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.
Best Answer
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.