656 Iterated and multiple integrals
10 Supposing that n > -2 and is 76 1, show that
1 x
f
- 1
dx fo (x I Y)" dY =
2n+1
o (n + , 1) (n + 2)
Investigate the case in which n = -1.
11 Show that
r1 2xy^3
r1 y
(a) Jo dx
fx
xdy 4 (b)Jo dxJ= xdy =
/1 dxr2zx log2 1 (lx i
(`) fo x Y
dy
=^2 (d) o dx fx ydy
12 Show that making one integration gives the formula
1 x (^1) r11 1
2fo dxfo 1 - xydy = 2f xlogl x2dx
which, so far as we know, has dubious validity because the last integrand is
meaningless when x = 0 and when x = 1. Show that
f
1
o 1 -xydy
has the value 0 when x = 0 and does not exist (or has the value + oo) when
x = 1. Then proceed to the next problem.
13 The integrals in
(1) r1
x
2Jo dxf
1
0 1 -xydy
cannot exist as iterated Riemann integrals because the integral in
(2) f(x) = fox 11 xydy
does not exist as a Riemann integral when x = 1. However, when 0 5 x < 1,
(3) f(x) =
fox
(I + xY + x2y2 + x3y3 + -- -) dy
Cy +
xy2
'Y' "'
= 2 + 3 + 4 ,+.. ..]y=ob=x
L x3 x6 x7
+2+3 +4+...
To each positive integer is there corresponds a positive number S such that
(4)
f(x)x+23+3s+...+xn>(1++...
+n/-1
when 1 - S < x < I (why?) and hence (why?)
(5) lim f(x) = m.
x-+1-