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Indefinite Integral

jacks

Well-known member
Apr 5, 2012
226
$\displaystyle \int\sqrt{\frac{x^2+1}{x^2-1}}dx$
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Mathematica says this doesn't have a solution in terms of elementary functions for what it's worth.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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
 
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topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
(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
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
(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
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).

CB
 
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chisigma

Well-known member
Feb 13, 2012
1,704
Setting $\displaystyle x= \cos t$ the integral becomes...

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

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

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
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


$\chi$ $\sigma$
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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


$\chi$ $\sigma$
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?

Mommy! My brain hurts. Give me a hug. (Hug)

-Dan
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
$\displaystyle \int\sqrt{\frac{x^2+1}{x^2-1}}dx$
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.

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.
 

chisigma

Well-known member
Feb 13, 2012
1,704
What Opalg told is of course 'all right', so that I have to clarify the 'little mystery'. The integral produce a real function only if is |x|>1 and in that case one have to write...

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

$\chi$ $\sigma$