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- Jan 26, 2012

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Mathematica says this doesn't have a solution in terms of elementary functions for what it's worth.

- Aug 30, 2012

- 1,157

Never mind. I was thinking backwards.

-Dan

Since Jameson had already thanked me I should quickly mention what went wrong with my original post.

I was thinking that even if the definite elliptic integral might not have closed form, it was possible that an indefinite elliptic integral might have a closed form solution. On second thought I decided this was wrong.

-Dan

-Dan

Since Jameson had already thanked me I should quickly mention what went wrong with my original post.

I was thinking that even if the definite elliptic integral might not have closed form, it was possible that an indefinite elliptic integral might have a closed form solution. On second thought I decided this was wrong.

-Dan

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- Aug 30, 2012

- 1,157

(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

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

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- Jan 26, 2012

- 890

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

CB

Last edited:

- Feb 13, 2012

- 1,704

$\displaystyle i\ \int \sqrt{1+\cos^{2} t}\ dt = i\ \sqrt{2}\ \text{E}\ (t|\frac{1}{2}) + c$

... where 'E' is an elliptic integral of the second kind and i is the imaginary unit...

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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

$\chi$ $\sigma$

- Aug 30, 2012

- 1,157

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?

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

$\chi$ $\sigma$

Mommy! My brain hurts. Give me a hug.

-Dan

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- Feb 7, 2012

- 2,738

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.$\displaystyle \int\sqrt{\frac{x^2+1}{x^2-1}}dx$

To put it another way, the domain of the integrand (as a real-valued function) excludes the interval [-1,1]. Outside that interval, the integral presumably has some sort of real-valued expression in terms of elliptic functions.

- Feb 13, 2012

- 1,704

$\displaystyle x= \cos t \implies x= \cosh (i\ t) \implies t=-i\ \cosh^{-1} x$ (1)

...and for |x|>1 the final result is...

$\displaystyle \int \sqrt{\frac{x^{2}+1}{x^{2}-1}}\ dx = i\ \sqrt{2}\ \text{E}\ (-i\ \cosh^{-1} x |\frac{1}{2}) + c$ (2)

Kind regards

$\chi$ $\sigma$