오늘의 계산 63 - Generalized Ahmed's Integral

드디어 몇 년동안 짬짬히 고민하던 적분 문제에 상당한 진척을 보였습니다. 일단은 급한 일이 있어서, 결과 요약이랑 작성중인 증명 파일만 올립니다. 나중에 좀 더 살을 붙여야지요.

 

Finally I made a significant progress in the integral problem I was struggling for some years. As I am busy now, I just present here a summarized result and the article containing the main proof.

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Theorem 1. For $ p, q, r > 0 $, define the generalized Ahmed's integral by \begin{equation} \label{int_ahmed_def} A(p,q,r) := p q r \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1} }{q \sqrt{p^{2}x^{2} + 1}} \frac{\, dx}{(r^{2}+1)p^{2}x^{2} + 1}. \end{equation} Then whenever $pqr = 1$, we have the following simple formula for $A(p, q, r)$: \begin{equation} \label{eq_wts} A(p, q, r) = \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} (1 / \tilde{p} ) - \arctan^{2} ( \tilde{q} ) - \arctan^{2} ( \tilde{r} ) \right\}, \end{equation} where $\tilde{p}$, $\tilde{q}$ and $\tilde{r}$ are defined by \begin{align*} \tilde{p} = (q^{2}+1)^{1/2}r, \quad \tilde{q} = (r^{2}+1)^{1/2}p, \quad \tilde{r} = (p^{2}+1)^{1/2}q. \end{align*}

Proof. Refer to the following file:

doc_026_Ahmed_Integral.pdf

다운로드

 

 

Theorem 2. Define $F(k, p)$ by \begin{equation} F(k, p) = \int_{0}^{p} \frac{\arctan(k / t)}{1 + t^{2}} \, dt. \end{equation} Also suppose that $p, q, r$ and $\tilde{p}, \tilde{q}, \tilde{r}$ be as in Theorem 1. If we put $k = pqr$, then \begin{equation} A(p, q, r) = F(k, k / \tilde{q} ) + F(k,k / \tilde{r}) - F(k, \tilde{p}). \end{equation}

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Revision History

1. Last updated at 09:02 12-01-2013, Modified the section 4 to the considerable amount.
2. Updated at 17:30 11-30-2013, Made amend to a critical typo in the definition (1.3) of the modified Ahmed integral.
3. Updated at 02:48 11-30-2013