Question 5.13: Calculation of the Gluonic Contribution to FL(x, Q²) Calcula......
Calculation of the Gluonic Contribution to F_{L}\left(x, Q^{2}\right)
Calculate in the same way as done in the foregoing text the contribution to F_{L} due to photon-gluon scattering via a quark loop. The corresponding diagrams are shown in the figure. A usefull trick is to express the product of two propagators involving an onshell momentum p^{2}=0 as
\frac{1}{(k+p)^{2}} \frac{1}{k^{2}}=\int\limits_{0}^{1} \mathrm{~d} u \frac{1}{(k+u p)^{4}}.
Prove this formula and find its generalization.
The color factor for the diagrams coming from the Gell-Mann matrices at the quark-gluon vertices can be written as
\frac{1}{\left(N^{2}-1\right)} \sum\limits_{A} \sum\limits_{B} \operatorname{tr}\left[\frac{\lambda^{A}}{2} \frac{\lambda^{B}}{2}\right]=\frac{1}{\left(N^{2}-1\right)} \sum\limits_{A} \sum\limits_{B} \frac{1}{2} \delta^{A B}=\frac{1}{2}, (1)
where we summed over outgoing gluon colors and averaged over the incoming ones. Together with a factor N_{\mathrm{f}}, which is due to the sum over flavors in the quark loop, this yields an overall factor T_{\mathrm{f}}=\frac{1}{2} N_{\mathrm{f}}.
We start with the first diagram which we name T_{\mu \nu}^{(1) g g} (direct). In the usual way the exchange term can be found by substituting \mu \leftrightarrow v and q \leftrightarrow-q. According to the Feynman rules we write
\begin{aligned} & T_{\mu \nu}^{(1) g g} \text { (direct) } \\ & =\mathrm{i} \frac{g^{2}}{2} T_{\mathrm{f}} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \operatorname{tr}\left[\gamma_{\mu} \frac{\mathrm{i}}{\not p+\not d+k k} \gamma_{\nu} \frac{\mathrm{i}}{\not p+\not k} \gamma_{\alpha} \frac{\mathrm{i}}{\not k} \gamma_{\beta} \frac{i}{\not p+\not k}\right]\left(-g^{\alpha \beta}\right), \end{aligned} (2)
where we have inserted an additional factor of \frac{1}{2} due to the averaging over the gluon spin and used \left(-g^{\alpha \beta}\right) for the polarization sum \frac{1}{2} \sum \varepsilon^{\alpha}(k) \varepsilon^{\beta}(k)=-g^{\alpha \beta} / 2. Again we compute i times the time-ordered product. Projecting with p^{\mu} p^{\nu} we find, with \gamma_{\alpha} \not k \gamma^{\alpha}=-2 \not k,
\begin{aligned} p^{\mu} p^{\nu} T_{\mu \nu}^{(1) g g}(\text { direct })= & \mathrm{i} g^{2} T_{\mathrm{f}} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{\operatorname{tr}[\not p(\not d+\not k) \not p \not k \not k \not k]}{(p+q+k)^{2}(p+k)^{4} k^{2}} \\ & =8 \mathrm{i} g^{2} T_{\mathrm{f}} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(q+k)(k p)}{(p+q+k)^{2}(p+k)^{4}} . & (3) \end{aligned}
The integral can be evaluated in the usual way
