16
Integration
16.1 The antiderivative function or the indefinite integral
The linearization process of differential calculus derives for function f (x) the linear differential relationship dy = f (x0 )dx. We will be concerned here with the inverse problem
of solving the differential equation dy = f (x)dx for function y(x). Function F (x) such
that F (x) = f (x) is called an antiderivative or a primitive function of f (x). For example,
dy = 2xdx is derived from f (x) = x2 since (x2 ) = 2x. Yet not only x2 solves the differential
equation y = x2 but any other function of the form x2 + c, where c is a constant, and
the converse is also generally true, namely, if F and G are the antiderivatives of the same
function, then F − G = c.
We state it formally.
Theorem: If F (x) is a primitive function of f (x), namely, F (x) = f (x) on some interval,
then the function G(x) is also a primitive function of f (x), if and only if, constant c exists
such that for each x in the interval G(x) = F (x) + c.
The anti-differentiation statement is written in the form
f (x)dx = F (x) + c, F (x) = f (x).
in which the elongated S (for sum) is the integral sign, f (x) the integrand, F (x) any antiderivative function of f (x), and c the constant of integration. Because c is arbitrary, the
act of finding F is called indefinite integration, and the totality of the primitive functions of
f (x) is said to be the indefinite integral of f (x).
For example
1 a 2
1 2
a
a
a
a
e xdx = e x + c,
e xda = xe + c,
dx = a ln x + c,
da =
a + c.
2
x
x
2x
a
If we think about x as time and about f (x) as velocity, then F (x) is the distance traveled,
which can be determined with certainty only with knowledge of the time traveled.
16.2 Some indefinite integrals
Below are listed the integrals of some common functions.
0dx = c
1dx = x + c
xn dx =
1
xn+1 + c, n = −1
n+1
1
dx = ln x + c, x > 0
x
1 x
ax dx =
a + c, a > 0
ln a
1
Chapter 16
1
x
1
dx
=
arctan
+c
x2 + a2
a
a
x
1
√
sin xdx = − cos x + c
dx = arcsin + c
2
2
a
a −x
1
1
cos xdx = sin x + c
a2 ± x2 dx = x a2 ± x2 ± ln(x + a2 ± x2 ) + c
2
2
16.3 Basic properties of the indefinite integral
ex dx = ex + c
It results from the definition of the indefinite integral that
cf (x)dx = c
f (x)dx, and
[f (x) ± g(x)]dx =
f (x)dx ±
g(x)dx.
But there is no product rule and no quotient rule for the indefinite integral. The (indefinite)
integrals
f
f gdx and
dx
g
can not be evaluated in terms of the integrals of f and g. For example
(1 − x)(1 + x)dx =
1 3
1
1+x
(1 − x )dx = x − x + c,
dx = ( + 1)dx = ln x + x + c
3
x
x
2
in which each integrand was first broken down into a sum and then integrated term wise.
Examples.
2
2
x2
x −1+1
x −1
1
1
1.
dx =
dx =
dx +
dx = (x − 1)dx +
dx
1+x
1+x
1+x
1+x
1+x
4
2
x4
x −1+1
x −1
1
2.
dx =
dx =
dx +
dx =
2
2
2
1+x
1+x
1+x
1 + x2
1
1
dx = x3 − x + arctan x + c.
= (x2 − 1)dx +
2
1+x
3
2
2
1
sin x + cos x
1
1
dx = − cot x + tan x + c
dx
=
dx
=
dx
+
3.
2
2
2
cos2 x
sin x cos2 x
sin x cos2 x
sin x
Lack of these rules makes the evaluation of the integral uncertain. Whereas the derivative
function of complicated functions comprising of elementary functions can be routinely written
in terms of elementary functions, the integral may be difficult or even impossible to express
in terms of elementary
√ functions. No function F (x) consisting of elementary functions exists
such that F (x) = 1 + x3 , or F (x) = sin x/x. The ubiquity of these, and similar, integrals
in applied mathematics led to the introduction of a good number of new such integral-defined
functions.
Crafty computer programs such as Mathematica or Maple, designed for doing symbolic
(as opposed to numerical) mathematics, have the capacity to do integration, and they do
it well. They are highly successful in expressing a given integral in terms of elementary, or
other, devised, functions, lifting from us the burden of an otherwise tedious, and often futile,
paper-and-pencil search.
