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Question 5.8: Calculation of Moments of Splitting Functions (Anomalous Dim......

Calculation of Moments of Splitting Functions (Anomalous Dimensions)

Calculate the moments of the splitting functions P_{G q}, P_{q q}, P_{q G}, and P_{G G}. The splitting functions are explicitly (see (5.70)-(5.72))

\begin{aligned} & P_{q G}(z)=\frac{1}{2}\left[z^2+(1-z)^2\right] & (5.70)\\ & P_{G q}(z)=\frac{4}{3}\left[1+(1-z)^2\right] \frac{1}{z} & (5.71) \\ & P_{G G}(z)=6\left[z\left(\frac{1}{1-z}\right)_{+} \frac{1-z}{z}+z(1-z)+\left(\frac{11}{12}-\frac{N_{\mathrm{f}}}{18}\right) \delta(1-z)\right]. & (5.72) \end{aligned}

(a) P_{G q}=\frac{4}{3} \frac{1+(1-x)^{2}}{x},

(b) \quad P_{q q}=-\frac{4}{3}(1+x)+2 \delta(1-x)+ \underset{\varepsilon \rightarrow 0}{\lim} \frac{8}{3}\left[\frac{1}{1-x+\varepsilon}-\delta(1-x) \int\limits_{0}^{1} \mathrm{~d} y \frac{1}{1-y+\varepsilon}\right] \text {, }

(c) P_{q G}=\frac{1}{2}\left[x^{2}+(1-x)^{2}\right],

(d) P_{G G}=6\left[\frac{1}{x}-2+x(1-x)\right]+\left(\frac{11}{2}-\frac{N_{\mathrm{f}}}{3}\right) \delta(1-x) +\underset{\varepsilon \rightarrow 0}{\lim} 6\left[\frac{1}{1-x+\varepsilon}-\delta(1-x) \int_{0}^{1} \mathrm{~d} y \frac{1}{1-y+\varepsilon}\right]

Step-by-Step
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(a) Inserting the explicit expression for the splitting function P_{G q} we get for the nth moment

\begin{aligned} d_{G q}(n) & =-\frac{6}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x x^{n-1} P_{G q}(x) \\ & =-\frac{6}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x x^{n-1} \frac{4}{3} \frac{1+(1-x)^{2}}{x} \\ & =-\frac{8}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x\left(2 x^{n-2}-2 x^{n-1}+x^{n}\right) \\ & =-\left.\frac{8}{33-2 N_{\mathrm{f}}}\left(\frac{2}{n-1} x^{n-1}-\frac{2}{n} x^{n}+\frac{1}{n+1} x^{n+1}\right)\right|_{0} ^{1} . & (1) \end{aligned}

Since n \geq 1 we get

\begin{aligned} d_{G q}(n) & =-\frac{8}{33-2 N_{\mathrm{f}}}\left(\frac{2}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right) \\ & =-\frac{8}{33-2 N_{\mathrm{f}}} \frac{n^{2}+n+2}{n\left(n^{2}-1\right)} . & (2) \end{aligned}

(b) The nth moment of the splitting function P_{q q} is given by

\begin{aligned} d_{q q}(n)= & -\frac{6}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x x^{n-1} P_{\mathrm{qq}}(x) \\ = & -\frac{6}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x x^{n-1}\left[-\frac{4}{3}(1+x)+2 \delta(1-x)\right. \\ & \left.+ \underset{\varepsilon \rightarrow 0}{\lim} \frac{8}{3}\left(\frac{1}{1-x+\varepsilon}-\delta(1-x) \int\limits_{0}^{1} \mathrm{~d} y \frac{1}{1-y+\varepsilon}\right)\right] \\ = & -\frac{6}{33-2 N_{\mathrm{f}}}\left[-\frac{4}{3} \int\limits_{0}^{1} \mathrm{~d} x\left(x^{n-1}+x^{n}\right)+2\right. \\ & \left.+ \underset{\varepsilon \rightarrow 0}{\lim} \frac{8}{3}\left(\int\limits_{0}^{1} \mathrm{~d} x \frac{x^{n-1}}{1-x+\varepsilon}-\int\limits_{0}^{1} \mathrm{~d} y \frac{1}{1-y+\varepsilon}\right)\right] . & (3) \end{aligned}

