A Calculating the Integral of $\sqrt{\frac{x}{x-a}}$

[felicja]

$\int\sqrt{\frac{x}{x-a}}dx=?$

 

[QuantumQuest]

What have you tried so far?

 

[felicja]

I have a problem with this integral, because there are several versions in the net, so, I'm very surprised... what is going on?
for example:
http://www.physicspages.com/2013/04/05/schwarzschild-metric-radial-coordinate/
and the proposed solution 6) is rather strange.

I can use the wolfram:
http://www.wolframalpha.com/input/?i=int+sqrt(x/(x-a))dx
and the proposed solution is even more ridiculous, strange and very complex.

 

[micromass]

I don't get it. I clicked the wolfram link. What is so strange and ridiculous about the solution?

 

[QuantumQuest]

The differences are due to simplifications and / or different ways to write the same thing.
Have you tried to do the integration? I give you the result if you want to try it out:

$\dfrac{a\left(\ln\left(\sqrt{x-a}+\sqrt{x}\right)-\ln\left(\left|\sqrt{x-a}-\sqrt{x}\right|\right)\right)}{2}+\sqrt{x}\sqrt{x-a}$

 

[felicja]

I easily compute this integral - just make a simple substitute:
$x = a\cosh^2(t)$
then:
$dx = 2a\cosh(t)\sinh(t)$
and
$x-a = a\sinh^2(t)$
so, the integral now is:
$\int cosh^2(t)dt = \int(\cosh(2t)+1)dt = \frac{1}{2}\sinh(2t)+t+C=\sinh(t)\cosh(t)+t+C$

finally:
$I = \sqrt{x(x-a)}+a\cdot arcosh{\sqrt{x/a}}$

 

[micromass]

You need to get back to the variable $x$.

 

[felicja]

So, I don't what is going on.

The proposed solutions are quite nonsensical - what it the reason?

$arcosh(x)=\ln(x+\sqrt{x^2-1})$

 

[micromass]

Can you tell us why they are nonsensical?

 

[felicja]

I showed this already: #3.
And there are much more idiotic versions in the net!

 

[micromass]

The links you posted in #3 are absolutely correct, so I have no idea what you're talking about.

 

[felicja]

Try to compute some definite integral using these 'alternative solutions' then You get it.

For example:
what is a correct distance, means: according to the GR, to the Sun from the Earth?

 

[micromass]

Can you please be more specific?

 

[felicja]

The question: 'what is a distance...', and with a given metric is rather very precise - there is no room for any more specification.

 

[micromass]

You're making no sense, sorry.

 

[felicja]

OK, I sorry.
The result is the same.
$2\ln(\sqrt{x}+\sqrt{x-1})=\ln(\sqrt{x}+\sqrt{x-1})^2=\ln(2\sqrt{x}\sqrt{x-1}+x+x-1)$

But look at this.
The distance with the Schwarzschild metric is equal to:
$s=\sqrt{r(r-a)} + a.arcosh(\sqrt{r/a})$
thus for the case of very big distances: $r >> a$, the distance is approximately:
$s\approx r + a\ln(4r/a)$

thus to the Sun it is some bigger distance, than the simple: r = 150 mln km,
because it's about: $ds = a\ln(4r/a)$ bigger.
a = 3km for the Sun, so this is:
$ds = 3\ln(4*150mln / 3) = 36.6 km$ more.