# Evaluating an indefinite integral at the boundaries

I have the following expression i need to integrate over $\alpha$ from 0 to 1 and $x$ from $x1$ to $\infty$ $$\frac{1}{2p1x(-p2(1-x1)x1(1-\alpha)+kp2(1-\alpha)\alpha+x^{2}\alpha\zeta2+\alpha\lambda2)}$$ where $\alpha$ is a Feynman parameter and $\lambda2,-p2\ll1$ and $kp2,\zeta2>0$. I first need to perform the $\alpha$ integration. In order to simplify this integration i apply a partial fraction decomposition and expand the resulting square roots to leading order in $p2$ and $\lambda2$. This yields $$\frac{kp2}{2p1x(-p2\left(1-x1\right)x1+kp2\alpha)\left(kp2-p2\left(1-x1\right)x1+x^{2}\zeta2+\lambda2\right)}-\frac{kp2}{2p1x\left(kp2\left(\alpha-1\right)-x^{2}\zeta2-\lambda2\right)\left(kp2-p2\left(1-x1\right)x1+x^{2}\zeta2+\lambda2\right)}$$ Performing an indefinite integration over $\alpha$ and evaluating the resulting expression at the boundaries gives $$\frac{-\log\left(\left(-p2\left(1-x1\right)x1\right)\left(x^{2}\zeta2+\lambda2\right)\right)}{2p1x\left(kp2-p2\left(1-x1\right)x1+x^{2}\zeta2+\lambda2\right)}+\frac{\log\left(\left(kp2-p2\left(1-x1\right)x1\right)\left(kp2+x^{2}\zeta2+\lambda2\right)\right)}{2p1x\left(kp2-p2\left(1-x1\right)x1+x^{2}\zeta2+\lambda2\right)}$$ I'm pretty sure that the integral over this expression over $x$ from $x1$ to $\infty$ does converge, since it goes like $\frac{\log\left(x^{2}\right)}{x^{3}}$. In order to perform the $x$ integration i apply Apart[expr,x] to the above expression and write the resulting terms as a list using Level[expr,1]. The indefinite $x$ integral of this expression is then, after summing the elements of the list using Total[expr], given by

(1/(4 p1 (kp2-p2 x1+p2 x1^2+λ2)))(2 Log[x] Log[1+(x^2 ζ2)/λ2]-2 Log[x] Log[p2 (-1+x1) x1 (x^2 ζ2+λ2)]+2 Log[x] Log[(kp2+p2 (-1+x1) x1) (kp2+x^2 ζ2+λ2)]-2 Log[x] Log[(kp2+x^2 ζ2+λ2)/(kp2+λ2)]-Log[(kp2+p2 (-1+x1) x1) (kp2+x^2 ζ2+λ2)] Log[(kp2+p2 (-1+x1) x1+x^2 ζ2+λ2)/(p2 (-1+x1) x1)]+Log[p2 (-1+x1) x1 (x^2 ζ2+λ2)] Log[(kp2+p2 (-1+x1) x1+x^2 ζ2+λ2)/(kp2+p2 (-1+x1) x1)]+PolyLog[2,-((x^2 ζ2)/λ2)]-PolyLog[2,-((x^2 ζ2)/(kp2+λ2))]+PolyLog[2,-((x^2 ζ2+λ2)/(kp2+p2 (-1+x1) x1))]-PolyLog[2,(kp2+x^2 ζ2+λ2)/(p2 x1-p2 x1^2)])


When I now try to evaluate this expression at the boundaries I have no problems with the lower boundary $x1$. However, I don't have any idea how to take the limit of $x$ to $\infty$ since the above expression diverges in this limit. What did I do wrong? Does anyone know how I could integrate my initial expression over $\alpha$ first and then over $x$ and then get a finite result? Thanks in advance, any help is welcome!

