오늘의 계산 61 - Legendre Chi Function, Revisited

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

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Proposition. The following holds:[1]

$$\begin{array}{}\text{(1)}& {\int }_0^{\frac{\pi }{2}}\arctan (r\sin \theta )\arctan (s\sin \theta )d\theta =\pi {\chi }_2(\alpha \beta ),\end{array}$$

where

$$\begin{array}{r}\alpha =\frac{\sqrt{r^2+1}-1}{r},\beta =\frac{\sqrt{s^2+1}-1}{s}\end{array}$$

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 $s\to \mathrm{\infty }$ to $\text{(1)}$.

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Mathematica also suggests that the following holds:

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Conjecture. Let

$$\begin{array}{r}A=\{1,3,5,7,\dots {\}}^4\end{array}$$

be the set of quadruples of odd natural numbers. Then the series expantion of the integral

$$\begin{array}{r}{\int }_0^{\frac{\pi }{2}}\arctan \left(\frac{2x\sin \theta }{1-x^2}\right)\arctan \left(\frac{2y\sin \theta }{1-y^2}\right)\arctan \left(\frac{2z\sin \theta }{1-z^2}\right)\arctan \left(\frac{2w\sin \theta }{1-w^2}\right)d\theta \end{array}$$

is given by

$$\begin{array}{r}\frac{\pi }{2}\sum _{\alpha \in A}(-1{)}^{|\alpha |/2}d(\alpha )\frac{x^{{\alpha }_1}}{{\alpha }_1}\frac{y^{{\alpha }_2}}{{\alpha }_2}\frac{z^{{\alpha }_3}}{{\alpha }_3}\frac{w^{{\alpha }_4}}{{\alpha }_4},\end{array}$$

where $d(\alpha )$ denotes the number of choices of signatures so that

$$\begin{array}{r}\pm {\alpha }_1\pm {\alpha }_2\pm {\alpha }_3\pm {\alpha }_4=0.\end{array}$$

Upon reordering $\alpha$ so that ${\alpha }_1\le {\alpha }_2\le {\alpha }_3\le {\alpha }_4$, this equals

$$\begin{array}{r}(-1{)}^{|\alpha |/2}d(\alpha )=\{\begin{array}{ll}6,& {\alpha }_1={\alpha }_2={\alpha }_3={\alpha }_4,\\ 4,& {\alpha }_1={\alpha }_2<{\alpha }_3={\alpha }_4,\\ 2,& {\alpha }_1+{\alpha }_4={\alpha }_2+{\alpha }_3\text{ and }{\alpha }_1<{\alpha }_2,\\ -2& {\alpha }_1+{\alpha }_2+{\alpha }_3={\alpha }_4.\end{array}\end{array}$$

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References

1. Zakharia Stanley, Integral ${\int }_0^1\frac{{\arctan }^{2}x}{\sqrt{1-x^2}}dx$ - Math StackExchange