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Question-190009




Question Number 190009 by Rupesh123 last updated on 26/Mar/23
Commented by Frix last updated on 26/Mar/23
Use t=x^(π/5)  to get (5/π)∫_0 ^∞ (dt/(t^(10) +1)) and now “just”  decompose etc.
$$\mathrm{Use}\:{t}={x}^{\frac{\pi}{\mathrm{5}}} \:\mathrm{to}\:\mathrm{get}\:\frac{\mathrm{5}}{\pi}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{10}} +\mathrm{1}}\:\mathrm{and}\:\mathrm{now}\:“\mathrm{just}'' \\ $$$$\mathrm{decompose}\:\mathrm{etc}. \\ $$
Commented by ARUNG_Brandon_MBU last updated on 27/Mar/23
which is equal to (5/π)∙(1/(10))β((1/(10)), (9/(10)))  =(1/(2π))Γ((1/(10)))Γ((9/(10)))=(1/(2π))∙(π/(sin((π/(10)))))  =(1/2)cosec((π/(10)))
$$\mathrm{which}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{5}}{\pi}\centerdot\frac{\mathrm{1}}{\mathrm{10}}\beta\left(\frac{\mathrm{1}}{\mathrm{10}},\:\frac{\mathrm{9}}{\mathrm{10}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\Gamma\left(\frac{\mathrm{1}}{\mathrm{10}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{10}}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\centerdot\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{10}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosec}\left(\frac{\pi}{\mathrm{10}}\right) \\ $$

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