With the expansion

\frac{x^{n-1}}{1-x+\varepsilon}=-\sum\limits_{i=0}^{n-2} x^{i}+\frac{1}{1-x+\varepsilon}     (4)

this equation can be put into a simple form:

\begin{aligned} d_{q q}(n) & =-\frac{6}{33-2 N_{\mathrm{f}}}\left[-\frac{8}{3} \sum\limits_{i=0}^{n} \int\limits_{0}^{1} \mathrm{~d} x x^{i}+\frac{4}{3} \int\limits_{0}^{1} \mathrm{~d} x\left(x^{n-1}+x^{n}\right)+2\right] \\ & =-\frac{6}{33-2 N_{\mathrm{f}}}\left[-\frac{8}{3} \sum\limits_{i=0}^{n} \frac{1}{i+1}+\frac{4}{3}\left(\frac{1}{n}+\frac{1}{n+1}\right)+2\right] \\ & =-\frac{6}{33-2 N_{\mathrm{f}}}\left[-\frac{8}{3} \sum\limits_{i=2}^{n} \frac{1}{i}-\frac{8}{3} \frac{1}{n+1}-\frac{8}{3}+\frac{4}{3} \frac{1}{n}+\frac{4}{3} \frac{1}{n+1}+2\right] \\ & =\frac{4}{33-2 N_{\mathrm{f}}}\left[4 \sum\limits_{i=2}^{n} \frac{1}{i}+1-\frac{2}{(n+1) n}\right] . & (5) \end{aligned}

(c) For the moment d_{q G}(n) we get

\begin{aligned} d_{q G}(n) & =-\frac{6}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x x^{n-1} P_{\mathrm{qG}}(x) \\ & =-\frac{6}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x x^{n-1} \frac{1}{2}\left[x^{2}+(1-x)^{2}\right] \\ & =-\frac{3}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x\left(2 x^{n+1}-2 x^{n}+x^{n-1}\right) \end{aligned}

\begin{aligned} & =-\left.\frac{3}{33-2 N_{\mathrm{f}}}\left(2 n+2 x^{n+2}-\frac{2}{n+1} x^{n+1}+\frac{1}{n} x^{n}\right)\right|_{0} ^{1} \\ & =-\frac{3}{33-2 N_{\mathrm{f}}} \frac{n^{2}+n+2}{(n+2)(n+1) n} . & (6) \end{aligned}

(d) The remaining moments d_{G G}(n) of the splitting function P_{G G} are obtained in the same way as in (b):

\begin{aligned} d_{G G}(n)= & -\frac{6}{33-2 N_{\mathrm{f}}} \int\limits_{0}^{1} \mathrm{~d} x x^{n-1} P_{G G}(x) \\ = & -\frac{6}{33-2 N_{\mathrm{f}}}\left[6 \int\limits_{0}^{1} \mathrm{~d} x x^{n-1}\left(\frac{1}{x}-2+x(1-x)\right)+\left(\frac{11}{2}-\frac{f}{3}\right)\right. \\ & \left.+6 \underset{\varepsilon \rightarrow 0}{\lim} \left(\int\limits_{0}^{1} \mathrm{~d} x \frac{x^{n-1}}{1-x+\varepsilon}-\int\limits_{0}^{1} \mathrm{~d} y \frac{1}{1-y+\varepsilon}\right)\right] . & (7) \end{aligned}

Again we employ the relation

\frac{x^{n-1}}{1-x+\varepsilon}=-\sum\limits_{i=0}^{n-2} x^{i}+\frac{1}{1-x+\varepsilon}   (8)

and get, after integrating and rearranging,

\begin{aligned} d_{G G}(n)= & -\frac{6}{33-2 N_{\mathrm{f}}}\left[6\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}-\frac{1}{n+2}\right)\right. \\ & \left.+\left(\frac{11}{2}-\frac{f}{3}\right)-6 \sum_{i=0}^{n-2} \int\limits_{0}^{1} \mathrm{~d} x x^{i}\right] \\ = & -\frac{6}{33-2 N_{\mathrm{f}}}\left[6\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}-\frac{1}{n+2}-\sum_{i=0}^{n-2} \frac{1}{i+1}\right)\right. \\ & \left.+\frac{11}{2}-\frac{N_{\mathrm{f}}}{3}\right] \\ = & -\frac{6}{33-2 N_{\mathrm{f}}}\left[-6 \sum_{i=2}^{n} \frac{1}{i}-6\right. \\ & \left.+6\left(\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n+1}-\frac{1}{n+2}\right)+\frac{11}{2}-\frac{N_{\mathrm{f}}}{3}\right] \\ = & \frac{9}{33-2 N_{\mathrm{f}}}\left[4 \sum_{i=2}^{n} \frac{1}{i}+\frac{1}{3}+\frac{2 N_{\mathrm{f}}}{9}-4\left(\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n+1}-\frac{1}{n+2}\right)\right] \\ = & \frac{9}{33-2 N_{\mathrm{f}}}\left[4 \sum_{i=2}^{n} \frac{1}{i}+\frac{1}{3}+\frac{2 N_{\mathrm{f}}}{9}-\frac{4}{n(n-1)}-\frac{4}{(n+2)(n+1)}\right] . & (9) \end{aligned}

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