\begin{aligned} \int\limits & \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{(k+q)_{\alpha} k_{\beta}}{(p+k+q)^{2}(p+k)^{4}} \\ & =\frac{\Gamma(3)}{\Gamma(1) \Gamma(2)} \int\limits_{0}^{1} \mathrm{~d} u \bar{u} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{(k+q)_{\alpha} k_{\beta}}{\left[u(p+k+q)^{2}+\bar{u}(p+k)^{2}\right]^{3}} \\ & =\Gamma(3) \int\limits_{0}^{1} \mathrm{~d} u \bar{u} \int\limits_{0} \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{\left.[k-(p+u q)]_{\alpha}[k-p+\bar{u} q)\right]_{\beta}}{\left[k^{2}-u \bar{u} Q^{2}\right]^{3}} \\ & =-q_{\alpha} q_{\beta} \Gamma(3) \int\limits_{0}^{1} \mathrm{~d} u \bar{u} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{1}{\left[k^{2}-u \bar{u} Q^{2}\right]^{3}}+\mathcal{O}\left(g_{\alpha \beta}, p_{\alpha} p_{\beta}\right) \\ & =-q_{\alpha} q_{\beta} \int\limits_{0}^{1} \mathrm{~d} u \bar{u} u \frac{\mathrm{i}}{(4 \pi)^{\frac{d}{2}}} \frac{1}{\left(u \bar{u} Q^{2}\right)}+\mathcal{O}\left(g_{\alpha \beta}, p_{\alpha} p_{\beta}\right) \\ & =\frac{-q_{\alpha} q_{\beta}}{Q^{2}} \frac{\mathrm{i}}{(4 \pi)^{2}} \cdot \frac{1}{2}+\mathcal{O}\left(g_{\alpha \beta}, p_{\alpha} p_{\beta}\right) . & (4) \end{aligned}
We performed the usual Feynman parametrization (see Exercise 4.7) and wrote the denominator as \left[u(p+k+q)^{2}+\bar{u}(p+k)^{2}\right]=\left[(k+p+u q)^{2}\right. \left.+u \bar{u} q^{2}\right], where \bar{u}=1-u. Inserting this into (3) yields
p^{\mu} p^{\nu} T_{\mu \nu}^{(1) g g}(\text { direct })=\frac{4 g^{2} T_{\mathrm{f}}}{(4 \pi)^{2}} \frac{(p \cdot q)^{2}}{Q^{2}}. (5)
Taking the other four diagrams of the same type into account, i.e. the exchange term and the opposite direction of the fermion loop, we get a contribution to F_{L}(5.251) of
\frac{\tilde{F}_{L}^{(1) g}}{2 x}=\frac{16 g^{2} T_{\mathrm{f}}}{(4 \pi)^{2}}, (6)
which is, however, independent of \omega, i.e. proportional to \omega^{0}. However, the sum in (5.258) starts as \omega^{2}, i.e. after transformation to Bjorken x these four diagrams do not contribute to the measurable structure function.
The remaining diagrams are sligthly more complicated to evaluate. They are shown in the last row of the figure at the beginnning of this exercise. We write according to the Feynman rules (see Fig. 5.25)
\begin{aligned} & T_{\mu \nu}^{(1) g g} \text { (direct) } \\ & =\frac{1}{2} T_{\mathrm{f}} g^{2} \mathrm{i} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \operatorname{tr}\left[\gamma_{\mu} \frac{\mathrm{i}}{\not q+\not k} \gamma_{\alpha} \frac{\mathrm{i}}{\not q+\not k+\not p} \gamma_{\nu} \frac{\mathrm{i}}{\not k+\not p} \gamma_{\beta} \frac{\mathrm{i}}{\not k}\right]\left(-g^{\alpha \beta}\right) . \end{aligned} (7)
Projecting onto F_{L} we write
\begin{aligned} p^{\mu} p^{v} T_{\mu \nu}^{(1) g g}(\text { direct })= & -\frac{1}{2} T_{\mathrm{f}} g^{2} \mathrm{i} \\ & \times \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{\operatorname{tr}\left[\not p(\not q+\not k) \gamma_{\alpha}(\not q+\not k+\not p) \not p(\not k+\not p) \gamma^{\alpha} k\right]}{(q+k)^{2}(q+k+p)^{2}(k+p)^{2} k^{2}} \\ = & +T_{\mathrm{f}} g^{2} \mathrm{i} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{\operatorname{tr}[\not p(\not q+\not k) \not k \not p(\not q+\not k) \not k]}{(q+k)^{2}(q+k+p)^{2}(k+p)^{2} k^{2}} , \quad (8) \end{aligned}
where we used \not p^{2}=p^{2}=0 and \gamma_{\alpha} \gamma^{\mu} \gamma^{\nu} \gamma^{\varrho} \gamma^{\alpha}=-2 \gamma^{\varrho} \gamma^{\nu} \gamma^{\mu}. The trace can be further simplified as
