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## Main Question or Discussion Point

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",

Help!

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!