#04 (08/28/2024)

Integration

Some of important integrals that you may have forgotten:


⌠
⌡ 		xα dx
xα+ 1
α+ 1
⌠
⌡ 		1
x
 dx
ln x
⌠
⌡ 		dx
1 + x2
arctan x
⌠
⌡ 		dx

√
a2 − x2
arcsin (x/a)
⌠
⌡ 		f(x) g′(x) dx
f(x)g(x) −⌠
⌡ 		f′(x) g(x) dx
⌠
⌡ 		f(x) dx
x f(x) − ⌠
⌡ 		x f′(x) dx

Exercises

Evaluate the following integrals:

(a) ⌠
⌡ 		dx
ex + e−x
(b)
⌠
⌡
√
 
a2 − x2
 
 dx
(c) ⌠
⌡ 		eax cos b x dx
(d)
⌠
⌡
√
 
x2 + 1
 
 dx
Solution for (d)  
I ≡ ⌠
⌡
√
 
x2 + 1
 
 dx.
Use integration by parts:
I
=
x
√
 
x2 + 1
 
 − ⌠
⌡ 		x ⎛
⎝
√
 
x2 + 1
 
 ⎞
⎠ 		′
 
 dx
=
x
√
 
x2 + 1
 
 − ⌠
⌡ 		x2

√
x2 + 1
 dx
=
x
√
 
x2 + 1
 
 − ⌠
⌡ 		x2 + 1 − 1

√
x2 + 1
 dx
=
x
√
 
x2 + 1
 
 − ⌠
⌡
√
 
x2 + 1
 
 dx + ⌠
⌡ 		dx

√
x2 + 1
=
x
√
 
x2 + 1
 
 − I + ⌠
⌡ 		dx

√
x2 + 1
 ,
(1)
so
I = 1
2
 ⎛
⎜
⎝ 		x
√
 
x2 + 1
 
 + ⌠
⌡ 		dx

√
x2 + 1
⎞
⎟
⎠ 		.
(2)
Now, the integral, J, defined as
J ≡ ⌠
⌡ 		dx

√
x2 + 1
 ,
(3)
is a challenging integral. The recipe is to change the variable from x to t as1

√
 
x2 + 1
 
 = t − x.
(4)
With this substitution,
1
=
t2 −2 t x,
(5)
dx
=
t − x
t
 dt,
(6)
so
J
=
⌠
⌡ 		1
t − x
 t − x
t
 dt
=
⌠
⌡ 		dt
t
=
lnt
=
ln⎛
⎝ 		x +
√
 
x2 + 1
 
⎞
⎠ 		,
(7)
and finally we obtain
I = 1
2
 ⎛
⎝ 		x
√
 
x2 + 1
 
 + ln ⎛
⎝ 		x +
√
 
x2 + 1
 
⎞
⎠ 		⎞
⎠ 		.
(8)
It is, however, more practical these days to carry out integrations like the one above via computer.
Enter the following line 
Sqrt[x^2 + 1]
exactly (case sensitive, square bracket) in the first box and x (lower case) in the second box, then press the query button.
Integrate with respect to .

Numerical Integration

You can differentiate almost any function exactly but very few functions can be integrated analytically which is why numerical integrations are more important than numerical differentiation.
numint.jpg
  • (Left) Rectangular rule:

    I  ∼  h (f0 + f1 + f2 + …+ fn−1).
    		(9)
  • Trapezoidal rule:
    I  ∼   h
    2
    		 (f0 + 2 f1 + 2f2 + …+ 2 fn−1 + fn).
    		(10)
  • Simpson rule:
    I  ∼   h′
    3
    		 (f0 +4 f1 + 2 f2 + 4 f3 + 2 f4 + …+ 4 f2n−1 + f2n).
    		(11)
where
h = b − a
n
 ,     h′ = b − a
2 n
 .
(12)
It is possible to estimate possible numerical errors for each scheme.
Derivation

Rectangular rule

The rectangular rule is to approximate ∫at f(x)dx by a rectangle with f(t) as the height and t − a ( = h) as the base.
num_rec.jpg
The error is then defined as