Here's my code of the above calculation, I didn't have it at hand when I asked the question

(* Integration of the Feynman parameter α *)

\$Assumptions =
kp2 >= 0 && -1 < p2 < 0 && 1 > λ2 > 0 && x1 >= x >= 0 &&
1 >= α >= 0 && p1 >= 0 && ζ2 >= 0

 kp2 >= 0 && -1 < p2 < 0 && 1 > λ2 > 0 && x1 >= x >= 0 &&
1 >= α >= 0 && p1 >= 0 && ζ2 >= 0

Denominator5 :=
2 p1*x (-kp2*α (α - 1) +
p2*x1 (α - 1) (1 - x1) + λ2*α + ζ2*
x^2*α)

Denominator5PartialFraction :=
2 p1*x*kp2*(-α + (kp2 + x^2 ζ2 + λ2)/
kp2) (α - (p2 (1 - x1) x1)/kp2)

StartingExpressionThirdTerm1 = 1/Denominator5

1/(2 p1 x (p2 (1 - x1) x1 (-1 + α) - kp2 (-1 + α) α +
x^2 α ζ2 + α λ2))

(*The same partial fraction decomposition as before yields*)

StartingExpressionThirdTerm2 =
Apart[1/Denominator5PartialFraction, α]

 kp2/(2 p1 x (-p2 x1 + p2 x1^2 + kp2 α) (kp2 - p2 x1 + p2 x1^2 +
x^2 ζ2 + λ2)) - kp2/( 2 p1 x (-kp2 + kp2 α - x^2 ζ2 - λ2) (kp2 - p2 x1 +
p2 x1^2 + x^2 ζ2 + λ2))

ResultThirdTermα =
Integrate[Level[StartingExpressionThirdTerm2, 1], α]

 {Log[p2 (-1 + x1) x1 + kp2 α]/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)), -(
Log[kp2 - kp2 α + x^2 ζ2 + λ2]/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)))}

(*Now evaluate this expression at the limits*)

ResultThirdTermαUpper =
Limit[ResultThirdTermα, α -> 1, Direction -> 1]

{Log[kp2 + p2 (-1 + x1) x1]/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)), -(
Log[x^2 ζ2 + λ2]/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)))}

ResultThirdTermαLower =
Limit[ResultThirdTermα, α -> 0, Direction -> -1]

 {Log[p2 (-1 + x1) x1]/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)), -(
Log[kp2 + x^2 ζ2 + λ2]/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)))}

 FinalResultThirdTermα =
FullSimplify[
Total[ResultThirdTermαUpper - ResultThirdTermαLower]]

 (-Log[p2 (-1 + x1) x1] +
Log[((kp2 + p2 (-1 + x1) x1) (kp2 + x^2 ζ2 + λ2))/(
x^2 ζ2 + λ2)])/(2 p1 x (kp2 + p2 (-1 + x1) x1 +
x^2 ζ2 + λ2))

 FinalResultThirdTermαList =
FullSimplify[ResultThirdTermαUpper - ResultThirdTermαLower]

 {(-Log[p2 (-1 + x1) x1] + Log[kp2 + p2 (-1 + x1) x1])/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)),
Log[1 + kp2/(x^2 ζ2 + λ2)]/(
2 p1 x (kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2))}

(*Before performing the x integration bring this to a different form*)

FinalResultThirdTermαSimplified1 =
FinalResultThirdTermα /. {Log[((kp2 + p2 (-1 + x1) x1) (kp2 +
x^2 ζ2 + λ2))/(x^2 ζ2 + λ2)] ->
Log[(kp2 + p2 (-1 + x1) x1) (kp2 + x^2 ζ2 + λ2)],
Log[p2 (-1 + x1) x1] -> Log[p2 (-1 + x1) x1 (x^2 ζ2 + λ2)]}

(-Log[p2 (-1 + x1) x1 (x^2 ζ2 + λ2)] +
Log[(kp2 + p2 (-1 + x1) x1) (kp2 +
x^2 ζ2 + λ2)])/(2 p1 x (kp2 + p2 (-1 + x1) x1 +
x^2 ζ2 + λ2))

Integration of the momentum fraction x

FinalResultThirdTermαPartialFraction =
Level[Apart[FinalResultThirdTermαSimplified1, x], 1]