\begin{aligned} \operatorname{tr}[\not p(\not q+\not k) \not k \not p(\not d+\not k) \not k] & =2 k \cdot p \operatorname{tr}[\not p(\not q+\not k)(\not q+\not k) \not k] \\ & -\operatorname{tr}[\not p(\not q+\not k) \not p \not k(\not q+\not k) \not k] \\ & =8(k \cdot p)^{2}(q+k)^{2}-2 p \cdot(q+k) \operatorname{tr}[\not p k(\not q+ \not k) \not k] \\ & =8(k \cdot p)^{2}(q+k)^{2}-16 p \cdot(q+k)(k \cdot q)(p \cdot k) \\ & -16 p(q+k) p \cdot k k^{2}+8[p \cdot(q+k)]^{2} k^{2}, & (9) \end{aligned}
so that we get
\begin{aligned} p^{\mu} p^{\nu} T_{\mu \nu}^{(1) g g}(\text { direct })= & 8 g^{2} \mathrm{i} T_{\mathrm{f}}\left[I_{1}(q, p)+I_{1}(-q, p)\right. \\ & \left.-2 I_{2}(q, p)-2 I_{3}(q, p)\right], & (10) \end{aligned}
where
\begin{aligned} I_{1}(q, p) & =\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{(k \cdot p)^{2}}{(q+k+p)^{2}(k+p)^{2} k^{2}}, \\ I_{1}(-q, p) & =\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{(p \cdot(q+k))^{2}}{(q+k)^{2}(q+k+p)^{2}(k+p)^{2}} \\ & =\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{(p \cdot k)^{2}}{(k+p)^{2}(k+p-q)^{2} k^{2}}, \\ I_{2}(q, p) & =\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(q+k) p \cdot k}{(q+k)^{2}(q+k+p)^{2}(k+p)^{2}}, \\ I_{3}(q, p) & =\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(q+k) k \cdot q p \cdot k}{(q+k)^{2}(q+k+p)^{2}(k+p)^{2} k^{2}} . \qquad (11) \end{aligned}
The integrals I_{1}(q, p) and I_{2}(q, p) are of the same type as we had before (see (5.265)).
\begin{aligned} I_{\alpha \beta}(p, q)= & \int \frac{\mathrm{d}^4 k}{(2 \pi)^4} \frac{k_\alpha k_\beta}{(p+k)^4(p+k+q)^2 k^2} \\ = & \frac{\Gamma(4)}{\Gamma(2) \Gamma(1) \Gamma(1)} \int_0^1 \mathrm{~d} u \int_0^{\bar{u}} \mathrm{~d} v u \\ & \times \int \frac{\mathrm{d}^4 k}{(2 \pi)^4} \frac{k_\alpha k_\beta}{\left[(p+k)^2 u+(p+k+q)^2 v+k^2(1-u-v)\right]^4}, \quad (5.265) \end{aligned}
Using suitable substitutions we can therefore deduce the corresponding results directly from the previous calculation of F_{L}^{q}. Comparing I_{1}(q, p) with (5.265) we see that (p+k)^{2} in the denominator comes now with a power of 1 instead of 2. Therefore we copy from (5.269)
\begin{aligned} & I_{\alpha \beta}(p, q) p^\alpha p^\beta \\ & =\Gamma(4)(p \cdot q)^2 \int_0^1 \mathrm{~d} v \bar{v}^2 v^2 \int_0^1 \mathrm{~d} u u \int \frac{\mathrm{d}^4 k}{(2 \pi)^4} \frac{1}{\left[k^2-v \bar{v} Q^2(1-\omega \bar{u})\right]^4} . \quad (5.269) \end{aligned}
with evident changes and get
I_{1}(q, p)=(p \cdot q)^{2} \Gamma(3) \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v \bar{v} v^{2} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{1}{\left(k^{2}-v \bar{v} Q^{2}(1-\omega \bar{u})\right)^{3}}. (12)
We changed \Gamma(4) \rightarrow \Gamma(3) corresponding to the different power and we omitted the Feynman parameter (u \bar{v}), which came from the parametrization of (p+k)^{4} in (5.265), which in our case now is (p+k)^{2}. Also we have taken care of the shift in the integration variable k_{\alpha} \rightarrow k_{\alpha}-(u \bar{v}+v) p_{\alpha}-v q_{\alpha} where due to the contraction with p^{\alpha} p^{\beta} only v^{2} q_{\alpha} q_{\beta} contributes.
