HP 15c - Manual (Page 48)

48 Section 2: Working with f
48
The approximation agrees with the value calculated In the previous problem for the same
integral.
Evaluating Difficult Integrals
Certain conditions can prolong the time required to evaluate an integral or can cause
inaccurate results. As discussed in the HP-15C Owner's Handbook, these conditions are
related to the nature of the integrand over the interval of integration.
One class of integrals that are difficult to calculate is improper integrals. An improper
integral is one that involves ∞ in at least one of the following ways:

One or both limits of integration are ±∞, such as
   
 

du e u 2
.

The integrand tends to ±∞ someplace in the range of integration, such as
1 ) ln(
1
0
 
du u .

The integrand oscillates infinitely rapidly somewhere in the range of integration,
such as
½ ) cos(ln
1
0
 
du u .
Equally troublesome are nearly improper integrals, which are characterized by

The integrand or its first derivative or its first derivative changes wildly within a
relatively narrow subinterval of the range of integration, or oscillates frequently
across that range.
The HP-15C attempts to deal with certain of the second type of improper integral by usually
not sampling the integrand at the limits of integration.
Because improper and nearly improper integrals are not uncommon in practice, you should
recognize them and take measures to evaluate them accurately. The following examples
illustrate techniques that are helpful.
Consider the integrand
2
2 ) cos( ln 2
) (
x
x
x f


.
This function loses its accuracy when x becomes small. This is caused by rounding cos(x
2
) to
1, which drops information about how small x is. But by using u = cos(x
2
), you can evaluate
the integrand as









. 1 if
cos
ln 2
1 if 1
) (
1
u
u
u
u
x f