Why can Mathematica solve integral(a)+integral(b), but not integral(a+b)?











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



$$
text{Integrate}left[frac{2 left(3 t epsilon text{Li}_2(t)+3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{t-1}{t}right)+6
epsilon text{Li}_2(-t)-6 epsilon text{Li}_2left(frac{t}{t+1}right)-pi ^2 t^2 epsilon +12 t^2 epsilon +3 t^2+12 t epsilon +3 epsilon
log (1-t) log (t)+3 epsilon log (t) log (t+1)+3 t-3 log (t+1)right)}{3 t (t+1)},{t,0,1}right]
$$



Nevertheless the splitted integral can be solved.



$$
text{Integrate}left[frac{2 left(3 t+3 t^2+12 t epsilon +12 t^2 epsilon -pi ^2 t^2 epsilon right)}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 (3 epsilon log (1-t) log (t)-3 log (1+t)+3 epsilon log (t) log (1+t))}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 left(3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{-1+t}{t}right)+6
epsilon text{Li}_2(-t)+3 t epsilon text{Li}_2(t)-6 epsilon text{Li}_2left(frac{t}{1+t}right)right)}{3 t (1+t)},{t,0,1}right] = epsilon left(-frac{5 zeta (3)}{2}+frac{1}{12} left(-105 zeta (3)-8 log ^3(2)+8 pi ^2 log (2)right)+frac{1}{3} left(24+pi
^2 (log (4)-2)right)+frac{1}{12} pi ^2 log (64)right)-frac{pi ^2}{6}+2+log ^2(2)
$$



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.










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



    $$
    text{Integrate}left[frac{2 left(3 t epsilon text{Li}_2(t)+3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{t-1}{t}right)+6
    epsilon text{Li}_2(-t)-6 epsilon text{Li}_2left(frac{t}{t+1}right)-pi ^2 t^2 epsilon +12 t^2 epsilon +3 t^2+12 t epsilon +3 epsilon
    log (1-t) log (t)+3 epsilon log (t) log (t+1)+3 t-3 log (t+1)right)}{3 t (t+1)},{t,0,1}right]
    $$



    Nevertheless the splitted integral can be solved.



    $$
    text{Integrate}left[frac{2 left(3 t+3 t^2+12 t epsilon +12 t^2 epsilon -pi ^2 t^2 epsilon right)}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 (3 epsilon log (1-t) log (t)-3 log (1+t)+3 epsilon log (t) log (1+t))}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 left(3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{-1+t}{t}right)+6
    epsilon text{Li}_2(-t)+3 t epsilon text{Li}_2(t)-6 epsilon text{Li}_2left(frac{t}{1+t}right)right)}{3 t (1+t)},{t,0,1}right] = epsilon left(-frac{5 zeta (3)}{2}+frac{1}{12} left(-105 zeta (3)-8 log ^3(2)+8 pi ^2 log (2)right)+frac{1}{3} left(24+pi
    ^2 (log (4)-2)right)+frac{1}{12} pi ^2 log (64)right)-frac{pi ^2}{6}+2+log ^2(2)
    $$



    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.










    share|improve this question
























      up vote
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      up vote
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      down vote

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      1





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



      $$
      text{Integrate}left[frac{2 left(3 t epsilon text{Li}_2(t)+3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{t-1}{t}right)+6
      epsilon text{Li}_2(-t)-6 epsilon text{Li}_2left(frac{t}{t+1}right)-pi ^2 t^2 epsilon +12 t^2 epsilon +3 t^2+12 t epsilon +3 epsilon
      log (1-t) log (t)+3 epsilon log (t) log (t+1)+3 t-3 log (t+1)right)}{3 t (t+1)},{t,0,1}right]
      $$



      Nevertheless the splitted integral can be solved.