For the \int\limits \mathrm{d} k integral we get
\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{1}{\left[k^{2}-v \bar{v} Q^{2}(1-\omega \bar{u})\right]^{3}}=\frac{-\mathrm{i}}{(4 \pi)^{2}} \frac{1}{\Gamma(3)} \frac{1}{v \bar{v} Q^{2}(1-\omega \bar{u})}. (13)
The \int\limits \mathrm{d} v integration gives
\int\limits_{0}^{1} \mathrm{~d} v \bar{v}^{2} v(v \bar{v})^{-1}=\frac{1}{2}, (14)
and therefore
I_{1}(q, p)=\frac{-\mathrm{i}}{(4 \pi)^{2}} \frac{(p \cdot q)^{2}}{Q^{2}} \cdot \frac{1}{2} \int\limits_{0}^{1} \mathrm{~d} u \frac{1}{(1-\omega \bar{u})} . (15)
Expanding
\frac{1}{1-\omega \bar{u}}=\sum\limits_{n=0}^{\infty}(\omega \bar{u})^{n}, (16)
we find, after trivial integration,
I_{1}(q, p)=\frac{-\mathrm{i}}{(4 \pi)^{2}} \frac{(p \cdot q)^{2}}{Q^{2}} \frac{1}{2} \sum\limits_{n=0}^{\infty} \frac{1}{(n+1)} \omega^{n}. (17)
In the same way we can evaluate I_{2}(q, p) :
\begin{aligned} I_{2}(q, p) & =\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(q+k) p \cdot k}{(q+k)^{2}(q+k+p)^{2}(k+p)^{2}} \\ & =\int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{(p \cdot k) p \cdot(k-q)}{k^{2}(k+p)^{2}(k+p-q)^{2}} \\ & =\Gamma(3) \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v \bar{v} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(k-v q) p \cdot(k-(1-v) q)}{\left[k^{2}-v \bar{v} Q^{2}(1+\omega \bar{u})\right]^{3}}, & (18) \end{aligned}
where we again copied from (5.269) and performed in addition to the changes carried out in (12) the substitution q \rightarrow-q, i.e. \omega \rightarrow-\omega. Also the shift of the integration variable now reads k \rightarrow k-(u \bar{v}+v) p+v q.
I_{2} therefore becomes
\begin{aligned} I_{2}(q, p) & =-\Gamma(3)(p \cdot q)^{2} \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v \bar{v}^{2} v \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{1}{\left[k^{2}-v \bar{v} Q^{2}(1+\omega \bar{u})\right]^{3}} & (19) \\ & =\frac{+\mathrm{i}}{(4 \pi)^{2}} \frac{(p \cdot q)^{2}}{Q^{2}} \frac{1}{2} \sum\limits_{n=0}^{\infty} \frac{1}{(n+1)}(-\omega)^{n}, & (20) \end{aligned}
where we used (13) and obtained for the \int\limits \mathrm{d} v integration
\int\limits_{0}^{1} \mathrm{~d} v \bar{v}^{2} v(\bar{v} v)^{-1}=\frac{1}{2}. (21)
We also expanded
\frac{1}{(1+\omega \bar{u})}=\sum\limits_{n=0}^{\infty}(-\omega \bar{u})^{n} (22)
Let us now proceed to calculate I_{3}(q, p), which is a little bit more tricky since we deal with four propagators, two of which contain off-shell momenta q^{2}=-Q^{2} \neq 0. To circumvent this problem we note the following relation:
