# Evaluating definite integral

1. Oct 13, 2014

### voko

I am trying to compute the value of $$\int\limits_{-1}^0 {dx \over \sqrt{\left(x + 1\right)\left(c^2 - x^2\right)}},$$ where $c^2 > 1$. This integral exists because the integrand is positive and $c^2 - x^2 > c^2 - 1 = a^2 > 0$, so $$\int\limits_{-1}^0 {dx\over \sqrt {\left(x + 1\right) \left(c^2 - x^2\right)}} < \int\limits_{-1}^0 {dx \over \sqrt {\left(x + 1\right) a^2}} = {1 \over a} \int\limits_{-1}^0 {dx \over \sqrt {x + 1} } = {2 \over a}$$

While the above gives me a useful estimate, I would like to obtain the exact value as a function of $c$. Unfortunately, my complex analysis technique is very rusty, and so I get absurd results. What I have tried: I set up a closed contour as follows: start at -1 "just below" the real line till 0; then infinitesimal circle about 0, then "just above" the real line till -1, complete the path with infinitesimal circle about -1. As far as I can tell, the only special point inside that contour is at -1. The point is special on two accounts. First, the integrand becomes infinite there; second, it is also a branch point of $\sqrt {z + 1}$.

The first absurdity I get is that because $z = 0$ is not a branch point of the integrand, the integral over "just below", and the integral over "just above", if starting at -1 as indicated above , cancel each out, so I arrive at no useful relationship with the original real-value integral. If I then start at $z = 0$, go "just above" to -1, taking the negative branch of $\sqrt {z + 1}$ there, then, switching to the positive branch after circling around at $z = -1$, the sum of the integral over "just above" with the integral over "just below" is twice the value of the original real-value integral. So far so good, but then another problem strikes: the residue of the integrand at $z = -1$ seems to be zero, giving zero as the end result, which is clearly wrong.

Help!

2. Oct 13, 2014

### zoki85

Uff...this is a nasty integral and can't be evaluated in the closed form of elementar functions.
Where did you get it from?

Last edited: Oct 13, 2014
3. Oct 13, 2014

### voko

Are you saying that the value of the definite integral is not expressible in elementary functions of $c$? Why?

I know that the indefinite integral is not expressible in elementary function of $x$; it is not my purpose to obtain it.

What I really need is an explanation or a hint at the mistake in my attempt.

4. Oct 13, 2014

### zoki85

Yes. For rational values of c=m/n (in the case the integral converges), the result can be expressed as combination of elliptic integrals in terms of c.
Elliptic integrals aren't elementar functions that can be expressed in finite form of elementary funtions (you must use infinite series)

5. Oct 13, 2014

### voko

I am not able to identify this integral with any of complete elliptic integrals listed at http://en.wikipedia.org/wiki/Elliptic_integral, but I guess it may be possible somehow.

Regardless, I would like to understand why my attempt results in the wrong value. Any clues on that?

6. Oct 13, 2014

### Staff: Mentor

Each integrand in the above integrals is undefined at x = -1. It looks to me like you haven't considered that.

7. Oct 13, 2014

### voko

I am not sure what you mean by "each integrand" and "the above integrals" in plural, as there is only one integral in the quotation. Are you also referring to $\int\limits_{-1}^0 {dx \over \sqrt {x + 1}}$ in the inequality? That integral most definitely exists, and equals 2.

8. Oct 13, 2014

### Staff: Mentor

Edit: What I quoted actually does appear now, so I don't understand why it didn't earlier.

I meant to quote this but didn't get all that I intended:
All of the integrals above are improper, because the integrand is undefined at the lower limit of integration. I didn't see anywhere that you had mentioned this or dealt with it (using limits). That was my concern.

I agree that the integral in post #7 exists and has a value of 2.

Last edited: Oct 13, 2014
9. Oct 13, 2014

### voko

I think we need to report the issue to admins because the LaTeX just does not get shown correctly when quoted.

So I think we agree that the (original) integral exists. Any idea where I went wrong afterwards?

10. Oct 13, 2014

### Staff: Mentor

I'll take another look and see if anything pops out at me.

11. Oct 13, 2014

### Staff: Mentor

The quoted LaTeX seems to be working now. Everything I quoted in post #6 is now appearing. Maybe refreshing the page causes it to appear correctly? I don't know, but I'm discussing it with Greg now.

12. Oct 13, 2014

### voko

I think I tried to refresh the page back then, and LaTeX was not displayed correctly. Now it is all good.

13. Oct 13, 2014

### Staff: Mentor

Greg said that LaTeX was breaking the "click to expand" capability, so he has disabled it. I'll take another look at your integrals in a couple of hours.

14. Oct 13, 2014

### zoki85

http://www.freesmileys.org/smileys/smiley-basic/popcorn.gif
We are waiting in suspense :D

15. Oct 13, 2014

### Staff: Mentor

This one is beyond my capabilities. I tried wolframalpha with the original integral, and it said that the computation was taking too long, but if I paid some \$, I could try again. Next, I replaced c2 with 4, and got what you see here: http://www.wolframalpha.com/input/?i=int 1/sqrt((x + 1)(4 - x^2)) dx, x=-1..0. WA reports that this is an elliptic integral of the first kind.

WA says that the value of my modified integral is about 1.07826.

16. Oct 13, 2014

### voko

Thanks Mark44. I have by now established the connection of this integral with the standard elliptic integrals.

However, I would still like to understand where I went wrong in my attempt. I understand that the attempt could not have resulted in any nice elementary formula, but it should not give random bogus results, either. Should I perhaps re-post this in the homework forums, where it could get more attention?

17. Oct 14, 2014

### Char. Limit

I replaced c^2 with various small square numbers, just to see if there was any sort of pattern. There... wasn't one, though it was interesting that 4 and 9 generated a result using only elliptic integrals of the first kind, while anything above that started throwing the imaginary unit and elliptic integrals of the second kind into the result. I'm not sure how you would get a result for arbitary c^2 > 1, but there /does/ seem to be a result out there.

That said, there's always a term involving the first elliptic integral of the arcsin of something. Maybe that'll help?

18. Oct 14, 2014

### Staff: Mentor

I don't know if it will get any more attention in the HW sections...