2
Chapter 16
As an example to indefinite integration we will prove the
Theorem: The only plane curves of constant curvature are the line and the circle.
Proof. The curvature κ of the plane curve y = y(x) is give by
κ=
y .
(1 + y 2 )3/2
If κ = 0, then y = 0 and we have
y(x) =
( 0dx)dx =
c1 dx = c1 x + c2
for constants c1 , c2 , which is the equation of a line. Assume that the curvature is a non zero
constant and put z = y to have it in the form
κ=
and
x=
1
1
z
or dx =
dz
3/2
2
κ (1 + z 2 )3/2
(1 + z )
1
1
1
z
dz = √
+ c.
3/2
2
κ (1 + z )
κ 1 + z2
We select the constant of integration c = 0, and have, after some algebra
z=y = √
κx
1
x
√
then y = κ
dx = −
1 − κ2 x2 + c.
κ
1 − κ2 x2
1 − κ2 x2
We take c = 0, square y, and are left with x2 + y 2 = 1/κ2 , which is the equation of a circle
specifically centered at the origin.
Exercises. √
x ln x dx = pxq (ln x − p) + c. Find numbers p and q.
1. Show that
1
x
dx = A(1 + x) 2 (x + B) + c. Find numbers A and B.
1+x
√
3. Show that x2 1 + x dx = A(1 + x)p + B(1 + x)q + C(1 + x)r + d.
2. Show that
√
Find numbers A, B, C, p, q, r.
4. Evaluate
ex · ex dx.
5. Show that
ex sin(x) dx = ex (A cos(2x) + B sin(2x)) + c. Find numbers A, B.
1
ln x dx = p(ln x)q + c. Find numbers p, q.
x
7. Show that (ln x)2 dx = x(A + B ln x + C(ln x)2 ) + d. Find numbers A, B, C.
6. Show that
8. Show that
ln x dx = x ln x + f (x) + c. Find function f (x).
3
Chapter 16
9. Show that
10. Show that
11. Show that
1
x ln x dx = x2 ln x + f (x) + c. Find function f (x).
2
xex ds = xex + f (x) + c. Find function f (x).
x sin x dx = −x cos x + f (x) + c. Find function f (x).
12. Knowing that (arcsin x) = √
1
show that arcsin x dx = x arcsin x + f (x) + c.
1 − x2
Find function f (x).
x3
dx = Ax3 + Bx2 + Cx + D ln(1 + x) + E. Find numbers A, B, C, D.
13. Show that
1√
+x
√
√
x
dx = A x + B arctan( x) + C. Find numbers A, B.
14. Show that
1 + x
√
√
x
√ dx = pxq + Ax + B x + C ln(1 + x) + D. Find numbers p, q,
15. Show that
1+ x
A, B, C.
√
√
√
1
√ dx = A ln(x − 1) + B x + C ln(−1 + x) + D ln(1 + x) + E.
16. Show that
1+ x
17. Show that if p(x) is a polynom, then
p(x)ex dx = ex (p − p + p − p + · · ·) + c.
16.4 Integration techniques
At first we will look at integration by algebraic reformations to bring the integrand into
a form from which its antiderivative can be directly deduced. This will give us a taste of the
clever algebraic rearrangements that are often needed with integrals.
Examples.
1. To find the antiderivative of x2 /(1 + x2 ) we write
2
x2 − 1
1
x2
x +1−1
1
dx =
dx = (
−
)dx = (x − 1)dx −
dx
1+x
1+x
1+x
1+x
1+x
1
= x2 − x + ln(1 + x) + c.
2
2. To find the antiderivative of 1/(1 − x2 ) we write
1
1
1
1
1
1
1
1
dx =
(
dx =
+
)dx = ln(1 + x) − ln(1 − x) + c.
2
1−x
1+x1−x
2
1+x 1−x
2
2
Substitutions or change of variables may also simplify an integral. Consider the integral
and the substitution u = u(x)
that becomes
x2
dx, u = 1 − x3 , du = −3x2 dx
1 − x3
1 1
−
du = ln u + c.