\frac{1}{(k+p)^{2 A}} \cdot \frac{1}{\left(k^{2}\right)^{B}}=\frac{\Gamma(A+B)}{\Gamma(A) \Gamma(B)} \int\limits_{0}^{1} \mathrm{~d} u \frac{u^{A-1} \bar{u}^{B-1}}{(k+u p)^{2(A+B)}} (23)
which holds for arbitrary k^{2} \neq 0 and p^{2}=0. It can be proven using Feynman parameters:
\frac{1}{(k+p)^{2 A}} \cdot \frac{1}{\left(k^{2}\right)^{B}}=\frac{\Gamma(A+B)}{\Gamma(A) \Gamma(B)} \int\limits_{0}^{1} \mathrm{~d} u \frac{u^{A-1} \bar{u}^{B-1}}{\left[u(k+p)^{2}+\bar{u} k^{2}\right]^{A+B}}, (24)
where \left[u(k+p)^{2}+\bar{u} k^{2}\right]=k^{2}+2 k u p=(k+u p)^{2} completes the proof. Using this twice, the integral I_{3}(q, p) can be written as
\begin{aligned} I_{3}(q, p)= & \int\limits_{0} \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(q+k) k \cdot q p \cdot k}{(q+k)^{2}(q+k+p)^{2}(k+p)^{2} k^{2}} \\ = & \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(q+k) k \cdot q p \cdot k}{(k+u p)^{4}(q+k+v p)^{4}} \\ = & \Gamma(4) \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v \int\limits_{0}^{1} \mathrm{~d} y y \bar{y} \\ & \times \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \frac{p \cdot(q+k) p \cdot k(k-u p) \cdot q}{\left[\bar{y} k^{2}+y(k+q+(v-u) p)^{2}\right]^{4}}, & (25) \end{aligned}
where in the last step k \rightarrow k-u p is shifted and a third Feynman parameter is introduced. The denominator can now be simplified in the usual way
\begin{aligned} {\left[\bar{y} k^{2}+y(k+q+(v-u) p)^{2}\right]=} & k^{2}+2 k \cdot(y q+(v-u) y p) \\ & +y(q+(v-u) p)^{2} \\ = & {[k+y q+(v-u) y p)^{2}+y \bar{y}(q+(v-u) p]^{2} } \\ = & k^{\prime 2}+y \bar{y}\left[q^{2}+2 q \cdot p(v-u)\right] \\ = & k^{\prime 2}-y \bar{y} Q^{2}[1-\omega(v-u)] & (26) \end{aligned}
The shift of the integration momenta k=k^{\prime}-y p-(v-u) y p has to be performed in a similar way in the numerator:
\begin{aligned} & (q+k)_{\alpha}(k-u p)_{\beta} k_{\sigma}\left[p^{\alpha} q^{\beta} p^{\sigma}\right] \\ & \quad=\left(k^{\prime}+\bar{y} q\right)_{\alpha}\left(k^{\prime}-y q-v y p-u \bar{y} p\right)_{\beta}\left(k^{\prime}-y q\right)_{\sigma}\left[p^{\alpha} q^{\beta} p^{\sigma}\right] \\ & \quad=k_{\alpha}^{\prime} k_{\beta}^{\prime}(-y q)_{\sigma}+k_{\beta}^{\prime} k_{\sigma}^{\prime}(\bar{y} q)_{\alpha}+\bar{y} q_{\alpha}(y q+v y p+u \bar{y} p)_{\beta} y q_{\sigma}\left[p^{\alpha} q^{\beta} p^{\sigma}\right] \\ & \quad=\frac{k^{\prime 2}}{4}(p \cdot q)^{2}[1-2 y]+\bar{y} y(p \cdot q)^{3}(v y+u \bar{y})+\bar{y} y^{2}(p \cdot q)^{2} q^{2}. &(27) \end{aligned}
Here we used the fact that under integration, contributions like \sim k_{\alpha} or k_{\alpha} k_{\beta} k_{\sigma} vanish and that k_{\alpha} k_{\beta}=\frac{1}{4} k^{2} g_{\alpha \beta}. (See our discussion on dimensional regularization before.)
