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[CHAP. 15
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15.2. Evaluate each of the following.
a 6 5! 5 4 3 2 30 2 3 2 2! 2 2 5 3 3 3 1 1 3 2 2 12 2 2 1 2 4 1 2 2 2 3 2:5 2! 1:5 0:5 0:5 16 5:5 4:5 3:5 2:5 1:5 0:5 0:5 315 6 8 6 5 2 2 4 3 3 3 2 3 3 5 2 5 3 3
b c d
15.3. Evaluate each integral.
1 a
0 1
x3 e x dx 4 3! 6 x6 e 2x dx: Let 2x 7. Then the integral becomes e y dy 1 7 2 2 1
1  6 y
y6 e y dy
7 6! 45 7 8 27 2
15.4. Prove that 1 2
p .
Letting x u2 this integral becomes
1 2
x 1=2 e x dx. 1 2
p  p 2  e u du 2  using Problem 12.31, 12 2
This result also is described in equation (11a,b) earlier in the chapter.
15.5. Evaluate each integral.
p y2 y e dy.
Letting y3 x, the integral becomes   p 1 p  1 1 1 1=2 x 1 1 x e dx x1=3 e x x 2=3 dx 3 3 0 3 2 3 0
1 b
3 4x dx
eln 3 4x dz e
e 4 ln 3 z dz.
Let 4 ln 3 z2 x and the integral becomes
! p 1  x1=2 1 1=2 d p p x 1=2 e x dx p p 2 4 ln 3 0 2 4 ln 3 4 ln 3 4 ln 3
dx p : ln x 0 becomes
Let ln x u. Then x e u . When x 1; u 0; when x 0; u 1. The integral 1
e u p du u
u 1=2 e u du 1=2
p 
1 15.6. Evaluate
xm e ax dx where m; n; a are positive constants.
CHAP. 15]
GAMMA AND BETA FUNCTIONS
Letting axn y, the integral becomes   &  ' 1 & 1=n 'm 1 y y 1=n 1 1 m 1 e y d y m 1 =n 1 e y dy m 1 =n m 1 =n a a n na na 0 0
15.7. Evaluate
(a) 1=2 ;
b 5=2 .
x 1 . x
We use the generalization to negative values de ned by x a Letting x 1 ; 2 1=2 p 1=2 2 : 1=2
b Letting x 3=2;
3=2
p p 1=2 2  4  ; using a : 3=2 3=2 3
Then 5=2
3=2 8 p : 5=2 15
1 15.8. Prove that
xm ln x n dx
1 n n! , where n is a positive integer and m > 1. m 1 n 1
Letting x e y , the integral becomes 1 n 1 1 n
yn e m 1 y dy. If m 1 y u, this last integral becomes un e u du 1 n 1 n n! n 1 n 1 m 1 m 1 n 1
un du 1 n e u m 1 m 1 n 1 m 1 n
Compare with Problem 8.50, 8, page 203.
15.9. A particle is attracted toward a xed point O with a force inversely proportional to its instantaneous distance from O. If the particle is released from rest, nd the time for it to reach O.
At time t 0 let the particle be located on the x-axis at x a > 0 and let O be the origin. Newton s law m d2x k x dt2 Then by
where m is the mass of the particle and k > 0 is a constant of proportionality. dx d 2 x dv dv dx dv Let v, the velocity of the particle. Then 2 v and (1) becomes dt dt dx dt dx dt mv upon integrating. dv k dx x or mv2 k ln x c 2 Then r r dx 2k a v ln dt m x 2
Since v 0 at x a, we nd c k ln a. mv a k ln x 2
where the negative sign is chosen since x is decreasing as t increases. We thus nd that the time T taken for the particle to go from x a to x 0 is given by r a m dx p T 4 2k 0 ln a=x
GAMMA AND BETA FUNCTIONS Letting ln a=x u or x ae u , this becomes r r 1 r m m 1 m T a u 1=2 e u du a 2 a 2k 0 2k 2k
[CHAP. 15
THE BETA FUNCTION 15.10. Prove that (a) B u; v B v; u ; b B u; v 2
=2
sin2u 1  cos2v 1  d.
(a) Using the transformation x 1 y, we have 1 1 1 B u; v xu 1 1 x v 1 dx 1 y u 1 yv 1 dy yv 1 1 y u 1 dy B v; u
0 2 0 0
(b) Using the transformation x sin , we have =2 1 sin2  u 1 cos2  v 1 2 sin  cos  d B u; v xu 1 1 x v 1 dx
=2
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