ϵ(t) = (t − a) f(t) −⌠
⌡ 		t

a 
 f(x) dx.
(13)
To see how this function behaves, take the derivative of ϵ(t) with respect to t as
ϵ′(t)
=
f(t) + (t − a) f′(t) − f(t)
=
f′(t) (t − a)
 ∼ 
f′(ξ) (t − a),
(14)
where ξ is some constant value assuming that f′(t) does not fluctuate much between a and t. By integrating Eq. (14) with respect to t, we have

ϵ(t)
 ∼ 
f′(ξ)
2
 (t − a)2
=
f′(ξ)
2
 h2,
(15)
where h = (t − a).
When adding up Eq. (15) over a finite interval of (a,b) for n times, the total error is

ϵtotal
 ∼ 
n h2 f′(ξ)
2
=
(b − a) f′(ξ)
2
 h,
(16)
where b − a = n h was used in Eq. (16). Equation (16) implies that when the rectangular rule is used to approximate
⌠
⌡ 		b

a 
 f(x) dx,
(17)
the numerical error involved is in the order of
ϵtotal  ∼ A h,
(18)
where A is a constant and h is the step size. In other words, to get 10−5 accuracy for b − a = 1, one would need n = (b − a)/h  ∼ 105 partitions over the interval.

Trapezoidal Rule

num_trap.jpg
The error involved in the trapezoidal rule is given

ϵ(t) = ⎛
⎝ 		f(t) + f(a)
2
⎞
⎠ 		(t − a) −⌠
⌡ 		t

a 
 f(x) dx.
(19)
Take the derivative of Eq. (19) as
ϵ′(t) = f′(t)
2
 (t − a) + f(t) + f(a)
2
 − f(t),
(20)
which does not yield much insight. Try taking another derivative of Eq. (20) as

ϵ"(t)
=
f"(t)
2
 (t − a) + f′(t)
2
 + f′(t)
2
 − f′(t)
 ∼ 
f"(ξ)
2
 (t − a).
(21)
Integrating Eq. (21) yields

ϵ′(t)  ∼  f"(ξ)
4
 (t − a)2.
(22)
Integrating Eq. (22) again yields

ϵ(t)
 ∼ 
f"(ξ)
12
 (t − a)3
=
f"(ξ)
12
 h3.
(23)
By adding up Eq. (23) over the interval of (a,b), we obtain
ϵtotal
 ∼ 
n h3 f"(ξ)
12
=
(b − a) f"(ξ)
12
 h2
 ∼ 
A h2,
(24)
where A is a constant. This conclusion implies that we need √{105}  ∼ 316 partitions to obtain 10−5 accuracy.

Simpson's Rule

From the result of error analysis of the trapezoidal rule, it was found that

I  ∼ Tn + A h2,
(25)
where I is the true value, Tn is the approximation by the trapezoidal rule, h is the step size and n is the number of partitions. If one makes the stepsize h one-half, Eq. (25) becomes

I  ∼ T2n + A⎛
⎝ 		h
2
⎞
⎠ 		2

 
 ,
(26)
or

I
 ∼ 
Tn + A h2,
(27)
4I
 ∼ 
4 T2n + A h2,
(28)
from which it follows

I  ∼  4T2n − Tn
3
 + (higher order terms than h2).
(29)
Actual computation of Eq. (29) is
I
 ∼ 
1
3
 ⎛
⎝ 		4 h
2
 (f0 + 2 f1 + 2 f2 + …+ 2 f2n−1 + f2n)− (2 h)
2
 (f0 + 2 f2 + 2 f4 + …+ f2n)⎞
⎠
=
h
3
 ⎛
⎝ 		f0 + 4 f1 + 2 f2 + 4 f3 + 2 f4 + …+2 f2n−2 + 4 f2n−1 + f2n⎞
⎠ 		.
(30)
Equation (30) is called the Simpson rule as you already figured it out.
An example of numerical integration with poor convergence:  

Using the relationship
4 ⌠
⌡ 		1

0 
√
 
1 − x2
 
 dx = π,
(31)
to compute an approximation of π.
numint_ex.jpg
Numerical integration of this function results in poor convergence. Why ?

Footnotes:

1For an integral of the type
⌠
⌡ 		F(x,
√
 
a x2 + b x + c
 
 ) dx     a > 0
the following change of variable usually works:

√
 
a x2 + b x + c
 
 = √a x − t.

File translated from TEX by TTH, version 4.03.
On 27 Aug 2024, 20:59.