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I have this integral $$\int\frac{dz}{\sqrt{(z^{2}-\rho^{2})(\lambda^{2} - z^{2})}}$$ and parameters obey the following conditions

$$z= \exp[k\varphi],$$ $$\lambda^{2} = \frac{b + \sqrt{b^{2} - 4ac}}{-2a} > 0,$$ $$\rho^{2} = \frac{b- \sqrt{b^{2} - 4ac}}{-2a} > 0,$$ $ \rho^{2} < \lambda^{2}$, $\rho \leq z \leq \lambda$, $a <0$, $c<0$, $b>0$.

I have tried to evaluate the integral using the elliptic integral of the first kind, the obtained result is complex.

Is there a way to find an analytical (real-value) solution to the integral?

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1 Answer

You can in fact bring it to the elliptic integral of the first kind (without complex numbers) $$F(\phi,k)=\int_0^{\sin\phi} \frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}}.\qquad(1)$$

Starting with ($0<\rho< z<\lambda$) $$I=\int\frac{dz}{\sqrt{(z^{2}-\rho^{2})(\lambda^{2} - z^{2})}}$$ You proceed via a Möbius transformation $z = \sqrt{\lambda \rho}(x - 1)/(x + 1)$. The integral then reads $$I=\int \frac{2 \sqrt{\lambda \rho} \,dx}{\sqrt{(A_- -A_+ x^2)(A_+-A_-x^2)}}$$ where $A_+=(\lambda+\sqrt{\lambda \rho})(\rho+\sqrt{\lambda \rho})$ and $A_- =(\lambda-\sqrt{\lambda \rho})(\sqrt{\lambda \rho}-\rho)$.

Making the substitution $x=t \sqrt{A_-/A_+}$ brings the integral finally in the desired form: $$I=\int\frac{2\sqrt{\lambda\rho}\, dt}{A_+\sqrt{(1-t^2)(1-k^2 t^2)} }$$ with $k=A_-/A_+<1$.

Using the definition (1), we obtain $$I=\frac{2\sqrt{\lambda \rho}}{A_+}F(\arcsin t,k)$$ with $$t=\sqrt{\frac{A_+}{A_-}}\frac{z-\sqrt{\lambda\rho}}{z+\sqrt{\lambda\rho}}$$

More information can be found here.

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Thanks for your answer! It is exactly what I have been looking for. However let me note that the transformation should be $z= \sqrt{\lambda\rho}\frac{x-1}{x+1}$. If $z = \sqrt{\lambda\rho}\frac{x+1}{x- 1}$, then $dz = -2\sqrt{\lambda \rho}\frac{1}{(x-1)^2}dx$ and one obtains the integral with sign $"-"$. –  Anastasia Krotoff Jan 19 '12 at 0:03
    
@AnastasiaKrotoff: as far as I understand, either of the two transformations does the job (I agree that I might have lost a minus sign along the way). But if you like your version better then I will change the answer accordingly... –  Fabian Jan 19 '12 at 7:14
    
@Anastasia: Any appropriate Möbius substitution should do the job. IIRC Byrd and Friedman take the route given by Fabian, but they mention that your route is also suitable. –  J. M. Feb 8 '12 at 13:09
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