## 오늘의 계산 63 - Generalized Ahmed's Integral

2013. 11. 28. 16:48

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.

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

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

## 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

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1. 리커리시 2013.12.01 02:06

흐아.. 존경스럽습니다 ㅎㄷ; 어느 문제든 몇 년이고 붙들면 결국 풀 수 있구나 하는 생각도 드네요.

2. Park 2013.12.06 16:03

I don't know what to say other than superb for you to have come up with that solution just in a few years! BTW, there appears to be a typo in "$a \leq b < c \text{ or } a \leq b < c$" part on the 4th page.