Question Number 159226 by cortano last updated on 14/Nov/21
$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} }−\sqrt{{x}}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\mathrm{ln}\:{x}}\:{dx}\:=? \\ $$
Answered by mindispower last updated on 14/Nov/21
![∫_0 ^∞ ((x^s −x^(1/2) )/((1+x^2 )ln(x))).dx=f(s) f′(s)=∫_0 ^∞ (x^s /(1+x^2 ))dx=∫_0 ^∞ (y^((s−1)/2) /(2(1+y)))dy=((β(((1−s)/2),((s+1)/2)))/2) =(π/(2sin(((π(1−s))/2))))=(π/(2cos(((sπ)/2)))) f(s)=∫_(1/2) ^s (π/(2cos(((sπ)/2))))ds =∫_(π/4) ^((sπ)/2) (du/(cos(u)))=∫_(π/4) ^((sπ)/2) ((cos(u)du)/((1−sin(u))(1+sin(u)))) =[(1/2).ln(((1+sin(u))/(1−sin(u)))]_(π/4) ^((sπ)/2) ]=((ln(((1+sin(((sπ)/2)))/(1−sin(((sπ)/2))))))/2) −((ln((((√2)+1)/( (√2)−1))))/2) Ω=f((5/6))=((ln(((1+sin(((5π)/(12))))/(1−sin(((5π)/(12))))))−ln(((1+(√2))/( (√2)−1))))/2) sin(((5π)/(12)))=cos((π/(12)))=((√(2+(√3)))/2)](https://www.tinkutara.com/question/Q159235.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}} −{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){ln}\left({x}\right)}.{dx}={f}\left({s}\right) \\ $$$${f}'\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{y}^{\frac{{s}−\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{1}+{y}\right)}{dy}=\frac{\beta\left(\frac{\mathrm{1}−{s}}{\mathrm{2}},\frac{{s}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi\left(\mathrm{1}−{s}\right)}{\mathrm{2}}\right)}=\frac{\pi}{\mathrm{2}{cos}\left(\frac{{s}\pi}{\mathrm{2}}\right)} \\ $$$${f}\left({s}\right)=\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}\overset{{s}} {\:}\frac{\pi}{\mathrm{2}{cos}\left(\frac{{s}\pi}{\mathrm{2}}\right)}{ds} \\ $$$$=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{{s}\pi}{\mathrm{2}}} \frac{{du}}{{cos}\left({u}\right)}=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{{s}\pi}{\mathrm{2}}} \frac{{cos}\left({u}\right){du}}{\left(\mathrm{1}−{sin}\left({u}\right)\right)\left(\mathrm{1}+{sin}\left({u}\right)\right)} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}.{ln}\left(\frac{\mathrm{1}+{sin}\left({u}\right)}{\mathrm{1}−{sin}\left({u}\right)}\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{{s}\pi}{\mathrm{2}}} \right]=\frac{{ln}\left(\frac{\mathrm{1}+{sin}\left(\frac{{s}\pi}{\mathrm{2}}\right)}{\mathrm{1}−{sin}\left(\frac{{s}\pi}{\mathrm{2}}\right)}\right)}{\mathrm{2}} \\ $$$$−\frac{{ln}\left(\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right)}{\mathrm{2}} \\ $$$$\Omega={f}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)=\frac{{ln}\left(\frac{\mathrm{1}+{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)}{\mathrm{1}−{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)}\right)−{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right)}{\mathrm{2}} \\ $$$${sin}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)={cos}\left(\frac{\pi}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$