This preview shows page 1. Sign up to view the full content.
Unformatted text preview: v x 2 dx
sin 2 x Þ tan u
tan v arctan t
t2 2 dt Therefore Þ0
v x 2 dx lim
u 0
sin 2 x Þu
v x 2 dx
sin 2 x
tan v arctan t 2
lim Þ
dt
u 0 tan u
t2
tan v arctan t 2
Þ
dt.
0
t2 Therefore
/2 Þ0 x 2 dx lim
v /2
sin 2 x Þ0
v x 2 dx
sin 2 x
tan v arctan t 2
lim Þ
dt
v /2
0
t2
Ý arctan t 2
Þ
dt log 2.
0
t2
then, for each such number x we have f x 1/ sin 2 x. Therefore,
If f x cot x for 0 x
2
integrating by parts, we obtain 386 /2 Þ0 x 2 dx lim
u 0
sin 2 x /2 Þu x 2 f x dx
/2 lim x 2 f x lim u 0 u u 0 /2 Þ0 /2 x 2 cos x
sin x lim
2Þ u 0 /2 /2 Þu 2xf x dx
/2 Þ
0 u lim
u u 2x cot xdx x cot xdx 0 x 2 dx 2 Þ /2 x cot xdx
0
sin 2 x and therefore
log 2 2 Þ /2 x cot xdx 0 from which we see that
/2 Þ0 log 2
.
2 x cot xdx Finally, applying integration by parts to the integral
/2 Þu
for 0 u
2 log sin xdx , we obtain
/2 Þu /2 log sin xdx x log sin x
u /2 Þu x cot xdx and so Þ /2
0 u /2 Þ
0 u log sin xdx lim lim u log sin u u 0 Þa Þc d log 2
2 1 log 2
.
2 7. Prove that if f is a function defined on a rectangle a, b
b log sin xdx c, d then the identity f x, y dydx Þc Þa
d b f x, y dxdy will hold as long as both sides exist as iterated Improper Riemann integrals and the left side converges
absolutely. Hint: Use the fact that f f  f f .
The desired result follows at once from the fact that the repeated integration can be reversed for
each of the functions f  f and f .
8. Given that f is improper Riemann integrable on 0, Ý , that a 0 and that g u f u
prove that g is improper Riemann integrable on the interval a, Ý and that Þ Ý 0 f x dx Þ Ý a whenever u a, g u du. a 9. In this exercise we suppose that f and g are nonnegative improper Riemann integrable functions on the
interval 0, Ý and that the function h is defined on the set 0, Ý
0, Ý by the equation
h x, y fx
0 a. Prove that 387 ygy if y x if y x . Ý Þ 0Þ Ý
0 Ý Þ 0Þ h x, y dxdy All we have to notice is that if w 0 then Þ Ý fx y Ý
y Ý Þ fx
0 y w y dx lim
w Þ y g y dxdy fx Þ
0 w y dx lim
w Ý Ý f 0 f t dt Þ
Þ Ý g. 0
Ý f. 0 b. Apply Fichtenholz’s theorem for improper integrals to the first integral in part a and deduce that the
integral is equal to
Ý Þ 0Þ x fx y g y dydx 0 To invert the integral we apply the Fichtenholz theorem for improper integrals to each of the
integrals Þ 0Þ
1 1 h x, y dxdy
0 and
Ý Þ1 Þ1 Ý h x, y dxdy to obtain
Ý Þ 0Þ Ý
0 h x, y dydx Ý Ý Þ 0Þ h x, y dxdy 0 This equation becomes
Ý Þ 0Þ x fx
0 Þ y g y dydx Ý
0 f Þ Ý g. 0 Some Exercises that Explore the Gamma Function
The exercises in this subsection can be used to develop most of the basic facts about the gamma function and
the related beta function that were defined earlier. In these exercises we assume that and are given
positive numbers. The expressions and B , are defined as follows:
Þ0 Ý B , Þ0 1 x 1 e x dx 1 t...
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