Page 337 - Schaum's Outline of Theory and Problems of Advanced Calculus

328 IMPROPER INTEGRALS [CHAP. 12

M ð M

ð
2 x 2 y 2
I M ¼ e dx e dy
0 0
ð M ð M
2 2
e ðx þy Þ dx dy
¼
0 0
ðð
2 2
e ðx þy Þ dx dy
¼
r M
where r M is the square OACE of side M (see Fig. 12-3). Since integrand is positive, we have
ðð ðð
2 2 2 2 2
e ðx þy Þ dx dy @ I M @ e ðx þy Þ dx dy ð1Þ
r 1 r 2
where r 1 and r 2 are the regions in the first quadrant bounded y
ffiffiffi
p
by the circles having radii M and M 2, respectively.
Using polar coordinates, we have from (1),
ffiffi
p
ð =2 ð M ð =2 ð M 2
2 2 2 D
e d d @ I M @ e d d ð2Þ
¼0 ¼0 ¼0 ¼0
or E C
M 2 2 2M 2
ð1 e Þ @ I M @ ð1 e M√2
4 4 Þ ð3Þ
Then taking the limit as M !1 in (3), we find M
2 2 p ffiffiffi
lim I M ¼ I ¼ =4 and I ¼ =2.
M!1
O A B
ð x
1 2
12.32. Evaluate e x cos xdx.
0 Fig. 12-3
ð
1 2
e x cos xdx. Then using integration by
Let Ið Þ¼
0
parts and appropriate limiting procedures,
dI ð 1 x 2 1 x 2 1 ð 1 x 2
xe sin xdx ¼ e sin xj 0 e cos xdx ¼ I
1
d ¼ 0 2 2 0 2
The differentiation under the integral sign is justified by Theorem 8, Page 314, and the fact that
ð
1 x 2 x 2 x 2
xe sin xdx is uniformly convergent for all (since by the Weierstrass test, jxe sin xj @ xe
0
ð
1 2
and xe x dx converges).
0
From Problem 12.31 and the uniform convergence, and thus continuity, of the given integral (since
ð
2 x 2 1 x 2
x
je cos xj @ e and e dx converges, so that that Weierstrass test applies), we have
0
1 p ffiffiffi
Ið0Þ¼ lim Ið Þ¼ 2 .
!0
ffiffiffi
dI p p ffiffiffi 2
=4
Solving e .
2
d ¼ I subject to Ið0Þ¼ 2 ,we find Ið Þ¼ 2
ð p ffiffiffi ð
2 2 2
1 1
e ðx =xÞ .(b) Evaluate e ðx þx Þ dx.
12.33. (a)Prove that Ið Þ¼ dx ¼
0 2 0
ð
1
2 2
(a)We have I ð Þ¼ 2 e ðx =xÞ ð1 =x Þ dx.
0
0
The differentiation is proved valid by observing that the integrand remains bounded as x ! 0 þ
and that for sufficiently large x,