With that we are left with two k^{\prime} integrations:
\begin{aligned} \frac{1}{4} \int\limits & \frac{\mathrm{d}^{4} k^{\prime}}{(2 \pi)^{4}} \frac{k^{\prime 2}}{\left[k^{\prime 2}-y \bar{y} Q^{2}(1-\omega(v-u))\right]^{4}} \\ & =\frac{-\mathrm{i}}{4(4 \pi)^{2}} \frac{\Gamma(3)}{\Gamma(2) \Gamma(4)} \frac{1}{\left(y \bar{y} Q^{2}(1-\omega(v-u))\right)} \\ & =\frac{-\mathrm{i}}{2(4 \pi)^{2}} \frac{1}{\Gamma(4)} \frac{1}{\left(y \bar{y} Q^{2}(1-\omega(v-u))\right)} & (28a) \end{aligned}
and
\begin{aligned} \int\limits \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} & \frac{1}{\left[k^{2}-y \bar{y} Q^{2}(1-\omega(v-u))\right]^{4}} \\ & =\frac{\mathrm{i}}{(4 \pi)^{2}} \frac{\Gamma(2)}{\Gamma(4)\left[y \bar{y} Q^{2}(1-\omega)(v-u)\right]^{2}} . & (28b) \end{aligned}
Inserting everything in (25) and performing the trivial \int\limits \mathrm{d} y integration we get
\begin{aligned} I_{3}(q, p)= & \frac{\mathrm{i}}{2(4 \pi)^{2}} \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v\left\{\frac{(p \cdot q)^{3}}{Q^{4}} \frac{(v+u)}{[1-\omega(v-u)]^{2}}\right. \\ & \left.+\frac{(p \cdot q)^{2}}{Q^{4}} q^{2} \frac{1}{[1-\omega(v-u)]^{2}}\right\} \\ = & \frac{\mathrm{i}}{4(4 \pi)^{2}} \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v \frac{(v+u) \omega-2}{[1-\omega(v-u)]^{2}} \frac{(p \cdot q)^{2}}{Q^{2}} . & (29) \end{aligned}
Expanding again in \omega yields
\frac{1}{[1-\omega(v-u)]^{2}}=\sum\limits_{n=0}^{\infty} \frac{\Gamma(n+2)}{\Gamma(2) \Gamma(n+1)}(v-u)^{n} \omega^{n}=\sum\limits_{n=0}^{\infty}(1+n)(v-u)^{n} \omega^{n} (30)
and the remaining integrations can be performed:
\begin{aligned} \int\limits_{0}^{1} \mathrm{~d} u \int\limits_{0}^{1} \mathrm{~d} v(v-u)^{n} & =\frac{1}{(1+n)} \int\limits_{0}^{1} \mathrm{~d} u\left[(1-u)^{1+n}+(-)^{n} u^{1+n}\right] \\ & =\frac{1}{(1+n)(2+n)}\left(1+(-)^{n}\right), & (31) \\ \int\limits_{0}^{1} \mathrm{~d} u & \int\limits_{0}^{1} \mathrm{~d} v(v+u)(v-u)^{n} \\ & =\int\limits_{0}^{1} \mathrm{~d} u\left[\frac{\bar{u}^{n+2}}{n+2}-(-)^{n} \frac{u^{n+2}}{n+2}+2 \frac{u \bar{u}^{n+1}}{(n+1)}-(-)^{n+1} \frac{2 u^{n+2}}{n+1}\right] \\ & =\frac{1}{(n+2)(n+3)}-\frac{(-)^{n}}{(n+2)(n+3)}+\frac{2}{(n+2)(n+3)(n+1)} \\ & -(-)^{n+1} \frac{2}{(n+1)(n+3)} \cdot & (32) \end{aligned}
Inserting (30), (31), and (32) into (29) yields
\begin{aligned} I_{3}(q, p) & \left\{\frac { \mathrm { i } } { 4 ( 4 \pi ) ^ { 2 } } \sum _ { n = 0 } ^ { \infty } \left\{\frac{1+n}{(n+2)(n+3)}\left(1-(-)^{n}\right)+\frac{2}{(n+2)(n+3)}\right.\right. \\ & \left.\left.+\frac{2(-)^{n}}{(n+3)}\right\} \omega^{n+1}-\frac{\mathrm{i}}{2(4 \pi)^{2}} \sum_{n=0}^{\infty} \frac{1}{(2+n)}\left(1+(-)^{n}\right) \omega^{n}\right\} \frac{(p \cdot q)^{2}}{Q^{2}} \\ = & \left\{\frac { \mathrm { i } } { 4 ( 4 \pi ) ^ { 2 } } \sum _ { n = 0 } ^ { \infty } \left\{\left[\frac{2}{(3+n)}-\frac{1}{(2+n)}\right]\left(1-(-)^{n}\right)\right.