I am trying to find a two term asymptotic expansion of the following elliptic integrals of first and second kind as $m\to 0$.

$$\int_{0}^{\pi/2} \frac{1}{\sqrt{1-m^2 \sin^2\theta}} d\theta$$

$$\int_{0}^{\pi/2} {\sqrt{1-m^2 \sin^2\theta}} d\theta$$

Using the local and global contributions, one can split the integral limits as $\int_{0}^{\pi/2}= \int_{0}^{\lambda} + \int_{\lambda}^{\pi/2}$. While the final result should be independent of $\lambda$, I get otherwise.

Please help. Thanks in advance.


We approximate the general integral $$\int_0^{\frac{\pi}{2}}\left(1-m^2 \sin^2 \theta \right)^\alpha d\theta \tag{1}$$

By Taylor developpement $$ \left(1-m^2 \sin^2 \theta \right)^\alpha = 1-\alpha \sin^2 (\theta). m^2 +\mathcal{O}(m^4) $$

As $\left(1-m^2 \sin^2 \theta \right)^\alpha >0$, we can change the order of integration and limit $$ \begin{align} \int_0^{\frac{\pi}{2}}\left(1-m^2 \sin^2 \theta \right)^\alpha d\theta &= \int_0^{\frac{\pi}{2}}\left(1-\alpha \sin^2 (\theta). m^2 +\mathcal{O}(m^4) \right) d\theta \\ &=\frac{\pi}{2}-\alpha \frac{\pi}{4} m^2 + \mathcal{O}(m^4) \tag{2} \end{align} $$

Replace $\alpha = -\frac{1}{2}$ and $\frac{1}{2}$ to $(2)$, you will get the results.

  • $\begingroup$ Thanks. Are you aware of the approach using local and global contributions? $\endgroup$
    – nchom
    Jul 13 '21 at 16:22
  • $\begingroup$ @nchom I don't know why it's necessary to split the integral into 2 parts for this case. It seems to me that this technique is used only when the integrand hasn't unnormal behavior (for example $f(x) \xrightarrow{x\to 0} +\infty$ ) at the lower bounds $x = 0$ $\endgroup$
    – NN2
    Jul 13 '21 at 16:37
  • $\begingroup$ You are right. Thanks. $\endgroup$
    – nchom
    Jul 13 '21 at 16:47

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