Γενικευμένο Ολοκλήρωμα

Tolaso J Kos · #1

Δείξτε ότι: $\displaystyle{\int_{0}^{\infty}\frac{dx}{\left [ x^4+\left ( 1+\sqrt{2} \right )x^2+1\right ]\left ( x^{100}-x^{99}+\cdots+1 \right )}=\frac{\pi}{2\left ( 1+\sqrt{2} \right )}}$ .

ΠΑΠΑΔΟΠΟΥΛΟΣ ΣΤΑΥΡΟΣ · #2

Επαναφορά.

pprime · #3

Tolaso J Kos έγραψε:Δείξτε ότι: $\displaystyle{\int_{0}^{\infty}\frac{dx}{\left [ x^4+\left ( 1+\sqrt{2} \right )x^2+1\right ]\left ( x^{100}-x^{99}+\cdots+1 \right )}=\frac{\pi}{2\left ( 1+\sqrt{2} \right )}}$ .

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{dx}{\left( x^{4}+\left( 1+2\sqrt{2} \right)x^{2}+1 \right)\left( x^{100}-x^{98}+...-x^{2}+1 \right)}}=\frac{\pi }{2\left( 1+\sqrt{2} \right)}}$

$\displaystyle{S=x^{100}-x^{98}+x^{96}-x^{94}+...+x^{4}-x^{2}+1}$

$\displaystyle{x^{2}S=x^{102}-x^{100}+x^{98}-x^{96}+...+x^{6}-x^{4}+x^{2}}$

$\displaystyle{x^{2}S+S=x^{102}+1\Rightarrow S=x^{100}-x^{98}+x^{96}-x^{94}+...+x^{4}-x^{2}+1=\frac{x^{102}+1}{x^{2}+1}}$

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{dx}{\left( x^{4}+\left( 1+2\sqrt{2} \right)x^{2}+1 \right)\left( x^{100}-x^{98}+...-x^{2}+1 \right)}}=\int\limits_{0}^{+\infty }{\frac{x^{2}+1}{\left( x^{4}+\left( 1+2\sqrt{2} \right)x^{2}+1 \right)\cdot \left( x^{102}+1 \right)}dx}}$

$\displaystyle{=\int\limits_{0}^{+\infty }{\frac{\frac{1}{x^{2}}+1}{\left( \frac{1}{x^{4}}+\left( 1+2\sqrt{2} \right)\frac{1}{x^{2}}+1 \right)\cdot \left( \frac{1}{x^{102}}+1 \right)}\frac{dx}{x^{2}}}=\int\limits_{0}^{+\infty }{\frac{\left( 1+x^{2} \right)x^{102}}{\left( x^{4}+\left( 1+2\sqrt{2} \right)x^{2}+1 \right)\cdot \left( 1+x^{102} \right)}dx}}$

$\displaystyle{=\frac{1}{2}\int\limits_{0}^{+\infty }{\frac{x^{2}+1}{x^{4}+\left( 1+2\sqrt{2} \right)x^{2}+1}dx}=\frac{1}{2}\int\limits_{0}^{+\infty }{\frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+1+2\sqrt{2}}dx}\underbrace{=}_{t=x-\frac{1}{x}}\frac{1}{2}\int\limits_{-\infty }^{+\infty }{\frac{dt}{t^{2}+2+1+2\sqrt{2}}}}$

$\displaystyle{=\frac{\pi }{2\sqrt{3+2\sqrt{2}}}=\frac{\pi }{2\sqrt{1+2\sqrt{2}+\left( \sqrt{2} \right)^{2}}}=\frac{\pi }{2\left( 1+\sqrt{2} \right)}}$
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$\displaystyle{\int\limits_{0}^{+\infty }{\frac{x^{2}+1}{x^{4}+2px^{2}+1}\cdot \frac{dx}{x^{q}+1}}=\frac{\pi }{2\sqrt{2p+2}}}$

$\displaystyle{\int\limits_{0}^{+\infty }{\frac{x^{2}+1}{x^{4}+2px^{2}+1}\cdot \frac{dx}{x^{q}+1}}\underbrace{=}_{x=\frac{1}{x}}\int\limits_{0}^{+\infty }{\frac{\frac{1}{x^{2}}+1}{\frac{1}{x^{4}}+\frac{2p}{x^{2}}+1}\cdot \frac{1}{\frac{1}{x^{q}}+1}\cdot \frac{dx}{x^{2}}}=\int\limits_{0}^{+\infty }{\frac{1+x^{2}}{1+2px^{2}+x^{4}}\cdot \frac{x^{q}}{1+x^{q}}dx}}$

