One-loop massless pentagon

The one-loop scalar pentagon with all five propagators and all five external legs massless. Its expansion through the finite part, a result of Bern, Dixon and Kosower (1994), is recomputed here without using the published answer, together with the symbol of its parity-odd part, which first appears one order higher.

The integral

All five momenta are incoming, with $p_i^2=0$ and $\sum_i p_i=0$. Writing $l$ for the loop momentum and $q_1=l$, $q_k=l+p_1+\dots+p_{k-1}$ for the propagator momenta, the integral in $d=4-2\varepsilon$ dimensions is

$$I_5(v_1,\dots,v_5)=\int\frac{d^{4-2\varepsilon}l}{i\pi^{2-\varepsilon}}\,\prod_{k=1}^{5}\frac{1}{q_k^2}\,,$$

with no factor of $e^{\gamma_E\varepsilon}$ in the measure. $I_5$ depends only on the five adjacent invariants

$$v_i=s_{i,i+1}=(p_i+p_{i+1})^2,\qquad i \text{ read mod } 5.$$

The quantity computed is the Laurent expansion of $I_5$ in $\varepsilon$ through the finite $\varepsilon^0$ coefficient, in the Euclidean region $v_i\lt0$ where it is real. At order $\varepsilon$ only the symbolthe iterated-integral skeleton of a polylogarithmic function: a sum of words in a finite set of letters (algebraic functions of the kinematics) that encodes its logarithmic branch structure and fixes it up to constants of the parity-odd part is computed: the part that changes sign under $\sqrt{\Delta_5}\to-\sqrt{\Delta_5}$, with $\Delta_5$ the Gram determinant of the external momenta.

Bern, Dixon and Kosower (1994) showed that in dimensional regularization the pentagon reduces to a sum over its five one-mass boxesthe box integrals obtained by contracting one of the five propagators, so that two neighboring legs merge into one leg of nonzero invariant mass plus a remainder proportional to $d-4$, and gave the result through order $\varepsilon^0$ in that form. The remainder first contributes at order $\varepsilon$, through the six-dimensional pentagon, which is finite and known in closed form as a one-dimensional integral from the work of Del Duca, Duhr, Glover and Smirnov (2010), Kniehl and Tarasov (2010) and Kozlov and Lee (2016).

The lettersthe finite list of algebraic functions of the kinematics whose logarithms build the answer as iterated integrals are organized by the cyclic symmetry $v_j\to v_{j+1}$. Fifteen rational letters, all linear in the invariants, form three orbits of five: the adjacent invariants $v_i$, the next-to-adjacent invariants $s_{i,i+2}=v_{i+3}-v_i-v_{i+1}$, and the differences $v_{i+2}-v_i$ that appear in the numerators of the dilogarithm arguments. The parity-odd letters, which go to their inverses when the sign of $\sqrt{\Delta_5}$ is flipped, are built on the five-point Gram determinant $\Delta_5=\det(2\,p_i\!\cdot\!p_j)_{i,j=1}^{4}$: five ratios $W_i=(a_i-\sqrt{\Delta_5})/(a_i+\sqrt{\Delta_5})$ with $a_i$ quadratic in the $v_j$. Counting $\Delta_5$ itself, the alphabet of the one-loop family has 21 letters. Through order $\varepsilon^0$ only the rational letters occur: with the first-entry conditionthe requirement that the first letter of every word be a Mandelstam invariant, so that branch cuts start only at physical thresholds and integrabilitythe condition that a word in the letters is the symbol of an actual function, i.e. that mixed partial derivatives commute imposed, no parity-odd function exists below weight three (three iterated logarithmic integrations), so $\sqrt{\Delta_5}$ first enters at order $\varepsilon$. At two loops the planar pentagon functions require the larger 26-letter alphabet of Gehrmann, Henn and Lo Presti (2016) and Chicherin, Henn and Mitev (2018).