3 u
4
Chapter 16
3. Consider the integral
1
1
x
dx, u = 1 + 2x, x = (u − 1), dx = du
1 + 2x
2
2
that becomes by the indicated transformation
1
1
1 u−1
du =
(1 − )du = u − ln u + c.
4
u
4
u
Such u-transformation shows the integral in the form
For example
f [u(x)]u (x)dx =
2
xex dx =
f (u)du.
2
1
1 2
2xex dx = ex + c.
2
2
4. An integral may be guessed up to some numerical factors, as for example
xex dx = Axex + Bex + c
with undetermined coefficients A and B that are obtained from
xex = (Axex + Bex + c) = Axex + (A + B)ex
as A = 1, B = −1 so that A + B = 0.
Integrating the derivative of the product given by the rule (f g) = f g + f g we obtain
(f g) dx =
gf dx +
f g dx or f g =
gdf +
f dg
which is known as integration by parts. Consider once more
xex dx.
For the choice f = x, g = ex we have f = 1, g = ex , and
xex =
1 · ex dx +
xex dx, so that
xex dx = xex − ex + c.
Integrating repeatedly by parts we obtain the interesting and useful integral identities
and
y
2
y 2 dx = −
dx =
y y dx + yy y y dx + y y − y y.
5
Chapter 16
Exercises.
1. Let Mathematica or Maple try to evaluate
√
√
√
√ x
√
1 + ex dx
1 + xex dx
1 + x sin x dx
xe dx
1 + e−x dx
√
√
√
sin x dx
1 + tan x dx
tan x2 dx
ln(1 + x3 ) dx
1 + xe−x dx
√
√
√
√
√
ln x dx
1 + ln x dx
1 + sin x dx
1 + sin x3 dx
tan x dx.
√
√
x tan x dx
ln sin x dx
ln tan x dx
tan ln x dx
tan x dx
16.5 The existence of the indefinite integral
The mathematical wizardry needed, the formidable expressions construed, and the imposibilities encountered with integrals stems from our insistence on expressing them in terms
of a limited number of essential functions. A simple function compounded of some elementary functions can often hardly, if at all, be integrated in terms of elemntary functions, to
wit, the integral of 1/x requires the novel function ln x. Yet there is no reason to doubt that
an anti derivative can be computed for any continuous function to any degree of accuracy.
Thinking about a traveling car this means that given the speed of the car as a function of
time, its distance from some starting point at any moment can be accurately computed.
One way of doing this is to divide the time interval between the start and the present into
small time subintervals in which the velocity of the car may be assumed constat, which it
nearly is by virtue of the continuity of speed function. The distance covered in each time
subinterval is easily computed, and when we add up all these sections we have the total
distance, approximately. In a more general mathematical context this method is known as
the Riemann mid-point method, and we prove its validity shortly.
Geometrically speaking what we are claiming is that if the slope function of a curve
passing through a fixed point is given, then any point on the curve can be computed to
within any prescribed accuracy. How this is may be systematically achieved is illustrated in
the figure below. Let the given continuous function f (x) be the slope function of function
F (x), F (x) = f (x). Let also the initial value of F (x) at x = a, F (a) be given.
F(x)
F(x)
F(a)
F(a)
h
a
h
h/2
h
a
x
x
In the figure above F (x) happens to be depicted increasing and concave, namely, such
6
Chapter 16
that it is above any of its tangent lines. Moving on the tangent line from x = a to x = a + h
we end at a point that we consider being on the integral of f (x) but for a lower constant
of integration. Continuing stepping forward along the rest of the tangent lines we construct
the approximation
F (a) + f (a)h + f (a + h)h + f (a + 2h)h
to F (x). As seen in the figure the approximating polygonal function is an uder estimation
of F (x)—the initial value F (a) plus the (here three) increments sum falls short of F (x).
Cutting the step size h in half and executing twice as many steps improves the accuracy
of the sum approximating F (x) as in the figure above to the right. Taking still smaller
(and many more) steps we achieve an even greater accuracy. Thus, we may consider the
antiderivative function F (x) to the slope function f (x) as existing by dint of the fact that
we may compute it to any degree of approximation.
If function f (x), for which we seek an anti derivative, is not only continuous but also
differentiable, then the accuracy of the approximating increment can be enhenced by computing it for an arc of the osculating parabola, instead of the osculating (tangent) line. For
the first step the parabolic rise would be
1
F (a) + f (a)h + f (a)h2
2
and so on for all subsequent steps carried out to reach an approximation to F (x).
