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freemind
Riemann Hypothesis
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Location: MIT or Moldova
Moldova, Republic of    United States
Post Posted: Mar 01, 2008, 7:35 am • # 1 


Find the exact value of E=\displaystyle\int_0^{\frac\pi2}\cos^{1003}x\text{d}x\cdot\int_0^{\frac\pi2}\cos^{1004}x\text{d}x\cdot.

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didilica
Yang-Mills Theory

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Romania
Post Posted: Mar 01, 2008, 7:56 am • # 2 


E=\frac{\pi}{2008}.

We have that

E=\frac{1}{2}\beta(502, 1/2)\frac{1}{2}\beta(1005/2,1/2)

=\frac{1}{4}\frac{\Gamma(502)\Gamma(1/2)}{\Gamma(1005/2)}\frac{\Gamma(1005/2)\Gamma(1/2)}{\Gamma(503)}

=\frac{\pi}{4\cdot 502}=\frac{\pi}{2008}.

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Yuriy Solovyov
Yang-Mills Theory
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Ukraine
Post Posted: Mar 03, 2008, 12:49 pm • # 3 


\int{\cos^{n}{x} dx}=\frac{\cos^{n-1}{x}\sin {x}}{n}+\frac{n-1}{n}\int{\cos^{n-2}{x} dx}
So we have:
\int_{0}^{\frac{\pi}{2}}{\cos^{n}{x} dx}=\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}{\cos^{n-2}{x} dx}
We have:
\int_{0}^{\frac{\pi}{2}}{\cos^{1003}{x} dx}=\frac{2\cdot 4\cdot 6\cdot ... \cdot 1002}{3\cdot 5\cdot 7\cdot ...\cdot 1003}\in...
\int_{0}^{\frac{\pi}{2}}{\cos^{1004}{x} dx}=\frac{3\cdot 5\cdot 7\cdot ... \cdot 1003}{2\cdot 4\cdot 5\cdot ...\cdot 1004}\in...
We will have:
E=\frac{2\cdot 4\cdot 6\cdot ...\cdot 1002}{3\cdot 5\cdot 7\cdot ... \cdot 1003}\cdot \frac{3\cdot 5\cdot 7\cdot ... \cdot 1...
 
 
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