Download a, b

CALCULUS HANDOUT 6 - RIEMANN-DARBOUX INTEGRAL. FOURIER SERIES
THE RIEMANN DARBOUX INTEGRAL
A partition P of the interval [a, b] is a finite set of points {x0 , x1 , .., xn } satisfying a = x0 < x1 < .. <
xn = b. Consider a function f defined and bounded on [a, b].
n
X
Upper Darboux sum of f related to P : Uf (P ) =
Mi (xi − xi−1 ) where Mi =
sup f (x).
xi−1 ≤x≤xi
i=1
Lower Darboux sum of f related to P : Lf (P ) =
n
X
mi (xi − xi−1 ) where mi =
i=1
inf
xi−1 ≤x≤xi
f (x).
Consider M = sup{f (x) | a ≤ x ≤ b} and m = inf{f (x) | a ≤ x ≤ b}.
For any partition P of [a, b] we have: m(b − a) ≤ Lf (P ) ≤ Uf (P ) ≤ M (b − a).
Lf = {Lf (P ) | P is a partition of [a, b]} and Uf = {Uf (P ) | P is a partition of [a, b]} are bounded sets.
So Lf = sup Lf and Uf = inf Uf exist. Moreover, Lf ≤ Uf .
A function defined and bounded on [a, b] is Riemann-Darboux integrable on [a, b] if Lf = Uf .
Z b
This common value is denoted by
f (x) dx = Lf = Uf .
a
Properties of the Riemann-Darboux integral:
If f and g are Riemann-Darboux integrable on [a, b] then all the integrals below exist and
Z b
Z b
Z b
(1)
(α f (x) + β g(x)) dx = α
f (x) dx + β
g(x) dx for any α, β ∈ R.
a
Zab
Z c
Z ab
(2)
f (x) dx =
f (x) dx +
f (x) dx for any a ≤ c ≤ b.
a
a
Zc b
Z b
(3) If f (x) ≤ g(x) on [a, b] then
f (x) dx ≤
g(x) dx.
a
a
¯Z
¯ Z
¯ b
¯
b
¯
¯
(4) ¯
f (x) dx¯ ≤
|f (x)| dx.
¯ a
¯
a
Classes of Riemann-Darboux integrable functions:
If f is continuous on [a, b], then f is Riemann-Darboux integrable on [a, b].
A function f is called piecewise continuous on [a, b] if there exists a partition P = {x0 , x1 , . . . , xn } of [a, b]
and continuous functions fi defined on [xi−1 , xi ], such that f (x) = fi (x) for x ∈ (xi−1 , xi ), i = 1, 2, . . . , n.
Z b
n Z xi
X
A piecewise continuous function is Riemann-Darboux integrable and
f (x) dx =
fi (x) dx.
a
i=1
xi−1
The integral mean value theorem:
If f and g are continuous on [a, b] and g(x) ≥ 0 for x ∈ [a, b], then there exists c between a and b such
Z b
Z b
that
f (x) · g(x) dx = f (c)
g(x) dx.
a
a
The fundamental theorem of calculus:
Z
x
f (t) dt, then F is continuous on [a, b].
If f is Riemann-Darboux integrable on [a, b] and F (x) =
a
Furthermore, if f is continuous on [a, b], then F is differentiable on [a, b] and F 0 = f .
Any function Φ such that Φ0 = f is called a primitive (antiderivative) of f .
Two primitives of the same function f differ by a constant.
Z b
If F is a primitive of f , then
f (x) dx = F (b) − F (a).
a
Integration by parts:
If the functions f and g are continuously differentiable on [a, b], then
Z
Z
0
f (x) · g (x) dx = f (x) · g(x) − f 0 (x) · g(x) dx
Z
Z
f 0 (x)g(x)dx represent the set of primitives of f g 0 and f 0 g, respectively.
¯b Z
Z b
¯
b
¯
0
Consequence:
f (x) · g (x) dx = f (x) · g(x)¯ −
f 0 (x) · g(x) dx
¯
a
a
where
f (x)g 0 (x)dx and
a
Change of variables:
If the function g : [α, β] → [a, b] is a continuously differentiable bijection having the property g(α) = a,
g(β) = b and f : [a, b] → R1 is continuous, then
µZ
¶
Z
f (x) dx ◦ g = (f ◦ g)(t) · g 0 (t) dt
Z
where
Z
f (x)dx and
Z
(f ◦ g)(t) · g 0 (t) dt represent the set of primitives of f and (f ◦ g) · g 0 , respectively.
Z
g(β)
Consequence:
β
f (x) dx =
g(α)
(f ◦ g)(t) · g 0 (t) dt
α
Improper integrals:
Let f be a function bounded on [a, ∞) and Riemann-Darboux integrable on [a, b] for every b > a.
Z b
Z ∞
If lim
f (x) dx exists, then the improper integral of first kind
f (x) converges.
b→∞
a
a
Let f be a function defined on (a, b] and Riemann-Darboux integrable on [a + ε, b], for ε ∈ (0, b − a).
