5.3 Dimensional regularisation
5.3.1 The Γ-function
The Γ-function lies at the heart of the dimensional regularisation technique, because its analytic properties allow to manage the problem of continue the dimension d of Euclidean space time from a few integer values where the Feynman integrals are convergent5 to the whole complexd-plane.
Indeed it was a quite similar task Euler and Gauß solved in the 18th century, namely the continuation of the factorial function, defined on the non-negative integer numbers, to the whole complex plane.
We start with Euler’s representation of the Γ-function:
Γ(z) = Z ∞
0
dtexp(−t)tz−1. (5.11)
Herein we understand the potential oft as
tz−1= exp[(z−1) lnt], (5.12)
where the logarithm along the positive real axis is defined as real (principal value of the logarithm). Now we show that the integral (5.12) is uniformly convergent in all compact areas in the right z-half-plane, i.e., for all Rez >0. This implies that the Γ-function is an analytic function in the rightz-half-plane.
For this purpose we split the integral in the following way:
Γ(z) = Z 1
0
dtexp(−t)tz−1+ Z ∞
1
dtexp(−t)tz−1. (5.13)
4The next much more involved task will be to show that the renormalised physical result is independent of the regularisation scheme, which will be done beginning with the next section.
5The whole dimensional regularisation program makes only sense if at least one integer space time dimension exists, where the integral under consideration is convergent.
5.3 · Dimensional regularisation
At first we look on the second integral:
ω(z) = Z ∞
1
dtexp(−t)tz−1 : (5.14)
For each t >1 the integrand is an analytic function of z∈C. For an arbitrary compact area B of the z-plane there exists x0∈R such that:
x0 = max
z∈B[Rez]. (5.15)
Since further for t≥1 the logarithm is non-negative, we find
∀z∈B: |exp(−t)tz−1|=|exp[−t+ (z−1) lnt]| ≤exp(−t)tx0−1. (5.16) Because the integral
Z ∞ 1
exp(−t)tx0−1dt (5.17)
is converging point-wise for allz∈B, due to Weierstraß’ convergence criterion this is also the case in the sense of uniform convergence in B and thusω is a analytic function in B. Thus ω is analytic in the whole complexz-plane.
A little more work is to do for the first integral in (5.13):
φ(z) = Z 1
0
dtexp(−t)tz−1. (5.18)
The modulus of the integrand is exp(−t)tx−1, and for x > 1 the integral converges. Thus (5.18) is an analytic function for Rez >1. We like to show that this is the case for all compact areas of the right z-half-plane. Because B is supposed to be compact, there exists
x1 = min
z∈BRez, (5.19)
and it is x1 >0. For 0< t≤1 we have lnt≤0. Thus also
∀z∈B : |exp(−t)tz−1| ≤exp(−t)tx1−1 (5.20) holds true.
Since the integral over this real function is converging, again applying Weierstraß’ criterion for uniform convergence shows that (5.18) is an analytical function inB.
Since B can be an arbitrary compact area in the right z-half-plane from (5.11) follows the analyticity of Γ in the whole open right z-half-plane.
The next step is to find a maximal analytic continuation of Γ to the left half plane. It is enough to do this for the first integral in (5.13), because the second one has been shown to be an analytic function of zin the whole complex plane.
Now the series
exp(−t) = X∞ n=0
(−t)n
n! (5.21)
is uniformly convergent for fixed t ∈ R. Plugging this into (5.18) by naive order by
Since the series at the right hand side of this equation is uniformly convergent in any compact subset of the right z-half-plane which does not contain any of the points {0;−1;−2;. . .}, the order by order-integration is justified and φ is analytically continued to a meromorphic function with simple poles at the non-positive integer numbers. This property is thus also true for the Γ-function itself. From this we read off Weierstraß’ expansion of the Γ-function:
Γ(z) =
In the following we understand this meromorphic function as Γ. Now we like to find some useful properties of the Γ-function.
Fur n∈Nwe can calculate the integral (5.11) analytically with the result
Γ(n+ 1) =n! (5.24)
This is easily shown inductively by integrating (5.11) by parts. For any real positive z this yields also the important functional relation
Γ(z+ 1) =zΓ(z). (5.25)
Since Γ is a meromorphic function this is valid for allz∈C\Z≤0. The next relation we like to show is
Γ(z)Γ(1−z) = π
Now setting 1−z forzand renaming the integration variable withv we find Γ(1−z) = 2
This can be read as an integral over the first quarter of theuv-plane and we transform it into plane polar coordinates:
5.3 · Dimensional regularisation To calculate this integral we substituteφ= arccot(√
x):
Γ(z)Γ(1−z) = Z ∞
0
dxxz−1
1 +x. (5.30)
The function
f(y) = (−y)z−1
1 +y (5.31)
has an essential singularity in y = 0 and we cut the complex y-plane along the positive real axis. Now we go to the sheet of the Riemannian surface for which
Imz→±0lim (−y)z−1 =|y|z−1exp[∓iπ(z−1)] (5.32) is valid. Now we integrate over the path shown in figure 5.4.
