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When solving for the angular deflection of a light ray passing close to a star in general relativity, one needs to compute an elliptic integral. By some clever factoring and application of a relevant approximation, one reduces the problem to $$\Delta \phi = -2\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du$$ I know (and have verified with Mathematica) that this has exact solution $$\Delta \phi = 2m\sqrt{-ab}+[4+2m(a+b)]\arctan\left(-\sqrt\frac{a}{b}\right),$$ but do not see how to replicate this solution. I've tried reverse engineering this in the case integration by parts was used but can't quite figure it out. Also, is there a complex contour integration approach - perhaps on the dogbone contour? Any hint or point to relevant literature would be appreciated. Thanks.

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    $\begingroup$ Use $u=a+(b-a)\sin^2t$. $\endgroup$
    – J.G.
    Jun 9, 2022 at 21:39
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    $\begingroup$ Typically, complex analysis techniques are only going to work when you have something like $\int_0^\infty$ or $\int_0^{2\pi}$. $\endgroup$ Jun 9, 2022 at 21:58
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    $\begingroup$ In general, integrals with square roots of a single quadratic polynomial, and some other polynomial stuff, are evaluable in elementary terms. $\endgroup$ Jun 9, 2022 at 21:58
  • $\begingroup$ @TedShifrin Not if you use a dogbone contour. $\endgroup$
    – J.G.
    Jun 10, 2022 at 18:07
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    $\begingroup$ @J.G. Agreed. Provided the endpoints of the integral — the doggy bone's ends — are at the branch points (not the case here); and then one must analyze points at infinity on the Riemann surface and apply the Residue Theorem appropriately. $\endgroup$ Jun 10, 2022 at 18:35

2 Answers 2

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Hmm. We have \begin{align*} -2\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du &=2\int_{0}^{a} \frac{1+mu}{\sqrt{(u-a)(b-u)}} \,du \\ &=2\int_{0}^{a} \frac{1+mu}{\sqrt{-(u^2-u(b+a)+ab)}} \,du \\ &=2\int_{0}^{a} \frac{1+mu}{\sqrt{-\left(\left(u-\frac{b+a}{2}\right)^{\!2}-\frac{(b-a)^2}{4}\right)}} \,du \\ &=2\int_{0}^{a} \frac{1+mu}{\sqrt{\frac{(b-a)^2}{4}-\left(u-\frac{b+a}{2}\right)^{\!2}}} \,du. \end{align*} From here, it seems advisable to substitute \begin{align*} v&=u-\frac{b+a}{2}\\ dv&=du \end{align*} to obtain \begin{align*} \Delta\phi &=2\int_{-(b+a)/2}^{(a-b)/2} \frac{1+m(v+(b+a)/2)}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv \\ &=2\int_{-(b+a)/2}^{(a-b)/2} \frac{(1+m(a+b)/2)+mv}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv \\ &=(2+m(a+b))\int_{-(b+a)/2}^{(a-b)/2} \frac{1}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv+2m\int_{-(b+a)/2}^{(a-b)/2} \frac{v}{\sqrt{\frac{(b-a)^2}{4}-v^2}} \,dv. \end{align*} The antiderivative of the first gives you the $\arctan$ function via trigonometric substitution, and the second is the square root function via substitution.

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Substitute $u=\frac{b y^2+a}{1+y^2}$ to remove the square-root in the denominator \begin{align} \Delta \phi =& -2\int_{a}^{0} \frac{1+mu}{\sqrt{u-a} \sqrt{b-u}} \,du\\ =& -4\int_0^{\sqrt{\frac{-a}b}}\frac{(bm+1)y^2+(am+1)}{(1+y^2)^2}dy\\ =&\ \bigg(\frac{2(b-a)my}{1+y^2}-[2(a+b)m+4]\tan^{-1} y\bigg)\bigg|_0^{\sqrt{\frac{-a}b}}\\ = &\ 2m\sqrt{-ab}-[2m(a+b)+4]\tan^{-1}\sqrt{\frac{-a}b} \end{align} (Note that the result cited in the post is incorrect.)

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  • $\begingroup$ How did you dream up that substitution? It's not the least bit obvious how anyone could come to that, and figuring out that that is a good substitution seems to me to be the main portion of the work. The rest is turning the crank. $\endgroup$ Jun 10, 2022 at 17:09
  • $\begingroup$ @AdrianKeister - Agree, not obvious. But, I did enough with $\sqrt{(b-u)(u-a)}$ in the denominator to know this sub would lead to $\arctan$ in its antiderivative, while yours leads to $\arcsin$. $\endgroup$
    – Quanto
    Jun 10, 2022 at 17:15
  • $\begingroup$ If you think there's an error in my derivation, please point out the incorrect step. The "did enough with ...", I think, should be included in your answer. Otherwise, it just looks like the stereotypical, "And then a miracle occurs...", and isn't terribly pedagogically useful. $\endgroup$ Jun 10, 2022 at 17:17
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    $\begingroup$ @AdrianKeister - I was pointing out that different subs lead to different anti’s, all could be correct. Pedagogically, the sub used removes the square root. $\endgroup$
    – Quanto
    Jun 10, 2022 at 17:50
  • $\begingroup$ So you essentially found the substitution that rendered what was under the square root to be a perfect square? If so, that algebra would be good to include in your answer. $\endgroup$ Jun 10, 2022 at 17:51

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