This preview shows page 1. Sign up to view the full content.
Unformatted text preview: , the integral Þ 1.
x2 Ý Since the integral Þ
j. Ý1
2
x2 1
exp 2 log x Þ3 Ý log x
Since 1
log log x Ý1
2
x2 dx is convergent, the integral Þ Ý 2 1
exp 2 log x
1
log log log x log x 1.
x2
dx is convergent. dx log log x 2
0
log x
1
log x whenever x is sufficiently large. For such x we have
we know that log log x 2
2
1
1
1
1.
x
exp log log x 2
exp 1 log x
log x log log x
2
lim
xÝ Since the integral Þ
l. Þ1 Ý exp
Since 1
log x Ý
3 1
x dx is divergent, the integral Þ Ý 3 1
log x log log x dx 1 log x
2 log x
whenever x is sufficiently large we have
1
exp log x 355 1
exp 1
2 log x 1
x dx is divergent. for x sufficiently large. Therefore, since the integral Þ
1
Ý
1
dx is divergent.
Þ1
exp log x
2. Ý 1
x dx is divergent, the integral a. Prove that the integral Þ1 Ý x 1 e x dx converges for every number .
Ý
The convergence of this integral follows at once from the convergence of the integral Þ
1
and the fact that
1 x
lim x e 0.
xÝ 1
x2 dx 1
x2 b. Prove that the integral Þ0 1 x 1 e x dx converges if and only if 0 and deduce that the integral Þ0 Ý x 1 e x dx converges if and only if 0. The latter integral defines the value at of the gamma function and is
denoted as . You will see more about this important function later.
1
The convergence of the integral Þ x 1 e x dx if and only if 0 follows at once from the
0 corresponding fact about the integral Þ 1
0 x 1 dx and the fact that
1
lim x e
x0
x1 x 1. 3. Prove that the integral Þ0 1 1 t 11
t dt converges if and only if both and are positive. This integral defines the value at the point , of the
beta function and is denoted as B , . Some Further Exercises on Improper Integrals
1. Determine the convergence or divergence of the following integrals:
a. Þ0 Ý sin x dx
x This integral converges by Diriclet’s test. We can use exactly the same argument that was
used in Subsection 13.4.7.
b. Þ2 Ý sin 3 x dx
x In order to use Dirichlet’s test for this integral we observe that if w 2 then
w
Þ 2 sin 3 xdx cos w 1 cos 3 w cos 2 1 cos 3 2 8 .
3
3
3
Ý
sin 3 x dx is convergent.
In this way we see that the integral Þ
x
2
c. Þ1 Ý e x sin e x
dx
x 356 e x sin e x
.
x It is instructive to ask Scientific Notebook to sketch the graph y
20 10 0 1.5 2 2.5 x 3 3.5 4 4.5 10 20 As we move from left to right, the function oscillates with a rapidly expanding amplitude but the
peaks, as they become higher, also become very narrow and, as we shall see in a moment,
Ý e x sin e x
the integral Þ
dx is convergent.
x
1
Given any number w 1 we have Þ 1 e x sin e x dx
w ew Þe sin udu and therefore Dirichlet’s test guarantees that the integral Þ 2
Ý 1 e x sin e x
dx is convergent.
x 2. Integrate by parts to obtain the identity
w
w
2
2
Þ 0 sin2 x dx sin w Þ 0 2 sin x cos x dx
w
x
x
and deduce that each of the following four improper integrals equals
Ý
Þ 0 sin x dx.
x
a. Þ0 Ý 2 sin x cos x dx
x Þ0 Ý 2 sin x cos x dx 2 Þ Ý...
View
Full
Document
This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details