Definition Let f be defined on [a, ∞) × [α, β] to R. Suppose that for each t ∈ J = [α, β] the infinite integral
F (t) = Z ∞
a
f (x, t)dx = lim
c→∞
Z c a
f (x, t)dx
exists. We say that this convergence is uniform on J if for every > 0 there exists a number M () such that
if c ≥ M () and t ∈ J, then |F (t) − Z c
a
f (x, t)dx| < .
Dominated Convergence Theorem Suppose that f is integrable over [a, c] for all c ≥ a and all t ∈ J = [α, β]. Suppose that there exists a positive function φ defined for x ≥ a such that
|f (x, t)| ≤ φ(x) for x ≥ a, t ∈ J, and such that the integral
Z ∞ a
φ(x)dx exists.
Then, for each t ∈ J, the integral
F (t) = Z ∞
a
f (x, t)dx is (absolutely) convergent and the convergence is uniform on J.
Dirichlet’s Test Let f be continuous on [a, ∞) × [α, β] and suppose that there exists a constant A such that
| Z c
a
f (x, t)dx| ≤ A for all c ≥ a and for all t ∈ J = [α, β].
Suppose that for each t ∈ J, the function φ(x, t) is monotone decreasing for x ≥ a, and converges to 0 as x → ∞ uniformly for t ∈ J. Then the integral
F (t) = Z ∞
a
f (x, t)φ(x, t)dx converges uniformly on J.
Examples (a) R∞
0
dx
x2+ t2 converges uniformly for |t| ≥ a > 0.
(b) R∞ 0
dx
x2+ t converges uniformly for t ≥ a > 0 and diverges when t ≤ 0.
(c) R∞
0 e−xcos txdx converges uniformly for t ∈ R by the dominated convergence theorem.
(d) R∞
0 e−x2−t2/x2dx converges uniformly for t ∈ R by the dominated convergence theorem.
Theorem Let f be continuous on K = [a, b] × [c, d] to R and F : [c, d] → R be defined by F (t) =
Z b a
f (x, t)dx.
Then F is continuous on [c, d] to R.
Proof. Let > 0, since f is uniformly continuous on K, there exists a δ() > 0 such that if t and t0 belong to [c, d] and |t − t0| < δ(), then |f (x, t) − f (x, t0)| < , for all x ∈ [a, b]. It follows that
|F (t) − F (t0)| = | Z b
a
f (x, t) − f (x, t0)dx| ≤ Z b
a
|f (x, t) − f (x, t0)|dx ≤ (b − a), which establishes the continuity of F.
Remark Suppose that f is continuous on [a, ∞) × [c, d] to R and F (t) =
Z ∞ a
f (x, t)dx converges uniformly on [c, d], we let
Fn(t) = Z a+n
a
f (x, t)dx.
Then Fn is continuous on [c, d] and F is continuous on [c, d] since Fn converges to F uniformly on [c, d].
Theorem Let f and its partial derivative ft be continuous on K = [a, b] × [c, d] to R. Then the function
F (t) = Z b
a
f (x, t)dx is differentiable on (c, d) and
F0(t) = Z b
a
ft(x, t)dx for all t ∈ (c, d).
Proof From the uniform continuity of ft on K we infer that if > 0, then there is a δ() > 0 such that
if |t − t0| < δ(), then |ft(x, t) − ft(x, t0)| < for all x ∈ [a, b].
Let t, t0 satisfy this condition and apply the Mean Value Theorem to obtain a t1 ( which may depend on x and lies between t and t0) such that
f (x, t) − f (x, t0) = (t − t0)ft(x, t1).
Combining these two relations, we infer that
if 0 < |t − t0| < δ() then |f (x, t) − f (x, t0)
t − t0 − ft(x, t0)| < ∀ x ∈ [a, b].
Thus, we obtain
|F (t) − F (t0) t − t0 −
Z b a
ft(x, t0)dx| ≤ Z b
a
|f (x, t) − f (x, t0)
t − t0 − ft(x, t0)|dx ≤ (b − a), which establishes F0(t) =Rb
a ft(x, t)dx.
Generalization. Let S be a measurable subset of Rn and T a subset of Rm. Suppose f (x, y) is a function on T × S that is integrable as a function of y ∈ S for each x ∈ T, and let F be defined on T by F (x) =R
Sf (x, y)dy for x ∈ T.
(a) If f (x, y) is continuous as a function of x ∈ T for each y ∈ S, and there exists a constant C such that |f (x, y)| ≤ C for all x ∈ T and y ∈ S, then F is continuous on T.
(b) Suppose T is open. If f (x, y) is of class C1 as a function of x ∈ T for each y ∈ S, and there is a constant C such that |∇xf (x, y)| ≤ C for all x ∈ T and y ∈ S, then F is of class C1 on T and ∂F
∂xj(x) =R
S
∂f
∂xj(x, y)dy for x ∈ T.
