0
|
11828 Views
|
3 Replies
|
3 Total Likes
View groups...
Share
GROUPS:

# Why can't Mathematica solve this integral in definite form?

Posted 11 years ago
 Hi guys, Could you help me out? Why can't Mathematica solve this integral in definite form? It's just stuck computing. Integrate[(1/Sqrt[(x - 1) (x^2 - a^2)]) - (1/ Sqrt[(x + 1) (x^2 - a^2)]), {x, a, b}] ->And it stuck Integrate[(1/Sqrt[(x - 1) (x^2 - a^2)]) - (1/ Sqrt[(x + 1) (x^2 - a^2)]] -> this gets solved Am I doing something wrong? Thanks, M.
3 Replies
Sort By:
Posted 11 years ago
 Even if you use Assumptions->1 b > a > 1] (2*((2*I)*Sqrt[(-1 + a^2)*(-1 + b)] - (2*I)*Sqrt[(-1 + a^2)*(1 + b)] + Sqrt[(1 + a)*(-1 + b^2)]*EllipticF[ArcSin[Sqrt[(a + b)/a]/Sqrt[2]], (2*a)/(-1 + a)] - Sqrt[(-1 + a)*(-1 + b^2)]* EllipticF[ArcSin[Sqrt[(a + b)/a]/Sqrt[2]], (2*a)/(1 + a)] - Sqrt[(1 + a)*(-1 + b^2)]*EllipticK[(2*a)/(-1 + a)] + Sqrt[(-1 + a)*(-1 + b^2)]*EllipticK[(2*a)/(1 + a)]))/ Sqrt[(-1 + a^2)*(-1 + b^2)] 
Posted 11 years ago
 Wow thanks, Marco. Thank you for the detailed answer. This does indeed seem to be the problem (I should have checked that myself).
Posted 11 years ago
 The problem might be the singularity at the lower integration boundary. When x equals a the denominators become zero. You can also see that when you plot the integrand for some values a and b: Plot[(1/Sqrt[(x - 1) (x^2 - a^2)]) - (1/Sqrt[(x + 1) (x^2 - a^2)]) /. a -> 1, {x, 1, 2}] You can also check that with Wolfram alpha:WolframAlpha["Domain of (1/Sqrt[(x - 1) (x^2 - a^2)]) - (1/Sqrt[(x + \ 1) (x^2 - a^2)])", {{"Result", 1}, "Output"}]This gives:HoldComplete[(a < -1 && (-Sqrt[a^2] < x < -1 || x > Sqrt[a^2])) || (-1 <= a <= 1 && x > 1) || (a > 1 && (-Sqrt[a^2] < x < -1 || x > Sqrt[a^2]))]When you solve the indefinite integral: Integrate[(1/Sqrt[(x - 1) (x^2 - a^2)]) - 1/Sqrt[(x + 1) (x^2 - a^2)], x] you obtain (2 I Sqrt[1 - (-1 + a)/(-1 + x)] Sqrt[1 + (1 + a)/(-1 + x)] (-1 + x)^(3/2) Sqrt[-a^2 + x^2] EllipticF[I ArcSinh[Sqrt[1 + a]/Sqrt[-1 + x]](1 - a)/(1 + a)])/(Sqrt[1 + a] Sqrt[1 - a^2 + 2 (-1 + x) + (-1 + x)^2]Sqrt[(-1 + x) (-a^2 + x^2)]) - (2 I (1 + x)^(3/2) Sqrt[-a^2 + x^2] Sqrt[1 + (-1 + a)/(1 + x)] Sqrt[1 - (1 + a)/(1 + x)] EllipticF[I ArcSinh[Sqrt[-1 - a]/Sqrt[1 + x]], (1 - a)/(1 + a)])/(Sqrt[-1 - a] Sqrt[(1 + x) (-a^2 + x^2)] Sqrt[1 - a^2 - 2 (1 + x) + (1 + x)^2]) `which is also clearly not defined at x==1. Cheers, Marco
Community posts can be styled and formatted using the Markdown syntax.