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r-1-0-pi-2-5-2-2-x-2r-dx-




Question Number 787 by 123456 last updated on 22/Mar/15
Σ_(r=1) ^(+∞) (√(∫_0 ^(π/2) 5+Γ((2/(2+x^(2r) )))dx))
$$\underset{{r}=\mathrm{1}} {\overset{+\infty} {\sum}}\sqrt{\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{5}+\Gamma\left(\frac{\mathrm{2}}{\mathrm{2}+{x}^{\mathrm{2}{r}} }\right){dx}} \\ $$
Commented by 123456 last updated on 13/Mar/15
Σ_(r=1) ^(+∞) (√(((5π)/2)+∫_0 ^(π/2) Γ((2/(2+x^(2r) )))dx))  Γ(z)=∫_0 ^(+∞) t^(z−1) e^(−t) dt     ℜ(z)>0  0<(2/(2+x^(2r) ))≤1,0≤x≤(π/2)  (2^(2r+1) /(2^(2r+1) +π^(2r) ))≤(2/(2+x^(2r) ))≪1,0≤x≤(π/2)
$$\underset{{r}=\mathrm{1}} {\overset{+\infty} {\sum}}\sqrt{\frac{\mathrm{5}\pi}{\mathrm{2}}+\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\Gamma\left(\frac{\mathrm{2}}{\mathrm{2}+{x}^{\mathrm{2}{r}} }\right){dx}} \\ $$$$\Gamma\left({z}\right)=\underset{\mathrm{0}} {\overset{+\infty} {\int}}{t}^{{z}−\mathrm{1}} {e}^{−{t}} {dt}\:\:\:\:\:\Re\left({z}\right)>\mathrm{0} \\ $$$$\mathrm{0}<\frac{\mathrm{2}}{\mathrm{2}+{x}^{\mathrm{2}{r}} }\leqslant\mathrm{1},\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}^{\mathrm{2}{r}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{r}+\mathrm{1}} +\pi^{\mathrm{2}{r}} }\leqslant\frac{\mathrm{2}}{\mathrm{2}+{x}^{\mathrm{2}{r}} }\ll\mathrm{1},\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$

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