Z b
Z b
If lim+
f (x) dx exists, then the improper integral of second kind
f (x) dx converges.
ε→0
a+ε
a
Comparison test for improper integrals of first kind:
Let f and g be defined on [a, ∞) and Riemann-Darboux
integrable on [a, b] for
every b > a.
Z ∞
Z ∞
Suppose that 0 ≤ f (x) ≤ g(x) for all x ≥ a and
g(x) dx converges. Then
f (x) dx converges too.
a
a
FOURIER SERIES
The Fourier series of a piecewise continuous function f defined on the interval [−π, π] is the series
∞
a0 X
+
(an · cos nx + bn · sin nx)
2
n=1
in which the Fourier coefficients an , bn are given by
Z
1 π
an =
f (x) · cos nx dx for n = 0, 1, 2, ...
π −π
Z
1 π
bn =
f (x) · sin nx dx for n = 1, 2, ...
π −π
Fourier Theorem:
Let f be a piecewise continuous function defined on the interval [−π, π] and extended by periodicity
outside it. We denote by Sn (x) the n-th partial sum of the Fourier series defined above.
If f (x) has finite left-hand and right-hand side derivatives at its points of discontinuity, then:
a) when x = x0 is a point of continuity of f , then lim Sn (x0 ) = f (x0 ).
n→∞
¤
1£
b) when x = x0 is a point of discontinuity of f , then lim Sn (x0 ) =
f (x+
) + f (x−
) .
0
0
n→∞
2
Change of the origin of the fundamental interval:
If f is a piecewise continuous function defined on the fundamental interval [−π, π] and by periodic
extension outside it, then for any α, the Fourier coefficients an , bn are given by
Z
1 α+π
f (x) · cos nx dx for n = 0, 1, 2, . . .
an =
π α−π
1
bn =
π
Z
α+π
f (x) · sin nx dx
for n = 1, 2, . . .
α−π
The Fourier series of f (x) converges at every point of continuity and:
∞
f (x) =
a0 X
+
(an cos nx + bn sin nx)
2
n=1
for x ∈ [α − π, α + π].
Change of the interval length:
If f is a piecewise continuous function defined on the interval [−L, L] and by periodic extension outside
it, then the Fourier coefficients are given by
an =
1
L
bn =
Z
1
L
L
f (x) · cos
−L
Z
nπx
dx
L
L
f (x) · sin
−L
nπx
dx
L
for n = 0, 1, 2, . . .
for n = 1, 2, . . .
The Fourier series (see below) of f (x) converges at every point of continuity and:
∞
f (x) =
a0 X
nπx
nπx
+
+ bn sin
)
(an cos
2
L
L
n=1
Odd and even functions:
If the function f is even, then bn = 0 for any n = 1, 2, . . ..
If the function f is odd, then an = 0 for any n = 0, 1, 2, . . ..
EXERCISES
1. Find the primitives of the following functions on their maximal domains of definition:
1
1
1 − x2
√
11. f (x) = √ + √
1. f (x) = x2 + 2x + 3
23. f (x) =
3
2
x
(1 + x2 ) 1 + x2
x
1
√
√
3
2. f (x) = x +
1
12. f (x) = x x + 2x x2
x
24. f (x) =
1 + x4
3. f (x) = sin x + cos x
13. f (x) = 1 + cos 3x
x2
1
14. f (x) = ex cosh(2x)
25. f (x) =
4. f (x) = √
1 + x4
1 − 4x2
15. f (x) = |x|
x2
1
1
26. f (x) =
5. f (x) = √
√
16. f (x) =
√
1 + x12
4 − x2
1+ x+ x+1
ex
2
2
arctan x
27. f (x) = 2x
x
e
+
6. f (x) =
e + e−2x
17. f (x) = p
cos2 x
sin2 x
2 )3
(1
+
x
1
1
28. f (x) =
7. f (x) =
2
2
1
+
|1
− ex |
f
(x)
=
sin(ln
x)
18.
sin x · cos x
√
x
1
|x|
19. f (x) = e
8. f (x) = 2
29. f (x) =
x +4
1 + |x|
20. f (x) = earcsin x
1
1
9. f (x) =
30. f (x) =
x4 arctan x
4x2 + 1
21. f (x) =
3
+
cos x
1 + x2
1
1
√
10. f (x) = 2
31. f (x) =
22. f (x) = tan x
x −1
1 + sin2 x
2. Compute the following integrals:
x−1
Z 1
e x+1
7.
1.
f (x) dx where f (x) =
(1 + x)3
0
Z π
2
8.
2.
f (x) dx where f (x) = ln(sin x)
0
Z π
2
3.
f (x) dx where f (x) = ln(cos x)
9.
Z0 1
4.
f (x) dx where f (x) = x ln(sin πx)
Z0 1
x earctan x
10.
5.
f (x) dx where f (x) = p
(1 + x2 )3
0
Z 3
1
√
6.
f (x) dx where f (x) =
(1 + x) 1 + x + x2
2
Z
1
x+1
x4 + x2 + 1
Z 3
√
f (x) dx where f (x) = x + 1 + (1 − 2x)[x − 1]
0