A function may be defined by a differential equation that relates the function to its
derivatives plus initial conditions, rather than in the form y (x) = f (x) for a given f (x). For
√
example the function y = ex is defined by y = y, y(0) = 1; the function y = x is defined
by y y = 1/2, y(1) = 1; the function y = ln x is defined by y = −y 2 , y(1) = 0, y (1) = 1;
and the function y = sin x is defined by y = −y, y(0) = 0, y (0) = 1.
Exercises.
1. Consider the IVP y = −y 2 , y(0) = 0, y (0) = 1. Repeatedly differentiate the differential
equation, then substitute the initial conditions to obtain the higher derivatives of y at x = 0,
and consrquently the polynomial approximation
y(x) =
x2
x3
x4
x5
x6
x
1
y(0) + y (0) + y (0) + y (0) + y (0) + y (5) (0) + y (6) (0).
0!
1!
2!
3!
4!
5!
6!
Theorem (The fundamental theorem of calculus:) The integral of a continuous function exists. In other words, a continuous function is the derivative function of some differentiable function on some interval.
Of course, the derivative function can not experience a jump discontinuity, and hence
the indefinite integral does not exist for such functions.
Proof. We will give a constructive proof to this theorem for an increasing integrand f (x).
Extension of the proof to decreasing functions or to continuous functions, in general, is
straightforward. Let F (x) be the differentiable function we are seeking to construct, being
such that F (x) = f (x) in the interval of interest. Then
F (x + h) = F (x) + hF (ξ), x < ξ < x + h
7
Chapter 16
if h > 0. Being continuos on [x, x + h], f (x) reaches its highest value, and its lowest value
in the interval. If f (x) is, as we have assumed it to be, increasing, then it is minimal at x
and maximal at x + h. With F (x) = f (x) it results that
F (x) + hf (x) < F (x + h) < F (x) + hf (x + h)
Thus, F (x + h) is located in the interval I1 = (F (x) + hf (x), F (x) + hf (x + h)) which we
call the full step interval in the figure below. Because f (x) is continuous, I1 → 0 as h → 0.
If point x + h is reached in two half steps, then
h
h
h
h
F (x + ) = F (x) + f (ξ1 ), and F (x + h) = F (x + ) + f (ξ2 )
2
2
2
2
where ξ1 is in the first interval, and ξ2 in the second. Now F (x + h) is in the interval
h
h
h
h
h
h
F (x) + f (x) + f (x + ) < F (x + h) < F (x) + f (x + ) + f (x + h)
2
2
2
2
2
2
which we call a half step in the figure below. We denote the half step interval by I2 , and it
is nested inside I1 . See the figure below.
full step
F(x+h)
half step
h/2
h/2
x
x+h
Say we are interested in the interval [a, x]. We may set F (a) = 0, reach point x by n steps
of size h = (x − a)/n, and obtain the interval In by the procedure we have just described
for the two half steps. Function F (x) is located in In . If we repeat this stepping-ahead
procedure with smaller steps, say with h/2, then we produce at x an interval I2n ⊂ In that
contains F (x). A succession of nested intervals is thus created with finer steps and F (x)
is trapped (and thus defined) by these ever shrinking intervals. In other words, F (x) can
be computed to any prescribed accuracy with a sufficiently large number of steps. End of
Proof.
The figure below shows the end points of the intervals created for
sin x
dx = Si(x) + c
x
8
Chapter 16
that can not be expressed in terms of elementary functions (the proof is not elementary,
though,) and which defines therefore a novel function denoted by Si x. The + sign on the
graphs marks the point of the ’exact’ Si(3).
