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282 November 7, 2013
in the spirit of dimensional regularization. That is, we shall assume that D, n
and m are such that the integral converges: where it does not, we define the
integral by analytical continuation from the convergence region. The number a 2
is not necessarily a positive real number, but again we shall reach other values
for a 2 by analytical continuation from positive real values.
In the first place, by scaling the vector ⃗q by a factor √ a 2 we find that
∫
I = a D+n−2m I ′ , I ′ d D q |⃗q| n
=
(2π) D (|⃗q| 2 + 1) m . (10.130)
Next, we compute W D (t), the number of D-dimensional Euclidean vectors ⃗q of
a given length t, as follows:
∫
W D (t) = d D q δ(|⃗q| − t)
∫
= 2t d D q δ(|⃗q| 2 − t 2 )
∫
= 2t dq 1 dq 2 · · · dq D δ ( (q 1 ) 2 + (q 2 ) 2 + · · · + (q D ) 2 − t 2)
∫∞
= (2t)2 D dq 1 dq 2 · · · dq D δ ( (q 1 ) 2 + (q 2 ) 2 + · · · + (q D ) 2 − t 2)
0
∫∞
= 2t D+1 −1/2 dy 1 · · · dy D y 1 · · · y −1/2 D δ ( t 2 (y 1 + · · · y D − 1) )
0
D−1 Γ(1/2)D
= 2t
Γ(D/2) = 2tD−1
πD/2
Γ(D/2) , (10.131)
where we have written q j = y j 1/2 t, and used Euler’s formula of sect.(10.14.2).
Hence,
I ′ =
1
(4π) D/2 Γ(D/2) I′′ , I ′′ =
∫ ∞
0
du u(D+n)/2−1
(u + 1) m , (10.132)
where we have used u = t 2 . Another application of Euler’s formula gives us
I ′′ =
=
=
∫ ∞
1
∫ 1
0
du u −m (u − 1) (D+n)/2−1 =
∫ 1
du u m−1−(D+n)/2 (1 − u) (D+n)/2−1
Γ (m − (D + n)/2) Γ ((D + n/2)
Γ(m)
0
du u m−2 ( 1
u − 1 ) (D+n)/2−1
. (10.133)