 {(x ζ2 (Log[p2 (-1 + x1) x1 (x^2 ζ2 + λ2)] -
Log[(kp2 + p2 (-1 + x1) x1) (kp2 + x^2 ζ2 + λ2)]))/(
2 p1 (kp2 - p2 x1 + p2 x1^2 + λ2) (kp2 - p2 x1 + p2 x1^2 +
x^2 ζ2 + λ2)), (-Log[
p2 (-1 + x1) x1 (x^2 ζ2 + λ2)] +
Log[(kp2 + p2 (-1 + x1) x1) (kp2 + x^2 ζ2 + λ2)])/(
2 p1 x (kp2 - p2 x1 + p2 x1^2 + λ2))}

 ResultThirdTermX =
Together[Total[Integrate[FinalResultThirdTermαPartialFraction, x]]]

(1/(4 p1 (kp2 - p2 x1 +
p2 x1^2 + λ2)))(2 Log[x] Log[1 + (x^2 ζ2)/λ2] -
2 Log[x] Log[p2 (-1 + x1) x1 (x^2 ζ2 + λ2)] +
2 Log[x] Log[(kp2 + p2 (-1 + x1) x1) (kp2 + x^2 ζ2 + λ2)] -
2 Log[x] Log[(kp2 + x^2 ζ2 + λ2)/(kp2 + λ2)] -
Log[(kp2 + p2 (-1 + x1) x1) (kp2 + x^2 ζ2 + λ2)] Log[(
kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)/(p2 (-1 + x1) x1)] +
Log[p2 (-1 + x1) x1 (x^2 ζ2 + λ2)] Log[(
kp2 + p2 (-1 + x1) x1 + x^2 ζ2 + λ2)/(
kp2 + p2 (-1 + x1) x1)] + PolyLog[2, -((x^2 ζ2)/λ2)] -
PolyLog[2, -((x^2 ζ2)/(kp2 + λ2))] +
PolyLog[2, -((x^2 ζ2 + λ2)/(kp2 + p2 (-1 + x1) x1))] -
PolyLog[2, (kp2 + x^2 ζ2 + λ2)/(p2 x1 - p2 x1^2)])

(*Now evaluate this expression at the boundaries*)

ResultThirdTermXLower = Limit[ResultThirdTermX, x -> x1, Direction -> -1]

 (1/(4 p1 (kp2 +
p2 (-1 + x1) x1 + λ2)))(2 Log[x1] Log[
1 + (x1^2 ζ2)/λ2] -
2 Log[x1] Log[p2 (-1 + x1) x1 (x1^2 ζ2 + λ2)] +
2 Log[x1] Log[(kp2 + p2 (-1 + x1) x1) (kp2 + x1^2 ζ2 + λ2)] -
2 Log[x1] Log[(kp2 + x1^2 ζ2 + λ2)/(kp2 + λ2)] -
Log[(kp2 + p2 (-1 + x1) x1) (kp2 + x1^2 ζ2 + λ2)] Log[(
kp2 + p2 (-1 + x1) x1 + x1^2 ζ2 + λ2)/(p2 (-1 + x1) x1)] +
Log[p2 (-1 + x1) x1 (x1^2 ζ2 + λ2)] Log[(
kp2 + p2 (-1 + x1) x1 + x1^2 ζ2 + λ2)/(
kp2 + p2 (-1 + x1) x1)] + PolyLog[2, -((x1^2 ζ2)/λ2)] -
PolyLog[2, -((x1^2 ζ2)/(kp2 + λ2))] +
PolyLog[2, -((x1^2 ζ2 + λ2)/(kp2 + p2 (-1 + x1) x1))] -
PolyLog[2, (kp2 + x1^2 ζ2 + λ2)/(p2 x1 - p2 x1^2)])

 ResultThirdTermXUpper = Limit[ResultThirdTermX, x -> Infinity, Direction -> 1]

 ComplexInfinity

-
Please, include any relevant code. –  Sektor Jun 11 '14 at 23:00