Rey Imy
x+i0 C
Figure 5.4: Path for the integral (5.30)
Letting the radius of the big circle go to infinity and this of the small one to zero these circles do not contribute to the integral, and we find
Z
C
dy(−y)z−1
1 +y = 2i sin(πz) Z ∞
0
dxxz+1
1 +x. (5.33)
On the other hand using the residuum theorem we see, that Z
C
dy(−y)z−1
1 +y = 2πi Res
y=−1
(−y)z−1
1 +y = 2πi. (5.34)
Both results prove together with (5.29) the conjecture (5.26) for z ∈ (0,1) and thus for all z∈C\ {0,−1,−2, . . .}, because Γ is a meromorphic function.
Especially plugging in z= 1/2 in (5.26) we find Γ
1 2
= Z ∞
0
dt√
texp(−t) =√
π. (5.35)
Further we needEuler’s Beta-function, defined by B(p;q) =
Z 1 0
dxxp−1(1−x)q−1. (5.36)
Substitution of t = 1−x yields the symmetry of the function under interchange of its two where we have introduced plane polar coordinates in the last step.
Substitution of t=r2 gives Γ(p)Γ(q) = 2 In the remaining integral we substitutex= cos2φand obtain its value to be B(p;q)/2. Thus we have
B(p;q) = Γ(p)Γ(q)
Γ(p+q). (5.41)
Now we want to give a proof for Gauß’s representation of the Γ-function as an infinite product:
1
γ is the so called Euler-Mascheroni-constant.
To prove (5.42) we use the following representation of the exponential function exp(−t) = lim
Naively looking on this definition we see that this series converges to Γ(z) in each regular pointz. We shall show that this is even the case in the sense of uniform convergence. But at first we show that we then also have proven the product representation (5.42):
5.3 · Dimensional regularisation
Substitution oft=nτ in (5.44) yields Pn(z) =nz
Here we have used the properties of the B-function given above as well as the functional property (5.25) of the Γ-function.
A little algebra of this result yields:
1
This shows that indeed the uniform convergence of the series Pn to Γ proves Gauß’ product representation (5.42).
From the principle of analytic continuation we know that it is sufficient to show this for real positivez. Differentiating with respect tot leads to the following relation:
1−
For 0< t < nthe integrand is positive. On the other hand we have Z t
From Euler’s definition of the Γ-function (5.11) we know that Γ(z)−Pn(z) = holds. Within the convergence proof of (5.11) we have shown that the second integral con-verges uniformly to 0 forn→ ∞. From the above inequality we read off
0 ≤
for all n ∈ N. Let > 0. Because of the uniform convergence of the last integral we may chose n0 so large that
Z ∞
n0
dtexp(−t)tz−1 <
2 (5.51)
holds. Then we have forn > n0 by using the inequality again:
0≤ From this we can read off immediately that the integral is uniformly convergent in each compact interval of the positive real axis. Thus we finished the proof of Gauß’ product representation for the Γ-function.
Deriving of this equation with respect to z leads to Ψ1(z) := d
Since the series converges uniformly on each compact subset of C which does not contain a negative integer number or 0, this is really the logarithmic derivative of the Γ-function.
Within the dimensional regularisation technique we shall also need the Laurent-expansion of the Γ-function around the simple poles atz∈Z≤0. It is enough to find the expansion up to the first order:
For proving this equation we state that from (5.54) follows Ψ1(1) =−γ−1 +
Now we look on the Taylor-expansion of the Γ-function around the regular point z= 1:
Γ(1 +) = 1 +Γ0(1) +O(2) = 1 +Ψ1(1) +O(2), (5.59)
5.3 · Dimensional regularisation
which is valid in the open disc of radius 1 around= 0, because the next pole of the Γ-function is at =−1. Dividing this equation throughyields:
Γ() = 1
Γ(1 +) = 1
−γ+O(). (5.60)
This is (5.55) forn= 0. For all othern∈N the equation can be shown by induction. Suppose it is true for n=k. Then we find making use of (5.25) again:
Γ[−(k+ 1) +] = Γ(−k+)
−(k+ 1) + = (−1)k+1 (k+ 1)!
1
+ Ψ1(k+ 1) + 1
k+ 1+O()
. (5.61) Comparing this with (5.55) we see that this is of course this equation forn=k+ 1, and this was to show.