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Mathematics Calculus Wolfram Language

Unexpected result of integration of Log function?

Максим Юлдашев

Максим Юлдашев, Tomsk Polytechnic University

Posted 2 years ago

Hello. I need to calculate the integral, but it turns out a strange answer. How to fix this problem? I'm new so don't judge strictly :) Integrate[Log[Sqrt[1 - x] + Sqrt[1 + x^2]], x]

POSTED BY: Максим Юлдашев

3 Replies

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J. M.

Posted 2 years ago

Yes, it looks like one needs elliptic integrals to express it. What kind of answer were you expecting for it?

POSTED BY: J. M.

Rohit Namjoshi

Posted 2 years ago

What version of Mathematica are you running? On "13.1.0 for Mac OS X ARM (64-bit) (June 16, 2022)" I get -x + (I (-Sqrt[1 - x] + Sqrt[1 - x] x^2 - Sqrt[1 + x^2] + x (-2 Sqrt[1 - x] + Sqrt[1 + x^2])) ((-1 - I) Sqrt[2] Sqrt[(-1 + I) (-1 + x)] (1 + x^2) EllipticE[ ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] - 2 Sqrt[2] Sqrt[(-1 + I) (-1 + x)] (1 + x^2) EllipticF[ ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] + (1 + I) Sqrt[2] Sqrt[(-1 + I) (-1 + x)] EllipticPi[1 - I, ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] + (1 + I) Sqrt[2] Sqrt[(-1 + I) (-1 + x)] x^2 EllipticPi[1 - I, ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] + I Sqrt[1 + x^2] Sqrt[1 - x + x^2 - x^3] Log[1 + x]))/(2 (-1 + x) (1 + x^2) (-1 - 2 x + x^2 - Sqrt[1 - x + x^2 - x^3])) + x Log[Sqrt[1 - x] + Sqrt[1 + x^2]]

What version of Mathematica are you running? On "13.1.0 for Mac OS X ARM (64-bit) (June 16, 2022)" I get

-x + (I (-Sqrt[1 - x] + Sqrt[1 - x] x^2 - Sqrt[1 + x^2] + 
      x (-2 Sqrt[1 - x] + Sqrt[1 + x^2])) ((-1 - I) Sqrt[2]
        Sqrt[(-1 + I) (-1 + x)] (1 + x^2) EllipticE[
        ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] - 
      2 Sqrt[2]
        Sqrt[(-1 + I) (-1 + x)] (1 + x^2) EllipticF[
        ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] + (1 + I) Sqrt[2]
        Sqrt[(-1 + I) (-1 + x)]
        EllipticPi[1 - I, ArcSin[Sqrt[1 - I x]/Sqrt[2]], 
        1 + I] + (1 + I) Sqrt[2] Sqrt[(-1 + I) (-1 + x)]
        x^2 EllipticPi[1 - I, ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] + 
      I Sqrt[1 + x^2] Sqrt[1 - x + x^2 - x^3] Log[1 + x]))/(2 (-1 + 
      x) (1 + x^2) (-1 - 2 x + x^2 - Sqrt[1 - x + x^2 - x^3])) + 
 x Log[Sqrt[1 - x] + Sqrt[1 + x^2]]

POSTED BY: Rohit Namjoshi

Hans Dolhaine

Hans Dolhaine, retired

Posted 2 years ago

The stone-age-old version V7 gives the following answer: 1/2 (-2 x - ((-1 - Sqrt[1 - x] Sqrt[1 + x^2] + x (1 - 2 Sqrt[1 - x] Sqrt[1 + x^2] + x (-1 + x + Sqrt[1 - x] Sqrt[1 + x^2]))) ((-1 + I) Sqrt[2] Sqrt[(-1 + I) (-1 + x)] (1 + x^2) (EllipticE[ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] + (1 - I) EllipticF[ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I] - EllipticPi[1 - I, ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I]) + Sqrt[1 + x^2] Sqrt[-(-1 + x) (1 + x^2)] Log[1 + x]))/((-1 + x) (1 + x^2)^( 3/2) (-1 + (-2 + x) x - Sqrt[-(-1 + x) (1 + x^2)])) + 2 x Log[Sqrt[1 - x] + Sqrt[1 + x^2]])

The stone-age-old version V7 gives the following answer:

1/2 (-2 x - ((-1 - Sqrt[1 - x] Sqrt[1 + x^2] + 
x (1 - 2 Sqrt[1 - x] Sqrt[1 + x^2] + 
x (-1 + x + Sqrt[1 - x] Sqrt[1 + x^2]))) ((-1 + I) Sqrt[2]
Sqrt[(-1 + I) (-1 + x)] (1 + 
x^2) (EllipticE[ArcSin[Sqrt[1 - I x]/Sqrt[2]], 
1 + I] + (1 - I) EllipticF[ArcSin[Sqrt[1 - I x]/Sqrt[2]], 
1 + I] - 
EllipticPi[1 - I, ArcSin[Sqrt[1 - I x]/Sqrt[2]], 1 + I]) + 
Sqrt[1 + x^2] Sqrt[-(-1 + x) (1 + x^2)] Log[1 + x]))/((-1 + 
x) (1 + x^2)^(
3/2) (-1 + (-2 + x) x - Sqrt[-(-1 + x) (1 + x^2)])) + 
2 x Log[Sqrt[1 - x] + Sqrt[1 + x^2]])

POSTED BY: Hans Dolhaine

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