Lectures on String Theory

– 126 –
where we have substituted γ = α 1 + α 2 and h = t(1 − t)p 2 . This is simplified to
I(α 1 , α 2 ) =
×
π d/2
Γ(α 1 )Γ(α 2 )
∫ 1
0
[k/2]
∑
m=0
Thus, the final formula reads as
I(α 1 , α 2 ) =
Ck
2m (2m)!
4 m m! (−ixp)k−2m (−x 2 ) m (p 2 ) m+d/2−α 1−α 2
dtt k−m+d/2−α 2−1 (1 − t) m+d/2−α 1−1 Γ(α 1 + α 2 − m − d/2).
π d/2 [k/2]
∑
Γ(α 1 )Γ(α 2 )
m=0
C 2m
k
(2m)!
(−ixp)
(− k−2m 1 ) m
m!
4 x2 p 2 (p 2 ) d/2−α 1−α 2
× Γ(k − m + d/2 − α 2)Γ(m + d/2 − α 1 )
Γ(α 1 + α 2 − m − d/2).
Γ(k + d − α 1 − α 2 )
For Q a,b (x, −ip) one therefore finds
k
Q a,b
k (x, −ip) = 1
Γ(a + b + d/2)
×
Γ(−a − b)
Γ(−a)Γ(−b)
[k/2]
∑
m=0
C 2m
k
Γ(k − m − a)Γ(m − b)
Γ(a + b + d/2 − m).
Γ(k − a − b)
By using the relation (a) −m = (−1)m
(1−a) m
Q a,b
k (x, ∂ y) =
[k/2]
1 ∑
(−a − b) k
m=0
(
(2m)!
(−ixp) k−2m − 1 ) m
m!
4 x2 p 2
it can be further simplified to give
(
k!(−a) m (−b) k−m
(x · ∂ y ) k−2m − 1 ) m
m!(k − 2m)!(−d/2 − a − b + 1) m 4 x2 ∆ y .
Further summation gives the conformal block of the scalar field and it is performed
by changing the order of the summation and the shift of the summation
variable:
∞∑ 1
k! Qa,b k (x, ∂ y) =
Since
k=0
∞∑
∞∑
m=0 k=2m
=
∞∑
m=0
1
(−a − b) k
we finally get
∞∑ 1
k! Qa,b k (x, ∂ y) =
k=0
=
m=0
(
(−a) m (−b) k−m
(x · ∂ y ) k−2m − 1 ) m
m!(k − 2m)!(−d/2 − a − b + 1) m 4 x2 ∆ y
(
(−a) m
− 1 ) m ∑ ∞
(−b) k+m (x∂ y ) k
m!(−d/2 − a − b + 1) m 4 x2 ∆ y .
(−a − b) k+2m k!
(−b) k+m
(−a − b) k+2m
=
k=0
(−b) m (−b + m) k
(−a − b) 2m (−a − b + 2m) k
∞∑
(
(−a) m (−b) m 1
1F 1 (−b + m; −a − b + 2m; x∂ y ) − 1 ) m
(−µ − a − b + 1) m m! (−a − b) 2m 4 x2 ∆ y ,