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For my physics research, I need to do some series expansions of complete elliptic integrals of the first kind. When I tell Mathematica to approximate it, I get...

$$\int_0^{\pi/2}d\theta \frac{1}{\sqrt{x^2 + \sin^2(\theta)}} = \frac{1}{2} \ln \frac{16}{x} + \frac{1}{8}x \ln \frac{x}{8} + x + \mathcal{O}(x^2)$$

I have no idea how Mathematica does this, and I'm going to need to figure this out because I have more complicated elliptic integrals coming down the line, and Mathematica gives up on those. I've tried to fiddle around with Taylor series-ing the integral for small $x$, but that doesn't work. I always get integrals like $\int_0^{\pi/2} \csc^n(\theta) d\theta$, and none of those converge. I believe the problem is, for any small but nonzero $x$, there'll always be a $\theta$ such that $\sin(\theta) \ll x$ on the interval I'm integrating over, and that screws up any series expansion of the integrand I make in $x$.

Any advice on good approximation schemes that would give me those logs, or any book chapters that might be helpful, would be greatly appreciated! Thanks so much.

Gary
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Ben Eustis-Guthrie
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2 Answers2

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Your integral $$f(x)=\int_0^{\pi/2}\frac{dt}{\sqrt{x^2+\sin^2t}}\tag{1}$$ equals $$\int_0^{\pi/2}\frac{dt}{\sqrt{1+x^{2}-\cos^2t}}=\frac {1}{\sqrt{1+x^2}} \int_0^{\pi/2}\frac{dt}{\sqrt{1-(1+x^2)^{-1}\cos^2t}} $$ And this is nothing but $$f(x) =(1+x^2)^{-1/2}K(k')\tag{2} $$ where $$k'=\sqrt{1-k^2}=(1+x^2)^{-1/2},k=x(1+x^2)^{-1/2}\tag{3}$$ When $x\to 0$ we have $k\to 0$ and we have the well known formula $$K(k') = \frac{2K(k)}{\pi}\log\left(\frac{4}{k}\right) - 2\left[\left(\frac{1}{2}\right)^{2}\left(\frac{1}{1\cdot 2}\right)k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}\left(\frac{1}{1\cdot 2} + \frac{1}{3\cdot 4}\right)k^{4} + \cdots\right]\tag{4}$$ Using $$\frac{2K(k)}{\pi}=1+\left(\frac{1}{2}\right)^2k^2+\left(\frac{1\cdot 3}{2\cdot 4}\right)^2k^4+\dots\tag{5}$$ together with $(3)$ in equation $(4)$ we can get the desired approximation given in question. We also need to use a bit of binomial theorem for general index to deal with $(1+x^2)^{-1/2}$ appearing in $(2),(3)$.

Paramanand Singh
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$$\int \frac{d\theta}{\sqrt{x^2 + \sin^2(\theta)}} =\frac 1x F\left(\theta \left|-\frac{1}{x^2}\right.\right)$$ $$\int_0^{\frac \pi 2} \frac{d\theta}{\sqrt{x^2 + \sin^2(\theta)}} =\frac 1x K\left(-\frac{1}{x^2}\right) $$

Expanding the result as series for small values of $x$, you should have $$\frac 1x K\left(-\frac{1}{x^2}\right)=\sum_{n=0}^\infty P_n\, x^{2n}$$ where, without simplifications, the $P_n$ are

$$\left( \begin{array}{cc} n & P_n \\ 0 & -\log (x)+2 \log (2) \\ 1 & -\frac{1}{4} (-\log (x)+2 \log (2)-1) \\ 2 & -\frac{3}{128} (-6 \log (x)+12 \log (2)-7) \\ 3 & -\frac{5}{1536} (-30 \log (x)+60 \log (2)-37) \\ \end{array} \right)$$

Truncated to this level, the approximation is good for $0\leq x \leq 1$.

Claude Leibovici
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