2 F(x)
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.5 1
F'(x)= sinx
x
h=0.2
x
1.5 2
F(x)
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
F'(x)= sinx
x
h=0.1
x
0
2.5 3
0.5 1
1.5 2
2.5 3
Looking at the above curves suggests the possibly better approximation
h
F (x + h) = F (x) + hf (x + )
2
for the stepwise integration of the initial value problem, IVP in short, F (x) = f (x), F (a) =
0. The graph of Si shown below for the interval 0 ≤ x ≤ 14 was computed by this method.
2
Si(x)
p/2
1
h=0.05
0
2
4
6
8
x
10
12
14
We will not dwell any longer on this interesting function but will press ahead with other
important issues of the integral
16.6 Differentiation of an integral with respect to a parameter
Consider the integral
f (x, a)dx = F (x, a) + c
of which we seek ∂F/∂a and which we denote, as before, Fa . Obviously, ∂F/∂x = Fx =
f (x, a). Wherever Fxa and Fax are continuous, Fxa = Fax holds. If this is the case, then
∂ ∂F
∂ ∂F
∂f
=
=
.
∂x ∂a
∂a ∂x
∂a
Integration with respect to x yields then
∂F
∂ ∂f
f dx + C
dx =
+C =
∂a
∂a
∂a
9
Chapter 16
which means that we are conditionally allowed to differentiate under the integral sign.
Examples.
1. We seek to approximate in the interval [0, 1] the cubic f (x) = x3 by a linear y = ax,
optimally, in the least squares sense. Or, we look for a such that
∂ 1 3
(x − ax)2 dx = 0.
∂a 0
Differentiating under the integral sign we obtain
1
2
0
1 a
(x3 − ax)(−x)dx = 2(− + ) = 0
5 3
and a = 3/5.
2. With the knowledge that
1
1
dx = ln(1 + ax) + c
1 + ax
a
we find, by differentiating both sides with respect to a
1
1 x
x
dx = 2 ln(1 + ax) −
+ c.
2
(1 + ax)
a
a 1 + ax
3. We know that
1
eax dx = eax + c.
a
Differentiating both sides with respect to a we obtain
3. From
xeax dx = −
xn dx =
1 ax x ax
e + e + c.
a2
a
1
xn+1 + c
n+1
we obtain by differentiation with respect to n
xn ln x dx =
1
1
xn+1 (ln x −
)+c
n+1
n+1
which we verify to be correct.
16.7 Power series representation
Consider the integral
x
F (x) =
10
1
1
dt
t
Chapter 16
for which we know that F (x) = ln(x). Otherwise we can attempt to generate the power
series representation
F (x) = F (1) + (x − 1)F (1) +
1
1
(x − 1)2 F (0) + (x − 1)3 F (0) + · · ·
2!
3!
for the integral. Here
F (x) =
1
1
2!
3!
, F (x) = − 2 , F (x) = 3 , F (x) = − 4 · · ·
x
x
x
x
and with F (1) = 0 the power series representation for F (x) becomes
1
1
1
F (x) = (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 ± · · ·
2
3
4
converging for any 1 ≤ x ≤ 2 and hence actually defining function F (x). Replacing x − 1 by
x we obtain
1
1
1
ln(x + 1) = x − x2 + x3 − x4 ± · · ·
2
3
4
good for any 0 ≤ x ≤ 1. In particular
ln 2 = 1 −
which we have seen before.
We define the function
1 1 1 1 1
+ − + − ± ···
2 3 4 5 6



f (x) = 

and consider the integral
sin x
x
x = 0
1
x=0
x
sin t
dt.
t
0
that is known not to be expressible in terms of elementary functions—the function F (x)
defined by the integral is a novel function denoted as Si(x). In order to express F (x) in as
a power series we will need the higher order derivatives of f (x) for x ≥ 0. We have that
F (0) = 0 and F (x) = f (x) so that F (0) = 1. Also
F (x) =
F (x) =
F (h) − F (0)
1 sin h
x cos x − sin x
and
F
(0)
=
lim
= lim (
− 1) = 0
2
h→0
h→0 h
x
h
h
obtained using L’Hôpital’s rule. In the same manner we obtain
1
1
−1
F (0) = − , F (0) = 0, F (5) (0) = , F (6) (0) = 0, F (7) (0) =
...
3
5
7
and
F (x) = x −
1 3
1 5
1 7
x +
x −
x ± ···
3!3
5!5
7!7
11
Chapter 16
with the general term
1 m
x , m = 2n − 1
m!m
such that an → 0 as n → ∞. The condition an ≥ an+1 occurs if
an =
2n
· (2n + 1) ≥ x
2n − 1
and the power series converges for any x. The graph of Si(x) is shown in section 12.x
12