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.
calculus-and-analysis
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up vote
3
down vote
favorite
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
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
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
calculus-and-analysis
asked Nov 23 at 9:49
Schnarco
182
182
add a comment |
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1 Answer
1
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]]
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
add a comment |
1 Answer
1
active
oldest
votes
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]]
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
add a comment |
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]]
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
add a comment |
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]]
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]]
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
add a comment |
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
add a comment |
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