186 Chapter 6Methods of integration
EXAMPLE 6.23Integrate.
The integral cannot be evaluated by the standard methods discussed in this chapter;
the method of integration by parts does not work. We consider instead the related
integral
Then
and the new integral can be integrated by parts as in Example 6.13. Then
(6.37)
and, integrating with respect to α,
To obtain the value of the constant of integration, we note thatI(α) 1 → 1 0asα 1 → 1 ∞,
so that
To retrieve the original integral we now setα 1 = 10 :
(6.38)
0 Exercise 76
When the limits of integration also depend on the parameter, the result of differen-
tiating the integral is given by Leibniz’s theorem: if a(α)and b(α)are continuous
functions of α,
(6.39)
d
d
fx dx
d
d
fx
a
b
a
b
α
α
α
α
ααααZZ
()
()
()
()
(),= (),
dx f b+, −,
db
d
fa
da
d
() ()α
α
α
α
I
x
x
() dx
sin
00 tan
22
01==−+=
−Z
∞ππ
C==
→−lim tan
αα
∞12
π
I
d
()α tan C
α
α
=− α
=− +
−Z
1
21d
d
I
α
α
α
()=−
1
1
2d
d
Iexdx
xα
α
α()=− sin
−Z
0∞I
ex
x
dx
x()
sin
α
α=
−Z
0∞Z
0∞sinx
x
dx