230 CHAPTER 10. IMPROPER INTEGRALS

Lemma 10.0.4 Suppose f ∈ L1 (a,b) . Then for g(t) any bounded continuous function,say |g(t)| ≤M, f g is improper Riemann integrable.

Proof: There is no issue about the existence of∫

β

α| f (t)| |g(t)|dt for [α,β ] ⊆ (a,b)

thanks to Theorem 9.3.13. Then

sup{∫

β

α

| f (t)| |g(t)|dt : [α,β ]

}≤ sup

{∫β

α

| f (t)|Mdt : [α,β ]

}

= M sup{∫

β

α

| f (t)|dt : [α,β ]

}= M

∫ b

a| f (t)|dt

Since | f g| ∈ L1 (a,b) , it follows that f g is improper Riemann integrable by Proposition10.0.2.

Note that, from the last claim about computing the improper integral, all the usual alge-braic properties of the Riemann integral carry over to these improper integrals of functionsin L1. For example, the integral is linear.

Sometimes it is convenient to define limR→∞

∫ R−R f (t)dt. This may exist even though

f (t) may not be improper integrable. Such limits are called the Cauchy principal valueintegrals.

Example 10.0.5 limR→∞

∫ R−R

t1+t2 dt = 0 but limn→∞

∫ R2

−Rt

1+t2 dt = ∞. Both −R→−∞ andR2→ ∞.

This is left as an exercise. Note that if∫

∞

−∞f (t)dt does exist, then you can find it as

a Cauchy principal value integral. It is just that sometimes the Cauchy principal valueintegral exists even though the function is not improper Riemann integrable. In the aboveexample, the function is not in L1. Indeed, you should show that limR→∞

∫ R−R

|t|1+t2 dt = ∞.

When functions have values in C, there is no extra problem. You simply consider thereal and imaginary parts. That is,∫ b

af (t)dt ≡

∫ b

aRe f (t)dt + i

∫ b

aIm f (t)dt

and define a function to be improper Riemann integrable if and only if this is true of its realand imaginary parts.

Here a complex valued function is in L1 (a,b) means both the real and imaginary partsare in L1 (a,b) . This is equivalent to saying | f | is in L1 (a,b) because

max(|Re f | , |Im f |)≤ | f | ≤ |Re f |+ |Im f |

Also for f ∈ L1 (a,b) , I use∣∣∣∫ b

a f (t)dt∣∣∣ ≤ ∫ b

a | f (t)|dt whenever convenient. This iscertainly true if f has real values. However, it is also true for complex valued f . Likelythe easiest way to see this is to note that it is true for sums. Since these approximate theintegrals, it will be true for the integrals also. You could also do the following. There existsθ ∈ C such that |θ |= 1 and

θ

∫ b

af (t)dt =

∣∣∣∣∫ b

af (t)dt

∣∣∣∣