3
$\begingroup$

I am new to Mathematica and came across the following problem. The integral at hand cannot be solved.

$$ \text{Integrate}\left[\frac{2 \left(3 t \epsilon \text{Li}_2(t)+3 \epsilon \text{Li}_2\left(-\frac{1}{t}\right)-3 \epsilon \text{Li}_2\left(\frac{t-1}{t}\right)+6 \epsilon \text{Li}_2(-t)-6 \epsilon \text{Li}_2\left(\frac{t}{t+1}\right)-\pi ^2 t^2 \epsilon +12 t^2 \epsilon +3 t^2+12 t \epsilon +3 \epsilon \log (1-t) \log (t)+3 \epsilon \log (t) \log (t+1)+3 t-3 \log (t+1)\right)}{3 t (t+1)},\{t,0,1\}\right] $$

Nevertheless the splitted integral can be solved.

$$ \text{Integrate}\left[\frac{2 \left(3 t+3 t^2+12 t \epsilon +12 t^2 \epsilon -\pi ^2 t^2 \epsilon \right)}{3 t (1+t)},\{t,0,1\}\right]+\text{Integrate}\left[\frac{2 (3 \epsilon \log (1-t) \log (t)-3 \log (1+t)+3 \epsilon \log (t) \log (1+t))}{3 t (1+t)},\{t,0,1\}\right]+\text{Integrate}\left[\frac{2 \left(3 \epsilon \text{Li}_2\left(-\frac{1}{t}\right)-3 \epsilon \text{Li}_2\left(\frac{-1+t}{t}\right)+6 \epsilon \text{Li}_2(-t)+3 t \epsilon \text{Li}_2(t)-6 \epsilon \text{Li}_2\left(\frac{t}{1+t}\right)\right)}{3 t (1+t)},\{t,0,1\}\right] = \epsilon \left(-\frac{5 \zeta (3)}{2}+\frac{1}{12} \left(-105 \zeta (3)-8 \log ^3(2)+8 \pi ^2 \log (2)\right)+\frac{1}{3} \left(24+\pi ^2 (\log (4)-2)\right)+\frac{1}{12} \pi ^2 \log (64)\right)-\frac{\pi ^2}{6}+2+\log ^2(2) $$

What ist the reason for this issue? I thought that Mathematica tries to solve as much as possible and gives the unsolved parts as an integral.

$\endgroup$
3
$\begingroup$

This is a known problem with the standard Mathematica Integrate function. Therefore I wrote (a long long time ago) Integrate2 in FeynCalc ( a package for High Energy Physics which you can easily install from http://www.feyncalc.org) :

Needs["FeynCalc`"]; 
AbsoluteTiming[
   li2 = PolyLog[2, #1] & ; 
   int = 2*((3*t*e*li2[t] + 3*e*li2[-t^(-1)] - 
       3*e*li2[(t - 1)/t] + 6*e*li2[-t] - 
       6*e*li2[t/(t + 1)] - Pi^2*t^2*e + 12*t^2*e + 
       3*t^2 + 12*t*e + 3*e*Log[1 - t]*Log[t] + 
       3*e*Log[t]*Log[t + 1] + 3*t - 3*Log[t + 1])/
      (3*t*(t + 1))); Collect[
    Integrate2[int, {t, 0, 1}] /. Zeta2 -> Zeta[2], e]]

enter image description here

$\endgroup$
  • $\begingroup$ Thanks a lot. I used FeynCalc some time ago but was not aware that it also provides improved integration routines. It is much faster than the standard function. $\endgroup$ – Schnarco Nov 23 '18 at 10:47
  • $\begingroup$ Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here) $\endgroup$ – Rolf Mertig Nov 23 '18 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.