1 x

Z 1

e
,x ≤ 0
2
√
f (x) dx where f (x) =
1+x−1

−1

,x > 0
x

3

Z 1

,x ≤ 0
− + ex
4
√
√
f (x) dx where f (x) =
x+4− 4

−1

,x > 0
x
f (x) dx where f (x) =
0
3. Find recurrence formulae for the following integrals:
Z π
Z π
Z π
2
2
4
sinn x dx
2.
sin2n x dx
3.
tgn x dx
1.
0
0
0
Z
1
4.
0
x2n
p
Z
1 − x2 dx
1
5.
0
xn sin πx dx
4. Study the convergence of the following improper integrals (a, b > 0, p, q ∈ R):
Z∞
1.
Z∞
1
dx
1 + x2
8.
0
0
Z∞
Z∞
2.
1
dx
x2
9.
1
0
Z∞
Z1
3.
1
1
√ dx
x
10.
0
Z∞
4.
Z∞
sin x dx
11.
π
1
√ dx
x
Z2
15.
0
Z∞
−x
e
dx
1 + x2
16.
e
√ dx
x
1
dx
xp
17.
6.
18.
0
Z1
7.
Z∞
13.
1
dx
x
14.
0
2
19.
Z∞
20.
1
dx
(ln x)ln x
ln(sin x)
√
dx
x
1
dx
xp (1 + xq )
0
24.
ln x
dx
xp
x ln x
dx
(1 + x2 )p
Z1
lnp
25.
1
dx
x
0
Z∞
26.
sin2 x
dx
xp
0
1
dx
1 + | sin x|
0
5. Find the Fourier series for:
1.
2.
3.
4.
5.
6.
7.
1
dx
xp ln x
0
0
e−x dx
−∞
Z∞
23.
π
0
1
√
dx
1 − x2
sin x
dx
x2
1
2
0
Z∞
2
Z∞
Z2
Z1
22.
0
a
1
dx
xp
1
dx
(b − x)p
a
Z∞
Z∞
Zb
12.
21.
2
Z∞
Z1
1
√ dx
x
arctan x
dx
x
Zb
0
−x
−∞
5.
1
√
dx
sin x
f (x) = π 2 − x2 , x ∈ [−π, π]
f (x) = x2 , x ∈ [−π, π]
f (x) = |x|, x ∈ [−π, π]
f (x) = x3 , x ∈ [−1, 1]
f (x) = cos(ax), x ∈ [−π, π]
f (x) = cosh(ax), x ∈ [−π, π]
f (x) = eax , x ∈ [−1, 1]
½
8. f (x) =




9. f (x) =
10. f (x) =



½
a
b
if − π ≤ x < 0
if 0 ≤ x ≤ π
π
−x
if − ≤ x ≤ 0
2
x
if 0 ≤ x < π
3π
2π − x if π ≤ x ≤
2
0 if − π ≤ x < 0
x if 0 ≤ x ≤ π