Page 106 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 5] INTEGRALS 97

SPECIAL METHODS OF INTEGRATION
1. Integration by parts.
Let u and v be differentiable functions. According to the product rule for differentials
dðuvÞ¼ udv þ vdu
Upon taking the antiderivative of both sides of the equation, we obtain
ð ð
uv ¼ udv þ vdu

This is the formula for integration by parts when written in the form
ð ð ð ð
udv ¼ uv vdu or f ðxÞg ðxÞ dx ¼ f ðxÞgðxÞ f ðxÞgðxÞ dx
0
0
where u ¼ f ðxÞ and v ¼ gðxÞ. The corresponding result for definite integrals over the interval
½a; bŠ is certainly valid if f ðxÞ and gðxÞ are continuous and have continuous derivatives in ½a; bŠ.
See Problems 5.17 to 5.19.
2. Partial fractions. Any rational function PðxÞ where PðxÞ and QðxÞ are polynomials, with the
QðxÞ
degree of PðxÞ less than that of QðxÞ, can be written as the sum of rational functions having the
A Ax þ B
form r , 2 r where r ¼ 1; 2; 3; ... which can always be integrated in terms of
ðax þ bÞ ðax þ bx þ cÞ
elementary functions.
3x 2 A B C D
EXAMPLE 1.
3 ¼ 4x 3 þ 3 þ 2 þ 2x þ 5
ð4x 3Þð2x þ 5Þ ð2x þ 5Þ ð2x þ 5Þ
2
5x x þ 2 Ax þ B Cx þ D E
EXAMPLE 2.
2
2 2 ¼ 2 2 þ x þ 2x þ 4 þ x 1
ðx þ 2x þ 4Þ ðx 1Þ ðx þ 2x þ 4Þ
The constants, A, B, C, etc., can be found by clearing of fractions and equating coefficients of like powers of x
on both sides of the equation or by using special methods (see Problem 5.20).
3. Rational functions of sin x and cos x can always be integrated in terms of elementary functions by
the substitution tan x=2 ¼ u (see Problem 5.21).
4. Special devices depending on the particular form of the integrand are often employed (see
Problems 5.22 and 5.23).

IMPROPER INTEGRALS
If the range of integration ½a; bŠ is not finite or if f ðxÞ is not defined or not bounded at one or more
points of ½a; bŠ, then the integral of f ðxÞ over this range is called an improper integral. By use of
appropriate limiting operations, we may define the integrals in such cases.
ð ð M M
1 dx dx
EXAMPLE 1. 2 ¼ lim 2 ¼ lim tan 1 ¼ lim tan 1 M ¼ =2
x
0 1 þ x M!1 0 1 þ x M!1 0 M!1
1 dx 1 dx ffiffiffi
ð ð 1
EXAMPLE 2. p ffiffiffi ¼ lim p ffiffiffi ¼ lim 2 x ¼ lim ð2 2 Þ¼ 2
ffiffiffi
p
p

0 x !0þ x !0þ !0þ
1 dx 1 dx
ð ð 1

EXAMPLE 3. ¼ lim ¼ lim ln x ¼ lim ð ln Þ

0 x !0þ x !0þ !0þ
Since this limit does not exist we say that the integral diverges (i.e., does not converge).