Proof Let {xj} be a sequence in T converging to x ∈ T. For each j ∈ N and y ∈ S, let fj(y) = f (xj, y) and let f (y) = f (x, y). Then each fj and f are integrable on S, and |fj(y)| ≤ C and fj(y) → f (y) for all j and all y ∈ S. The bounded convergence theorem implies that lim F (xj) = limR
Sf (xj, y)dy = limR
Sfj = R
Slim fj = R
Slim f (xj, y) = R
Sf (x, y) = F (x).
Hence, F is continuous and this proves (a).
Part (b) is proved by applying the bounded convergence theorem to the sequence of difference quotients
f (x + hjei, y) − f (x, y)
hj ,
where ei denotes the unit vector in the xi−coordinate and {hj} is a sequence of numbers tending to zero. The uniform bound on these quotients is obtained by applying the mean value theorem .
Examples
(a) Let f (x, t) = cos tx
1 + x2 for x ∈ [0, ∞) and t ∈ (−∞, ∞). Then R∞
0 f (x, t) converges uniformly for t ∈ R by Dominated Convergence Theorem.
(b) Let f (x, t) = e−xxt for x ∈ [0, ∞) and t ∈ [0, ∞). For any β > 0, the integral R∞
0 f (x, t) converges uniformly for t ∈ [0, β] by Dominated Convergence Theorem. Similarly, the Laplace transform of xn, n = 0, 1, 2, . . . , defined by
L {xn}(t) = Z ∞
0
xne−txdx
also converges uniformly for t ≥ γ > 0 to n!
tn+1. For t ≥ 1, define the gamma function Γ by Γ(t) =
Z ∞ 0
xt−1e−xdx.
Then it is uniformly convergent on an interval containing t. Note that Γ(t + 1) = tΓ(t) and hence Γ(n + 1) = n! for any n ∈ N.
(c) Let f (x, t) = e−txsin x for x ∈ [0, ∞) and t ≥ γ > 0. Then the integral F (t) =
Z ∞ 0
e−txsin xdx
the laplace transform of sin x, denoted by L {sin x}(t). Note that an elementary calculation shows that
L {sin x}(t) = 1 1 + t2. (d) Let f (x, t, u) = e−txsin ux
x for x ∈ [0, ∞) and t, u ∈ [0, ∞). By taking φ = e−tx/x and by applying the Dirichlet’s test, one can show that R∞
γ f (x, t, u) converges uniformly for t ≥ γ ≥ 0. Note that if
F (t, u) =L {sin ux x }(t) =
Z ∞ 0
e−txsin ux x dx,
then ∂F
∂u(t, u) = Z ∞
0
e−txcos uxdx = t
t2+ u2, and F (t, u) = tan−1u t. By setting u = 1 and by letting t → 0+, we obtain that
Z ∞ 0
sin x
x dx = π 2.
(e) Let G(t) =R∞
0 e−x2−t2/x2dx for t > 0. Then G0(t) = −2G(t) and G(t) =
√π 2 e−2t. (f) Let F (t) =R∞
0 e−x2cos txdx for t ∈ R. Then F0(t) = −t
2F (t) and F (t) =
√π 2 e−t2/4.
Leibiniz’s Formula Let f and its partial derivative ft be continuous on K = [a, b] × [c, d] to R and α and β be differentiable functions on [c, d] and have values in [a, b]. Then the function
φ(t) = Z β(t)
α(t)
f (x, t)dx is differentiable on (c, d) and
φ0(t) = f (β(t), t)β0(t) − f (α(t), t)α0(t) + Z β(t)
α(t)
ft(x, t)dx for t ∈ (c, d).
Proof Let H be defined for (u, v, t) by
H(u, v, t) = Z u
v
f (x, t)dx, where u, v belong to [a, b] and t belongs to [c, d]. Then
φ(t) = H(β(t), α(t), t).
Applying the Chain Rule to obtain the result.
Interchange Theorem Let f be continuous on K = [a, b] × [c, d] to R. Then Z d
c
Z b a
f (x, t)dx
dt =
Z b a
Z d c
f (x, t)dt
dx.
Proof Since f is uniformly continuous on K, if > 0 there exists a δ() > 0 such that if |x − x0| < δ() and |t − t0| < δ() then |f (x, t) − f (x0, t0)| < .
Let n be chosen so large that (b − a)/n < δ() and (d − c)/n < δ() and divide K into n2 equal rectangles by dividing [a, b] and [c, d] each into n equal parts. For j = 0, 1, . . . , n, we let
xj = a + (b − a)j/n, tj = c + (d − c)j/n.
Then Z d
c
Z b a
f (x, t)dx
dt =
n
X
i=1 n
X
j=1
Z ti
ti−1
(Z xj
xj−1
f (x, t)dx )
dt =
n
X
i=1 n
X
j=1
f (x0j, t0i)(xj−xj−1)(ti−ti−1).
