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Applications of Integration
Adrian Down 16779577
July 28, 2005
1 Sequences of integrable functions
Theorem: if (fn) is a sequence of functions that are integrable on [a, b], and (fn) converges to f on [a, b], then f is integrable on [a, b] and
∫ (^) b
a
f (x)dx = lim n→∞
∫ (^) b
a
fn(x)dx
Notes: 1) The proof is easy if fn is continuous ∀n
- Theorem is false if fn converges to f pointwise. Proof: simply follow the definitions. Let > 0 be given. By the definition of uniform convergence, ∃N such that
n > N → |fn(x) − f (x)| <
4(b − a)
, ∀x
Fix n > N. f (n) is integrable → ∃ a partition P such that U (fn, P ) − L(fn, P ) < 2. Now we want to compare upper and lower sums for f for the same par- tition. For each subinterval [tk− 1 , tk] ⊆ P ,
M (f, [tk− 1 , tk]) ≤ M (fn, [tk− 1 , tk]) +
4(b − a)
Likewise,
m(f, [tk− 1 , tk]) ≥ m(fn, [tk− 1 , tk]) −
4(b − a)
Now we examine the upper and lower sums for f.
U (f, P ) ≤
∑^ n
k=
M (fn, [tk− 1 , tk]) · (tk − tk− 1 ) +
∑^ n
k=
4(b − a)
(tk − tk− 1 )
≤ U (fn, P ) +
4(b − a)
(b − a) = U (fn, p) +
Can also get a lower bound for L(f, P ) similarly.
L(f, P ) ≥ L(fn, P ) −
Put these two facts together to examine the difference,
U (f, P ) − L(f, P ) ≤ U (fn, P ) +
− (L(f, P ) −
= U (fn, P ) − L(fn, P ) +
Thus f is integrable. Now we need to compute the value of the integral. Given n > N ,
∫ (^) b
a
f (x)dx −
∫ (^) b
a
fn(x)dx| = |
∫ (^) b
a
(f (x) − fn(x))dx|
∫ (^) b
a
|f (x) − fn(x)|dx ≤
∫ (^) b
a
4(b − a)
dx =
4(b − a)
(b − a) =
Thus
lim n→∞
∫ (^) b
a
fn(x)dx =
∫ (^) b
a
f (x)dx
2 Power Series
As usual, the type of series of functions that we’re interested in is power series. In particular, if
fn(x) =
∑^ n
k=
akxk^ with f (x) =
∑^ ∞
k=
akxk
Proof: Let
g(x) =
∑^ ∞
k=
ak · kxk−^1
g has the same radius of convergence as
x · g(x) =
∑^ ∞
k=
ak · kxk
Let β = lim sup n→∞
|akk|
(^1) k = lim k→∞
k
(^1) k · lim sup |ak|
(^1) k
limk→∞ k
1 k (^) = 1 → x · g(x) has the same radius of convergence as f (x). Thus we can differentiate a power series term by term. These two tactics are useful if you see a power series that looks similar to one for which you know the formula for.
3 Improper Integrals
So far, we can only integrate functions on finite closed intervals. Improper integrals generalize this. Suppose f is defined on [a, b), where b = +∞ is possible. Then if f is integrable on [a, d] for any d < b, consider
lim d→b−
∫ (^) d
a
f (x)dx =
∫ (^) b
a
f (x)dx
This definition holds if the limit exists. It is called the improper integral. Can do the same thing for half-open intervals on the other side. If f is defined on (a, b], where a could be −∞, and f is defined and integrable on [c, d], ∀c > a,
lim c→a+
∫ (^) b
c
f (x)dx =
∫ (^) b
a
f (x)dx
If f is defined on (a, b), then define ∫ (^) b
a
f (x)dx =
∫ (^) α
a
f (x)dx +
∫ (^) b
α
f (x)dx
for some α ∈ (a, b), assuming both limits exist. Note that we can’t take ∫ (^) ∞
−∞
f (x)dx = lim a→∞
∫ (^) a
−a
f (x)dx
This implies the integral is symmetric, which might not be the case. Examples:
∫ (^) ∞
1
dx x
= lim d→∞
∫ (^) d
1
dx x
= lim d→∞
(log d) = +∞
0
dx x
= lim c→ 0 +
c
dx x
= lim c→ 0 +
(− log c) = +∞
1
x−pdx = lim d→∞
∫ (^) d
1
x−pdx = lim d→∞
x−p+ −p + 1
|d 1 = lim d→∞
d−p+ −p + 1
−p + 1
If p > 1, then −p + 1 < 0 → d−p+1^ converges to 0 as d approaches ∞. So ∫ (^) ∞
0
x−pdx =
p − 1
If p < −1, then −p + 1 > 0 → d−p+1^ diverges to +∞ as d approaches ∞. So ∫ (^) ∞
1
x−pdx
diverges to +∞.
∫ (^) ∞
0
e−λxdx = lim d→∞
∫ (^) d
0
e−λxdx = lim d→∞
λ
e−λx)|d 0 = lim d→∞
λ
e−λd^ +
λ
λ > 0 → e−λd^ → 0 as d → 0 ∫ (^) ∞
0
e−λxdx =
λ
