Page 390 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 15] GAMMA AND BETA FUNCTIONS 381

n
Letting ax ¼ y,the integral becomes
ð m ð
y y y 1=n 1 ðmþ1Þ=n 1 y 1 m þ 1
1 1=n 1
e d y e
a a ¼ na ðmþ1Þ=n dy ¼ na ðmþ1Þ=n n
0 0
15.7. Evaluate (a) ð 1=2Þ; ðbÞð 5=2Þ.

.
ðx þ 1Þ
We use the generalization to negative values defined by ðxÞ¼
x
1
ðaÞ Letting x ¼ ; ð1=2Þ ¼ 2 :
ffiffiffi
p
2 ð 1=2Þ¼ 1=2
ffiffiffi
ffiffiffi
2 4
p
p
ðbÞ Letting x ¼ 3=2; ð 1=2Þ ; using ðaÞ:
3=2 ¼ 3=2 ¼ 3
ð 3=2Þ¼
ffiffiffi
:
ð 3=2Þ 8 p
Then ð 5=2Þ¼ ¼
5=2 15
n
ð 1 ð 1Þ n!
n
m
15.8. Prove that x ðln xÞ dx ¼ nþ1 , where n is a positive integer and m > 1.
0 ðm þ 1Þ
ð
1
y n n ðmþ1Þy
Letting x ¼ e ,the integral becomes ð 1Þ y e dy.If ðm þ 1Þy ¼ u,thislast integral becomes
0
ð n n ð n n
1 u u du 1 ð 1Þ n!
n ð 1Þ n u ð 1Þ
ð 1Þ n e m þ 1 ¼ nþ1 u e du ¼ nþ1 ðn þ 1Þ¼ nþ1
0 ðm þ 1Þ ðm þ 1Þ 0 ðm þ 1Þ ðm þ 1Þ
Compare with Problem 8.50, Chapter 8, page 203.
15.9. Aparticle is attracted toward a fixed point O with a force inversely proportional to its instanta-
neous distance from O.If the particle is released from rest, find the time for it to reach O.
At time t ¼ 0 let the particle be located on the x-axis at x ¼ a > 0 and let O be the origin. Then by
Newton’s law
2
d x k
m 2 ¼ ð1Þ
dt x
where m is the mass of the particle and k > 0isa constant of proportionality.
2
dx d x dv dv dx dv
Let ¼ v,the velocity of the particle. Then ¼ ¼ ¼ v and (1)becomes
dt dt 2 dt dx dt dx
dv k mv 2
mv ¼ or ¼ k ln x þ c ð2Þ
dx x 2
upon integrating. Since v ¼ 0at x ¼ a,we find c ¼ k ln a. Then
2k
mv 2 a dx r ffiffiffiffiffi r ffiffiffiffiffiffiffiffiffi a
¼ k ln or ln
2 x v ¼ dt ¼ m x ð3Þ
where the negative sign is chosen since x is decreasing as t increases. We thus find that the time T taken for
the particle to go from x ¼ a to x ¼ 0isgiven by
r ffiffiffiffiffi ð a
m dx
p ffiffiffiffiffiffiffiffiffiffiffiffiffi
T ¼ ð4Þ
2k 0 ln a=x