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 $$A(p,q,r):=pqr{\int }_0^1\frac{\arctan q\sqrt{p^2{x}^2+1}}{q\sqrt{p^2{x}^2+1}}\frac{dx}{(r^2+1)p^2{x}^2+1}.$$ Then whenever $pqr=1$, we have the following simple formula for $A(p,q,r)$: $$A(p,q,r)=\frac{{\pi }^2}{8}+\frac{1}{2}\{{\arctan }^2(1/\tilde{p})-{\arctan }^2(\tilde{q})-{\arctan }^2(\tilde{r})\},$$ where $\tilde{p}$, $\tilde{q}$ and $\tilde{r}$ are defined by $$\begin{array}{r}\tilde{p}=(q^2+1{)}^{1/2}r,\tilde{q}=(r^2+1{)}^{1/2}p,\tilde{r}=(p^2+1{)}^{1/2}q.\end{array}$$

Proof. Refer to the following file:

doc_026_Ahmed_Integral.pdf

다운로드

 

 

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 $$A(p,q,r)=F(k,k/\tilde{q})+F(k,k/\tilde{r})-F(k,\tilde{p}).$$

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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