$\displaystyle{=\frac{1}{2}\int\limits_{0}^{+\infty }{\frac{x^{2}+1}{x^{4}+2px^{2}+1}dx}=\frac{1}{2}\int\limits_{0}^{+\infty }{\frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+2p}dx}\underbrace{=}_{t=x-\frac{1}{x}}\frac{1}{2}\int\limits_{-\infty }^{+\infty }{\frac{dt}{t^{2}+2+2p}}=\frac{\pi }{2\sqrt{2+2p}}}$
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$\displaystyle{\int\limits_{0}^{+\infty }{\left( \frac{x^{2}}{x^{4}+2px^{2}+1} \right)^{r}\cdot \frac{x^{2}+1}{x^{q}+1}\cdot \frac{dx}{x^{2}}}=\frac{\left( 1+p \right)^{\frac{1}{2}-r}}{2^{\frac{1}{2}+r}}\cdot \frac{\Gamma \left( r-\frac{1}{2} \right)\sqrt{\pi }}{\left( r-1 \right)!}}$

$\displaystyle{\int\limits_{0}^{+\infty }{\left( \frac{x^{2}}{x^{4}+2px^{2}+1} \right)^{r}\cdot \frac{x^{2}+1}{x^{q}+1}\cdot \frac{dx}{x^{2}}}=\int\limits_{0}^{+\infty }{\left( \frac{\frac{1}{x^{2}}}{\frac{1}{x^{4}}+\frac{2p}{x^{2}}+1} \right)^{r}\cdot \frac{\frac{1}{x^{2}}+1}{\frac{1}{x^{q}}+1}\cdot x^{2}\cdot \frac{dx}{x^{2}}}}$

$\displaystyle{=\int\limits_{0}^{+\infty }{\left( \frac{x^{2}}{1+2px^{2}+x^{4}} \right)^{r}\cdot \frac{1+x^{2}}{1+x^{q}}\cdot \frac{x^{q}}{x^{2}}dx}=\frac{1}{2}\int\limits_{0}^{+\infty }{\left( \frac{x^{2}}{1+2px^{2}+x^{4}} \right)^{r}\cdot \left( \frac{1+x^{2}}{x^{2}} \right)dx}}$

$\displaystyle{=\frac{1}{2}\int\limits_{0}^{+\infty }{\left( \frac{1}{x^{2}+\frac{1}{x^{2}}+2p} \right)^{r}\cdot \left( 1+\frac{1}{x^{2}} \right)dx}\underbrace{=}_{t=x-\frac{1}{x}}\frac{1}{2}\int\limits_{-\infty }^{+\infty }{\frac{1}{\left( t^{2}+2+2p \right)^{r}}dt}=\int\limits_{0}^{+\infty }{\frac{1}{\left( t^{2}+2+2p \right)^{r}}dt}}$

$\displaystyle{=\frac{1}{\left( \sqrt{2p+2} \right)^{2r}}\int\limits_{0}^{+\infty }{\frac{1}{\left( \left( \frac{t}{\sqrt{2p+2}} \right)^{2}+1 \right)^{r}}dt}=\frac{1}{\left( \sqrt{2p+2} \right)^{2r-1}}\int\limits_{0}^{+\infty }{\frac{1}{\left( t^{2}+1 \right)^{r}}dt}}$

$\displaystyle{\underbrace{=}_{t^{2}=u}\frac{1}{2\left( \sqrt{2p+2} \right)^{2r-1}}\int\limits_{0}^{+\infty }{\frac{u^{1-\frac{1}{2}-1}}{\left( u+1 \right)^{1-\frac{1}{2}+r-1+\frac{1}{2}}}du}=\frac{\beta \left( \frac{1}{2},r-\frac{1}{2} \right)}{2^{1+\frac{2r-1}{2}}\left( p+1 \right)^{\frac{2r-1}{2}}}=\frac{\left( 1+p \right)^{\frac{1}{2}-r}}{2^{\frac{1}{2}+r}}\cdot \frac{\Gamma \left( r-\frac{1}{2} \right)\sqrt{\pi }}{\left( r-1 \right)!}}$

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