Lectures on String Theory

– 125 –
where Ck 2m
k!
= . There appear only even powers 2m since for odd powers
(2m)!(k−2m)!
the internal integral is zero. Thus, we now need to evaluate the integral
∫
L =
dq (xq)2m
[q 2 + h] γ .
Under the integral the symmetric product q i1 ...q i2m
may be substituted for
q i1 ...q i2m =
(q 2 ) m
d(d + 2)...(d + 2(m − 1)) (δ i 1 i 2
...δ i2m−1 i 2m
+ permutations),
(B.6)
where on the r.h.s. all non-trivial permutations are present. The total number of
this permutations is (2m)! . Therefore, under the integral one may substitute
2 m m!
(xq) 2m = x i 1
...x i 2m
q i1 ...q i2m =
(x 2 ) m (q 2 ) m (2m)!
d(d + 2)...(d + 2(m − 1)) 2 m m! .
Now we have
∫
dq
(q2 ) m
[q 2 + h] = πd/2
γ Γ(d/2)
∫ ∞
0
dq 2 (q2 ) m+d/2−1
[q 2 + h] γ ,
Performing the change of variables t =
∫
h , we arrive at
q 2 +h
dq
(q2 ) m
∫ 1
[q 2 + h] = πd/2
γ Γ(d/2) hm+d/2−γ dtt γ−m−d/2−1 (1 − t) m+d/2−1
= πd/2
Γ(d/2)
Thus, we evaluate the integral L
∫
L =
dq (xq)2m
[q 2 + h] γ =
0
Γ(γ − m − d/2)Γ(m + d/2)
h m+d/2−γ .
Γ(γ)
(x 2 ) m (2m)! π d/2 Γ(γ − m − d/2)Γ(m + d/2)
=
h m+d/2−γ
d(d + 2)...(d + 2(m − 1)) 2 m m! Γ(d/2)
Γ(γ)
d/2 (2m)! Γ(γ − m − d/2)
= π (x 2 ) m h m+d/2−γ .
4 m m! Γ(γ)
Finally, one gets
I(α 1 , α 2 ) = π d/2 Γ(α 1 + α 2 )
Γ(α 1 )Γ(α 2 )
×
∫ 1
0
[k/2]
∑
m=0
dtt α 1+k−2m−1 (1 − t) α 2−1 (2m)!
4 m m!
Ck
2m (−ixp) k−2m (−1) m
Γ(α 1 + α 2 − m − d/2)
(x 2 ) m (t(1 − t)p 2 ) m+d/2−α 1−α 2
,
Γ(α 1 + a 2 )