## 오늘의 계산 61 - Legendre Chi Function, Revisited

2013. 11. 12. 10:01

나흘 연휴의 절반이 지나가는 동안 한 거라곤 폐인짓밖에 없네…. 공부해야지~

Proposition. The following holds:[1] $$\label{eq_wts} \int_{0}^{\frac{\pi}{2}} \arctan ( r \sin\theta ) \arctan ( s \sin\theta ) \, d\theta = \pi \chi_{2}(\alpha \beta),$$ where \begin{align*} \alpha = \frac{\sqrt{r^{2} + 1} - 1}{r}, \quad \beta = \frac{\sqrt{s^{2} + 1} - 1}{s} \end{align*} and $\chi_{2}$ is the Legendre chi function.

Proof. See the reference [1] below.

Remark. This result extends my previous result. You can check this by taking the limit $s \to \infty$ to \eqref{eq_wts}.

Mathematica also suggests that the following holds:

Conjecture. Let \begin{align*} A = \{ (2i+1, 2j+1, 2k+1, 2l+1) : i, j, k, l \in \Bbb{N}_{0} \} \end{align*} be the set of quadruples of odd natural numbers. Then the series expantion of the integral \begin{align*} \int_{0}^{\frac{\pi}{2}} \arctan \left( \frac{2x \sin\theta}{1-x^{2}} \right) \arctan \left( \frac{2y \sin\theta}{1-y^{2}} \right) \arctan \left( \frac{2z \sin\theta}{1-z^{2}} \right) \arctan \left( \frac{2w \sin\theta}{1-w^{2}} \right) \, d\theta \end{align*} is given by \begin{align*} \frac{\pi}{2} \sum_{\alpha \in A} (-1)^{|\alpha|/2} d(\alpha) \frac{x^{\alpha_{1}}}{\alpha_{1}} \frac{y^{\alpha_{2}}}{\alpha_{2}} \frac{z^{\alpha_{3}}}{\alpha_{3}} \frac{w^{\alpha_{4}}}{\alpha_{4}}, \end{align*} where $d(\alpha)$ denotes the number of choices of signatures so that \begin{align*} \pm \alpha_{1} \pm \alpha_{2} \pm \alpha_{3} \pm \alpha_{4} = 0. \end{align*} Upon reordering $\alpha$ so that $\alpha_{1} \leq \alpha_{2} \leq \alpha_{3} \leq \alpha_{4}$, this equals \begin{align*} (-1)^{|\alpha|/2} d(\alpha) = \begin{cases} 6, & \alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}, \\ 4, & \alpha_{1} = \alpha_{2} < \alpha_{3} = \alpha_{4}, \\ 2, & \alpha_{1} + \alpha_{4} = \alpha_{2} + \alpha_{3} \text{ and } \alpha_{1} < \alpha_{2}, \\ -2 & \alpha_{1} + \alpha_{2} + \alpha_{3} = \alpha_{4}. \end{cases} \end{align*}