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If the indefinite integral has a closed form in terms of elementary functions, so does any definite integral with the same integrand (since the definite integral is just the difference of the indefinite evaluated at the end point on the interval of integration).(sigh) I can't get this thing out of my mind and I keep thinking in circles. This'll be my last post on the topic. I promise!
Okay, to be specific about the beef I originally had with this problem is that Mathematica says this is an elliptic integral of the second kind. But elliptic integrals are definite integrals. Is it possible that this indefinite integral might have a closed form solution even though the definite integral does not.
-Dan
The obvious implication is that the elliptic integral must be imaginary. But by its definition I don't think that the elliptic integral can be anything but real?That's very interesting the results supplied by 'Monster Wolfram'...
Wolfram Mathematica Online Integrator
Apart the 'odd identities'...
$\displaystyle \frac{\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}=1$
$\displaystyle \frac{\sqrt{1-x^{2}}}{\sqrt{x^{2}-1}}=i$
... the result seems to be...
$\displaystyle \int \sqrt{\frac{x^{2}+1}{x^{2}-1}}\ dx = i\ \text{E}\ (\sin^{-1} x|-1) + c $ (1)
... and also in this case the imaginary unit appears... what's Your opinion about that?...
Kind regards
$\chi$ $\sigma$
Notice that the denominator of the fraction is negative unless $|x|>1$. So you should only expect to get a real value for the integral if you avoid the interval [-1,1]. In particular, the substitution $x=\cos t$ implicitly assumes that $|x|\leqslant1$ and hence inevitably leads to a non-real answer.$\displaystyle \int\sqrt{\frac{x^2+1}{x^2-1}}dx$