9.2.1 Integrating to Infinity

In our study of expected lifetime, we considered integrals of the form

where `g text[(] t text[)]` was the specific function `t e^(-r t)` and `T` was "large." What we really wanted to describe is the integral "from zero to infinity," i.e., the limiting case as `T` increases to `oo`. For convenience, we use the abbreviated notation

to describe this two-step process:

1. Integrate from `0` to a large (but unspecified) upper limit `T`.
2. Find the limiting value as `T` grows infinitely large.

Such an integral is called improper to distinguish it from a "proper" definite integral.

Definition   If `g text[(] t text[)]` is a continuous function for `t ≥ a` then the improper integral ∫ a ∞ g ( t )   d t is the limiting value as `T rarr oo` of integrals of the form ∫ a T g ( t )   d t . If the limiting value exists as a finite number, we say the improper integral converges. Otherwise, it diverges.

Example 1

1. For light bulbs with cumulative failure fraction function

   F ( t ) = 1 - e - r ⁢ t ,

   show that the expected time to failure is exactly `1 text[/] r`.

2. Find the expected lifetime of a light bulb for which `r = 0.032` when time is measured in days.

Solution   

1. As we saw in the preceding section, the expected lifetime is the integral from `0` to `oo` of `t f text[(] t text[)]`, where `f text[(] t text[)] = F' text[(] t text[)] = re^(-rt)`. As in Activity 3 in Section 9.1, we may integrate by parts to find that
   ∫ t e - r ⁢ t   d t = - 1 r t e - r ⁢ t - 1 r 2 e - r ⁢ t = - 1 r 2 ( r ⁢ t + 1 ) e - r ⁢ t ,
   so
   r ∫ 0 T t e - r ⁢ t   d t = - 1 r ( r ⁢ t + 1 ) e - r ⁢ t ⁢ | 0 T = - 1 r ( r ⁢ T + 1 ) e - r ⁢ t + 1 r .
   Note that `T` appears only in the first of the two terms on the right — and the second term is our announced answer. To complete the argument that
   r ∫ 0 ∞ t e - r ⁢ t   d t = 1 r ,
   we need to show that, as `T rarr oo`,
   ( r ⁢ T + 1 ) e - r ⁢ T = r ⁢ T + 1 e r ⁢ T → 0.
   In the next chapter we will establish firmly that exponential functions grow much more rapidly than linear functions — indeed, than any polynomial functions. For now, observe that `rT + 1` grows at a constant rate (slope`= r`) and `e^(rT)` grows at an exponentially increasing rate. Since the denominator grows much faster than the numerator, the quotient approaches `0`.
2. If `r=0.032`, then the expected lifetime is `1 text[/] r = 31.25` days or `750` hours. Note that this agrees with the manufacturer's claim on the package, and not with our preliminary calculation of `741` hours.

Note 1 – On the discrepancy in average lifetimes

Checkpoint 1