Closed form

With $I_4^{(i)}$ the one-mass box obtained by contracting propagator $i$, the reduction of Bern, Dixon and Kosower reads

$$I_5=-\tfrac12\sum_{i=1}^{5}c_i\,I_4^{(i)}+(d-4)\,(\cdots),\qquad c_i=\sum_{j}(S^{-1})_{ij},\quad S_{ij}=-\tfrac12\,(q_i-q_j)^2 .$$

The nonzero entries of $S$ are $-\tfrac12 v_k$ (adjacent $q_i-q_j$ are lightlike), so the coefficients $c_i$ are rational functions of the $v_j$. Inserting the one-mass box in $d$ dimensions gives, through order $\varepsilon^0$, the Bern–Dixon–Kosower form

$$I_5=r_\Gamma\sum_{j=1}^{5}\frac{1}{v_{j-2}\,v_{j-1}\,v_j}\left[\frac{(-v_{j-2})(-v_{j-1})(-v_j)}{(-v_{j+1})(-v_{j+2})}\right]^{-\varepsilon}\left[\frac{1}{\varepsilon^2}+2\,\mathrm{Li}_2\!\Big(1-\tfrac{v_{j+3}}{v_{j+1}}\Big)+2\,\mathrm{Li}_2\!\Big(1-\tfrac{v_j}{v_{j+2}}\Big)-\frac{\pi^2}{6}\right]+O(\varepsilon),\qquad r_\Gamma=\frac{\Gamma(1+\varepsilon)\Gamma^2(1-\varepsilon)}{\Gamma(1-2\varepsilon)}$$

with indices read mod 5 and $\mathrm{Li}_2$ the dilogarithm. For $v_i\lt0$ every ratio inside the brackets is positive, every dilogarithm argument is real and less than one, and $I_5$ is real. The double pole is $\sum_j 1/(v_{j-2}v_{j-1}v_j)$, negative throughout the Euclidean region. Because the five prefactors $1/(v_{j-2}v_{j-1}v_j)$ are distinct rational functions, $I_5$ is not an algebraic factor times a single pure function of weight two: no single rational or algebraic normalization turns its finite part into a weight-two function with rational-number coefficients. Within the span of the five box functions $B_i=\mathrm{Li}_2(1-v_i/v_{i+2})+\mathrm{Li}_2(1-v_i/v_{i+3})+\tfrac12\log^2(v_{i+2}/v_{i+3})$, cyclic symmetry allows only the unit-coefficient sum $\sum_i B_i$, in which the weight-one parts of the boxes cancel because $\sum_i\log(v_{i+2}/v_{i+3})=0$, and that sum differs from the finite part of the pentagon in every normalization. At the symmetric point $v_i=-1$ both dilogarithm arguments and the logarithm in each $B_i$ vanish, so every $B_i$, and their sum, is zero there, a check that can be done by hand.

Order $\varepsilon^1$

The order-$\varepsilon$ coefficient of $I_5$ is the sum of the boxes' order-$\varepsilon$ terms, each weighted by $-\tfrac12 c_i$, plus the $(d-4)$ remainder, proportional at this order to the six-dimensional pentagon $I_5^{(d=6)}$ at $\varepsilon=0$. Over the 21 letters, the weight-three symbols that satisfy the first-entry condition and integrability and are odd under parity span a space of dimension 100, and exactly one combination in it is invariant under the cyclic shift $v_j\to v_{j+1}$. Up to normalization that combination is the symbol of $\sqrt{\Delta_5}\,I_5^{(d=6)}$; its 40 words have coefficients $\pm1$ and each ends in one of the parity-odd letters $W_i$.

Checks

Evaluator

Same family

The other polylogarithmic integrals without internal masses computed by the same method are the two-loop Sudakov form factor, a single-scale vertex whose closed form is exact in $d$; the three-loop ladder, a weight-six function of two ratios of external invariants; and the massless planar double box, a two-letter function of the ratio of its two Mandelstam invariants. The other five-point integral is the non-planar hexa-box with two massive legs, a two-loop topology for diboson-plus-jet production.

The paper

• Polylogarithmic, elliptic, and Calabi–Yau Feynman integrals from a hybrid bootstrap (PDF) — Matthew D. Schwartz.

References

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