infty}^{\infty}{73\frac{x^2}{9 + x^6} dx}\] |
HELPFUL HINTIf evaluating an improper integral on \([a,b]\) and there is a discontinuity at a point \(c \in (a,b)\), then you must split the improper integral into two pieces about \(c\). Moreover, if either of these resulting improper integrals are divergent, the entire improper integral is divergent! |

**EXAMPLE 4 | An original sin(e)**

Determine if the integral converges or diverges. If it converges, determine its value. \[\int_0^{\sqrt{2}}{\frac{1}{\sqrt{2 - x^2}} dx}\]

**EXAMPLE 5 | This integral has got issues**

Consider the following integral. \[\int_1^3{\frac{1}{x^2 - 2} dx}\]

- State why this is an improper integral.
- You decide to use partial fraction decomposition, while your friend decides to tackle this using a trigonometric substitution. Explain why your friend's tactic is a losing one.
- Determine if this integral converges or diverges. If it converges, determine its value.