      $$
      text{Integrate}left[frac{2 left(3 t+3 t^2+12 t epsilon +12 t^2 epsilon -pi ^2 t^2 epsilon right)}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 (3 epsilon log (1-t) log (t)-3 log (1+t)+3 epsilon log (t) log (1+t))}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 left(3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{-1+t}{t}right)+6
      epsilon text{Li}_2(-t)+3 t epsilon text{Li}_2(t)-6 epsilon text{Li}_2left(frac{t}{1+t}right)right)}{3 t (1+t)},{t,0,1}right] = epsilon left(-frac{5 zeta (3)}{2}+frac{1}{12} left(-105 zeta (3)-8 log ^3(2)+8 pi ^2 log (2)right)+frac{1}{3} left(24+pi
      ^2 (log (4)-2)right)+frac{1}{12} pi ^2 log (64)right)-frac{pi ^2}{6}+2+log ^2(2)
      $$



      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.










      share|improve this question













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



      $$
      text{Integrate}left[frac{2 left(3 t epsilon text{Li}_2(t)+3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{t-1}{t}right)+6
      epsilon text{Li}_2(-t)-6 epsilon text{Li}_2left(frac{t}{t+1}right)-pi ^2 t^2 epsilon +12 t^2 epsilon +3 t^2+12 t epsilon +3 epsilon
      log (1-t) log (t)+3 epsilon log (t) log (t+1)+3 t-3 log (t+1)right)}{3 t (t+1)},{t,0,1}right]
      $$



      Nevertheless the splitted integral can be solved.



      $$
      text{Integrate}left[frac{2 left(3 t+3 t^2+12 t epsilon +12 t^2 epsilon -pi ^2 t^2 epsilon right)}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 (3 epsilon log (1-t) log (t)-3 log (1+t)+3 epsilon log (t) log (1+t))}{3 t (1+t)},{t,0,1}right]+text{Integrate}left[frac{2 left(3 epsilon text{Li}_2left(-frac{1}{t}right)-3 epsilon text{Li}_2left(frac{-1+t}{t}right)+6
      epsilon text{Li}_2(-t)+3 t epsilon text{Li}_2(t)-6 epsilon text{Li}_2left(frac{t}{1+t}right)right)}{3 t (1+t)},{t,0,1}right] = epsilon left(-frac{5 zeta (3)}{2}+frac{1}{12} left(-105 zeta (3)-8 log ^3(2)+8 pi ^2 log (2)right)+frac{1}{3} left(24+pi
      ^2 (log (4)-2)right)+frac{1}{12} pi ^2 log (64)right)-frac{pi ^2}{6}+2+log ^2(2)
      $$



      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.







      calculus-and-analysis






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      asked Nov 23 at 9:49









      Schnarco

      182




      182






















          1 Answer
          1






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          oldest

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          up vote
          2
          down vote



          accepted










          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






          share|improve this answer





















          • 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.
            – Schnarco
            Nov 23 at 10:47










          • Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here)
            – Rolf Mertig
            Nov 23 at 11:22











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          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






          share|improve this answer





















          • 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.
            – Schnarco
            Nov 23 at 10:47










          • Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here)
            – Rolf Mertig
            Nov 23 at 11:22















          up vote
          2
          down vote



          accepted










          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






          share|improve this answer





















          • 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.
            – Schnarco
            Nov 23 at 10:47










          • Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here)
            – Rolf Mertig
            Nov 23 at 11:22













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          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






          share|improve this answer












          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







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 23 at 10:26









          Rolf Mertig

          13.6k13366




          13.6k13366












          • 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.
            – Schnarco
            Nov 23 at 10:47










          • Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here)
            – Rolf Mertig
            Nov 23 at 11:22


















          • 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.
            – Schnarco
            Nov 23 at 10:47










          • Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here)
            – Rolf Mertig
            Nov 23 at 11:22
















          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.
          – Schnarco
          Nov 23 at 10:47




          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.
          – Schnarco
          Nov 23 at 10:47












          Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here)
          – Rolf Mertig
          Nov 23 at 11:22




          Integrate2 uses Integrate3 which is basically just a table lookup function (find the implemented list here)
          – Rolf Mertig
          Nov 23 at 11:22


















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