\right. \\ & \left.+2\left[\frac{1}{(2+n)}-\frac{1}{(3+n)}\right]+\frac{2(-)^{n}}{(n+3)}\right\} \omega^{n+1} \\ & \left.-\frac{\mathrm{i}}{2(4 \pi)^{2}} \sum_{n=0}^{\infty} \frac{1}{(2+n)}\left(1+(-)^{n}\right) \omega^{n}\right\} \frac{(p \cdot q)^{2}}{Q^{2}} \\ = & \left\{\frac{\mathrm{i}}{4(4 \pi)^{2}} \sum_{n=0}^{\infty} \frac{1}{(2+n)} \omega^{n+1}\left(1+(-)^{n}\right)\right. \\ & \left.-\frac{\mathrm{i}}{2(4 \pi)^{2}} \sum_{n=0}^{\infty} \frac{1}{(2+n)}\left(1+(-)^{n}\right) \omega^{n}\right\} \frac{(p \cdot q)^{2}}{Q^{2}} \\ = & \left\{\frac{\mathrm{i}}{2(4 \pi)^{2}} \sum_{n=0,2,4,}^{\infty} \frac{1}{(2+n)} \omega^{n+1}-\frac{\mathrm{i}}{(4 \pi)^{2}} \sum_{n=0,2,4}^{\infty} \frac{1}{(2+n)} \omega^{n}\right\} \frac{(p \cdot q)^{2}}{Q^{2}} . & (33) \end{aligned}
Taking into account the contribution coming from the exchange term (q \rightarrow-q), which in this case cancels odd powers of \omega and doubles even powers, we find
I_{3}(q, p)+I_{3}(-q, p)=\left(\frac{-2 \mathrm{i}}{(4 \pi)^{2}} \sum\limits_{n=0,2,4}^{\infty} \frac{1}{(2+n)} \omega^{n}\right) \frac{(p \cdot q)^{2}}{Q^{2}}. (34)
To get the final answer we have to insert (34), (17), and (20), into (10), which yields
\begin{aligned} & p^{\mu} p^{\nu} T_{\mu \nu}^{(2) g g}(\text { direct })+p^{\mu} p^{\nu} T_{\mu \nu}^{(2) g g}(\text { exchange }) \\ & =8 g^{2} \mathrm{i} T_{\mathrm{f}}\left[2 I_{1}(q, p)+2 I_{1}(-q, p)-2\left(I_{2}(q, p)\right.\right. \\ & \left.\left.\quad+I_{2}(-q, p)+I_{3}(q, p)+I_{3}(-q, p)\right)\right] \\ & =\frac{8}{(4 \pi)^{2}} g^{2} T_{\mathrm{f}} \sum_{n=0,2,4,}\left\{\frac{2}{(n+1)}+\frac{2}{(n+1)}-\frac{4}{(n+2)}\right\} \omega^{n} \frac{(p \cdot q)^{2}}{Q^{2}} \\ & =\frac{32}{(4 \pi)^{2}} g^{2} T_{\mathrm{f}} \sum_{n=0,2,4} \frac{1}{(1+n)(2+n)} \omega^{n}\left(\frac{(p \cdot q)^{2}}{Q^{2}}\right) . & (35) \end{aligned}
Finally we therefore get
\frac{\tilde{F}_{2}}{2 x}=32 \frac{\alpha_{s}}{4 \pi} T_{\mathrm{f}} \cdot \sum\limits_{n=0,2,4}^{\infty} \frac{1}{(1+n)(2+n)} \omega^{n}. (36)
Repeating the steps we did after (5.277), and observing that
\begin{aligned} \frac{1}{(1+n)(2+n)}=\frac{1}{(1+n)}-\frac{1}{(2+n)} & =\int\limits_{0}^{1} \mathrm{~d} x x^{n-1} x-\int\limits_{0}^{1} \mathrm{~d} x x^{n-1} x^{2} \\ & =\int\limits_{0}^{1} \mathrm{~d} x x^{n-1} x(1-x), & (37) \end{aligned}
the desired result emerges as
F_{L}^{g}\left(x, Q^{2}\right)=T_{\mathrm{f}}\left(\frac{\alpha_{s}}{4 \pi}\right) 16 \int\limits_{x}^{1} \frac{\mathrm{d} y}{y}\left(\frac{x}{y}\right)\left(1-\frac{x}{y}\right) F_{2}^{g}(y), (38)
which has been quoted in (5.282).
F_L^g\left(x, Q^2\right)=T_{\mathrm{f}}\left(\frac{\alpha_s}{4 \pi}\right) 16 \int\limits_x^1 \frac{\mathrm{d} y}{y}\left(\frac{x}{y}\right)\left(1-\frac{x}{y}\right) F_2^g(y), (5.282)
