St. Louis University
Spring 2008 Solutions: Worksheet 4 Math 143–03
Prof. G. Marks
1. Is Z∞
1
ln x
1 + x2dx convergent or divergent? Substitute u= 1/x in this integral to equate it with a different
improper integral. Then determine Z∞
0
ln x
1 + x2dx. Illustrate your answer with a graph.
Solution. To avoid the difficulty of integrating ln x
1 + x2,we use the Comparison Theorem. For all xin [1,∞),
06ln x
1 + x26ln x
x2;
furthermore,∗
Z∞
1
ln x
x2dx =−1 + ln x
xx=∞
x=1
= 1 −lim
x→∞
1 + ln x
x
L′H
= 1 −lim
x→∞
1
x
1= 1
converges. So Z∞
1
ln x
1 + x2dx must also converge, by the Comparison Theorem.†
Now substitute u= 1/x to obtain
Zx=∞
x=1
ln x
1 + x2dx =Zu=0
u=1
ln (1/u)
1 + (1/u)2·−1
u2du =Zu=0
u=1
ln u
u2+ 1 du =−Zu=1
u=0
ln u
u2+ 1 du.
This last integral is equal to −Zx=1
x=0
ln x
x2+ 1 dx. (We’ve just changed the dummy variable from “u” to “ x. ”) Hence,
Zx=∞
x=0
ln x
1 + x2dx =Zx=1
x=0
ln x
1 + x2dx +Zx=∞
x=1
ln x
1 + x2dx = 0.
(We just showed that these last two integrals are each other’s negatives, and that each converges, i.e. equals a finite
quantity. Therefore they cancel to give 0.) Graphing the integrand is left to you.
2. Suppose we want to estimate Z1
0p1 + x3dx.
(a) Write a formula for the Trapezoid Rule approximation with 100 subdivisions. Find a bound on the error. How
many subdivisions would suffice to compute the integral to within ±10−12?
∗Here and subsequently, the symbol L′H
= will b e used to mean “equals, by L’Hospital’s Rule.” One must note in each case that the
conditions of L’Hospital’s Rule are satisfied.
†This argument illustrates the proper explanation to give when proving convergence of an improper integral using the Comparison
Theorem. Statements like
“for large x, the 1 becomes insignificant compared with the x2,so ln x
1 + x2≈ln x
x2”
and
“the 1 b ecomes insignificant, so as far as convergence is concerned, Z∞
1
ln x
1 + x2dx behaves like Z∞
1
ln x
x2dx”
do not belong in a mathematical argument. Mathematics is based on precise, provable assertions, not vague, subjective claims about
something being much larger than something else.
That having been said, it should be added that the intuitive mental process, which enables you to guess with some degree of accuracy
whether the integral converges, does involve thoughts such as: “If we replace 1 + x2by x2in the denominator, it should not affect
convergence.” Such thoughts guide us toward formulating a correct solution, but they are too imprecise to belong in a solution.
Hughes-Hallett provides dubious guidance in this matter. In Examples 1 and 3 on pp. 334–335, each solution begins with a vague,
intuitive argument, which admirably illustrates how one might think up a solution, but which does not belong in the solution itself.
The solution to Example 1 should not have included anything prior to: “[W]e observe that for 0 6x36x3+ 5,we have.. .” The
solution to Example 3 should not have included anything prior to: “We know that (ln t)−1<ln t < t for all positive t. ” (Actually,
this statement requires some justification. Note that if we define the function g(t) = t−ln twith domain t > 0,then the derivative
g′(t) = 1 −1/t is negative whenever 0 < t < 1 and positive whenever t > 1; therefore, g(1) = 1 is the absolute minimum of g. Hence
g(t) = t−ln t>1⇒t>1 + l n tfor all t > 0.)
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