Similarly,
Z b a
Z d c
f (x, t)dt
dx =
n
X
i=1 n
X
j=1
f (x00j, t00i)(xj − xj−1)(ti− ti−1).
The uniform continuity of f implies that two iterated integrals differ by at most (b − a)(d − c).
Since > 0 is arbitrary, the equality of these integrals is confirmed.
Example Let A ⊆ R2 be the set consisting of all pairs (i/p, j/p) where p is a prime number, and i, j = 1, 2, . . . , p − 1. (a) Show that each horizontal and each vertical line in R2 intersects A in a finite number (often zero) of points and that A does not have content. Let f be defined on K = [0, 1] × [0, 1] by f (x, y) = 1 for (x, y) ∈ A and f (x, y) = 0 otherwise. (b) Show that f is not integrable on K. However, the iterated integrals exist and satisfy
Z 1 0
Z 1 0
f (x, y)dx
dy =
Z 1 0
Z 1 0
f (x, y)dy
dx.
Example Let K = [0, 1] × [0, 1] and let f : K → R be defined by
f (x, y) =
0 if either x or y is irrational, 1
n if y is rational and x = m
n where m and n > 0 are relatively prime integers.
Show that
Z
K
f = Z 1
0
Z 1 0
f (x, y)dx
dy = 0, but that R1
0 f (x, y)dy does not exist for rational x.
Fubini’s Theorem Let f be continuous on K = [a, b] × [c, d] to R. Then Z
K
f = Z d
c
Z b a
f (x, y)dx
dy =
Z b a
Z d c
f (x, y)dy
dx.
Proof Let F be defined for y ∈ [c, d] by F (y) =
Z b a
f (x, y)dx.
Let c = y0 < y1 < · · · < yr = d be a partition of [c, d], let a = x0 < x1 < · · · < xs = b be a partition of [a, b], and let P denote the partition of K obtained by using the cells [xi−1, xi] × [yj−1, yj]. Let yj∗ be any point in [yj−1, yj] and note that
F (yj∗) = Z b
a
f (x, yj∗)dx =
s
X
i=1
Z xi
xi−1
f (x, y∗j)dx.
The Mean Value Theorem implies that for each j, there exists a x∗ji ∈ [xi−1, xi] such that
F (yj∗) =
s
X
i=1
f (x∗ji, yj∗)(xi− xi−1).
We multiply by (yj − yj−1) and add to obtain
r
X
j=1
F (yj∗)(yj − yj−1) =
r
X
j=1 s
X
i=1
f (x∗ji, y∗j)(xi− xi−1)(yj− yj−1).
We have shown that an arbitrary Riemann sum for F on [c, d] is equal to a particular Riemann sum of f on K corresponding to the partition P. Since f is integrable on K, the existence of the iterated integral and its equality with the integral on K is established.
A minor modification of the proof given for the preceding theorem yields the following, slightly stronger, assertion.
Generalization Theorem Let f be integrable on K = [a, b] × [c, d] to R and suppose that for each y ∈ [c, d], the function x 7→ f (x, y) of [a, b] into R is continuous except possibly for a finite number of points, at which it has one-sided limits. Then
Z
K
f = Z d
c
Z b a
f (x, y)dx
dy.
Corollary Let A ⊆ R2 be given by A = {(x, y) : α(y) ≤ x ≤ β(y), c ≤ y ≤ d}, where α and β are continuous functions on [c, d] with values in [a, b]. If f is continuous on A 7→ R, then f is integrable on A and
Z
A
f = Z d
c
(Z β(y) α(y)
f (x, y)dx )
dy.
Proof Let K ⊇ A be a closed cell and fK be the extension of f to K. Since ∂A has content zero, fK is integrable on K. Now for each y ∈ [c, d] the function x 7→ fK(x, y) is continuous except possibly at the two points α(y) and β(y), at which it has one-sided limits. It follows from the preceding theorem that
Z
A
f = Z
K
fK = Z d
c
Z b a
fK(x, y)dx
dy =
Z d c
(Z β(y) α(y)
f (x, y)dx )
dy.
Example Let R denote the triangular region in the first quadrant bounded by the lines y = x, y = 0, and x = 1. ThenR1
0
R1 y
sin x
x dxdy =R1 0
Rx 0
sin x x dydx.
Example Z 2
0
Z 1 y/2
ye−x3 dxdy = Z 1
0
Z 2x 0
ye−x3 dydx = Z 1
0
y2
2e−x3|2x0 dx = Z 1
0
2x2e−x3dx = −2e−x3
3 |10 = 2
3(1−e−1).
Example For β > α > 0, let R = [0, ∞) × [α, β] and f (x, t) = e−tx. Then log β
α = Z β
α
1 tdt =
Z β α
Z ∞ 0
e−txdxdt = Z ∞
0
Z β α
e−txdtdx = Z ∞
0
e−αx− e−βx
x dx.
