MAT187 L20 - Improper Integrals with Infinite Bounds

Improper Integrals

For a function $f (x)$ that is continuous on $[a, \infty)$ for some $a \in R$ , we can define an improper integral to be:
$\int_{a}^{\infty} f (x) d x = lim_{t \to \infty} \int_{a}^{t} f (x) d x$

If this limit exists, then the improper integral converges.
If it doesn't exist, the improper integral diverges.

If the improper integral diverges, that doesn't necessarily mean that the function "blows up"—rather, it means that the integral is not finite, i.e. it doesn't converge.

For example, the function $lim_{t \to \infty} \int_{0}^{t} \sin (x) d x$ diverges, but does not "blow up". Instead, it oscillates to infinity.

Note that you cannot say

\int_{0}^{\infty} f (x) d x

This is not valid because technically $\infty$ isn't a defined number.

Applications

A very common use case for improper integrals is the normal distribution curves (e.g., the bell curve). This can be represented by the function $f (x) = e^{- x^{2}}$ .

If we want to find the probability that something falls below a certain value, whose distribution is represented by $f (x) = e^{- x^{2}}$ , we need to integrate from $- \infty$ to some value.
Thus, we should use an improper integral to evaluate this.

	left=-2; right=2;
    top=2; bottom=-1;
	---
	y=e^{-x^2}

Example: Car Warranties

Imagine, if you will, a company that produces car parts. The failure rate of the parts is represented by a continuous function $f (t)$ , so the cost of the parts replacements is given by $\int_{a}^{b} f (t) d t$ , from time $t = a$ to $t = b$ .

If we want to find the total cost of car failures that will happen after a value $t = a$ , we need to evaluate to "infinity".
- This is represented by $lim_{t \to \infty} \int_{a}^{t} f (t) d t$ .

Example: CPU Failure Rate

Consider Intel, a CPU company. Suppose their Core i9-14900K CPUs fail at a rate that increases as time goes on, and is thus represented by $\int_{a}^{b} 0.2 e^{- 0.2 t} d t$ .

The number of CPUs that fail between year 1 and year 2 is: $\int_{1}^{2} 0.2 e^{- 0.2 t} d t = 0.148 \dots$
- Thus, the percentage of CPUs that fail between year 1 and 2 is $14.8 %$ .
We can also evaluate the number of CPUs that fail over an entire lifetime. This is represented by $lim_{a \to \infty} \int_{0}^{a} 0.2 e^{- 0.2 t} d t$ .

Note that this rate is abnormally high for CPUs, which generally have a failure rate <5%.

This is because the 14th and 13th gen of Intel CPUs are prone to degrading. Oops!