Lectures on String Theory

– 124 –
where the operator Q k (x, ∂ y ) is defined by the relation
and a, b are given by
(y 2 ) a (x · ∂ y ) k (y 2 ) b = Q k (x, ∂ y )(y 2 ) a+b (B.2)
a = − 1 2 (∆ B + ∆ O − ∆ A ), b = − 1 2 (∆ A + ∆ O − ∆ B ).
The explicit form of the operator Q k (x, ∂ y ) is found by the Fourier transform. Indeed,
we have
(y 2 ) a = 2 2a+d π d/2 Γ(a + d)
∫
2
1 e −ip·y
(B.3)
Γ(−a) (2π) d (p 2 ) a+d/2
and similar for the others. Substituting the Fourier transform of every function of
y 2 one gets
1 Γ(a + d)Γ(b + d)
∫
2 2
Γ(−a − b)
π d/2 Γ(a + b + d) Γ(−a)Γ(−b)
2
∫
e −ipy
= dp
(p 2 Qa,b
)
a+b+d/2 k
(x, −ip),
where Q a,b (x, −ip) is defined by
k
Q a,b
k
(x, ∂ y)e ipy = Q a,b
k (x, −ip)eipy ,
and we have used the change of variables p → p − q.
e −ipy
From here one gets that Q a,b (x, −ip) is given by the integral
k
dpdq
(p 2 ) a+d/2 (q 2 ) b+d/2 (−ixq)k (B.4)
Q a,b
1 Γ(a + d
k
(x, −ip) = )Γ(b + d)
∫
2 2
Γ(−a − b)
π d/2 Γ(a + b + d) Γ(−a)Γ(−b) (p2 ) a+b+d/2
2
(−ixq) k
dq
((p − q) 2 ) a+d/2 (q 2 ) . b+d/2
Thus, the problem is reduced to evaluation of the integral
∫
(−ixq) k
I(α 1 , α 2 ) = dq
((p − q) 2 ) α 1 (q2 ) . α 2
(B.5)
One has
I(α 1 , α 2 ) = Γ(α ∫
1 + α 2 ) 1
∫
dtt α1−1 (1 − t) α2−1 (−i) k (xq) k
dq
Γ(α 1 )Γ(α 2 ) 0
[(q − tp) 2 + t(1 − t)p 2 ] α 1+α 2
= Γ(α ∫
1 + α 2 ) 1
∫
dtt α1−1 (1 − t) α2−1 (−i) k (xq + txp) k
dq
Γ(α 1 )Γ(α 2 ) 0
[q 2 + t(1 − t)p 2 ] α 1+α 2
= Γ(α ∫
1 + α 2 ) 1
[k/2]
∑
∫
dtt α1−1 (1 − t) α2−1 (−i) k Ck
2m (txp) k−2m (xq) 2m
dq
Γ(α 1 )Γ(α 2 )
[q 2 + t(1 − t)p 2 ] α 1+α 2
,
0
m=0