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Question Number 126203 by frc2crc last updated on 18/Dec/20

$${\stackrel{\mathrm{\infty }}{\int }}_0\frac{1}{1+x^s+x^{2s}}$$

Answered by Olaf last updated on 18/Dec/20

$$$$$$1+x^s+x^{2s}=0$$$$(x^s-e^{i\frac{2\pi }{3}})(x^s-e^{-i\frac{2\pi }{3}})=0$$$$x=e^{i(\frac{2\pi }{3s}+\frac{2k\pi }{s})}\mathrm{or}x=e^{i(-\frac{2\pi }{3s}+\frac{2k\pi }{s})},k=0...s-1$$$$x=e^{i\frac{2\pi }{s}(2k+\frac{1}{3})}\mathrm{or}x=e^{i\frac{2\pi }{s}(2k-\frac{1}{3})},k=0...s-1$$$$\mathrm{Let}{x}_k^{\oplus }=e^{i\frac{2\pi }{s}(2k+\frac{1}{3})}\mathrm{and}{x}_k^{⊝}=e^{i\frac{2\pi }{s}(2k-\frac{1}{3})}$$$$\mathrm{R}\left(x\right)=\frac{1}{1+x^s+x^{2s}}$$$$\mathrm{R}\left(x\right)=\frac{1}{\underset{k=0}{\prod ^{s-1}}(x-x_k^{\oplus })\times \underset{k=0}{\prod ^{s-1}}(x-x_k^{⊝})}$$$$\mathrm{R}\left(x\right)=\underset{p=0}{\sum ^{s-1}}\frac{{\mathrm{A}}_p^{\oplus }}{x-x_p^{\oplus }}+\underset{p=0}{\sum ^{s-1}}\frac{{\mathrm{A}}_p^{⊝}}{x-x_p^{⊝}}$$$${\mathrm{A}}_p^{\oplus }=\frac{1}{\underset{\underset{k\ne p}{k=0}}{\prod ^{s-1}}(x_p^{\oplus }-x_k^{\oplus })\times \underset{k=0}{\prod ^{s-1}}(x_p^{\oplus }-x_k^{⊝})}$$$${\mathrm{A}}_p^{⊝}=\frac{1}{\underset{k=0}{\prod ^{s-1}}(x_p^{⊝}-x_k^{\oplus })\times \underset{\underset{k\ne p}{k=0}}{\prod ^{s-1}}(x_p^{⊝}-x_k^{⊝})}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\mathrm{R}\left(x\right)dx$$$$\mathrm{I}=\underset{p=0}{\sum ^{s-1}}{\int }_0^{\mathrm{\infty }}[\frac{{\mathrm{A}}_p^{\oplus }}{x-x_p^{\oplus }}+\frac{{\mathrm{A}}_p^{⊝}}{x-x_p^{⊝}}]dx$$$$\mathrm{I}=\underset{p=0}{\sum ^{s-1}}{[{\mathrm{A}}_p^{\oplus }\ln \mid x-x_p^{\oplus }\mid +{\mathrm{A}}_p^{⊝}\ln \mid x-x_p^{⊝}\mid ]}_0^{\mathrm{\infty }}$$$$...\mathrm{to}\mathrm{be}\mathrm{continued}...$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{good}\mathrm{way}...$$

Answered by Dwaipayan Shikari last updated on 18/Dec/20

$${\int }_0^{\mathrm{\infty }}\frac{1}{1+x^s+x^{2s}}dx{x}^s=t$$$$=\frac{1}{s}{\int }_0^{\mathrm{\infty }}{t}^{\frac{1-s}{s}}\frac{1}{1+t+t^2}dt=\frac{1}{s}{\int }_0^{\mathrm{\infty }}{t}^{\frac{1-s}{s}}.\frac{1}{(t-e^{\frac{2i\pi }{3}})(t-e^{-\frac{2\pi i}{3}})}dt$$$$=\frac{1}{s(e^{\frac{2i\pi }{3}}-e^{-\frac{2i\pi }{3}})}{\int }_0^{\mathrm{\infty }}\frac{t^a}{t-e^{\frac{2i\pi }{3}}}-\frac{1}{s(e^{\frac{2i\pi }{3}}-e^{\frac{-2i\pi }{3}})}{\int }_0^{\mathrm{\infty }}\frac{t^a}{t-e^{-\frac{2i\pi }{3}}}$$$$=\frac{1}{\mathrm{\Lambda }}{\int }_0^{\mathrm{\infty }}\frac{t^a}{t+m}-\frac{t^a}{t+b}dtt=my\Rightarrow 1=m\frac{dy}{dt}t=bj\Rightarrow 1=b\frac{dj}{dt}$$$$=\frac{m^a}{\mathrm{\Lambda }}{\int }_0^{\mathrm{\infty }}\frac{y^a}{y+1}-\frac{b^a}{\mathrm{\Lambda }}\int \frac{j^a}{j+1}dj$$$${\int }_0^{\mathrm{\infty }}\frac{y^a}{y+1}dy\frac{y}{y+1}=g\Rightarrow \frac{1}{{(1+y)}^2}=\frac{dg}{dy}$$$$={\int }_0^1{y}^{a-1}\frac{g}{{(1-g)}^2}dg={\int }_0^1{g}^a{(1-g)}^{-a-1}dg=-\pi cosec\left(\pi a\right)$$$$Similarly{\int }_0^{\mathrm{\infty }}\frac{j^a}{j+1}=-\pi cosec\left(\pi a\right)$$$$I=-\frac{\pi cosec\pi a}{\mathrm{\Lambda }}(m^a-b^a)(a=\frac{1-s}{s},\mathrm{\Lambda }=2issin\frac{2\pi }{3},m=-e^{\frac{2\pi i}{3}}b=-e^{\frac{-2\pi i}{3}})$$$$Notsureifitistrueornot...$$

Answered by mathmax by abdo last updated on 19/Dec/20

$${\mathrm{A}}_{\mathrm{s}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^{\mathrm{s}}+{\mathrm{x}}^{2\mathrm{s}}}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}{\mathrm{x}}^{\mathrm{s}}=\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{t}}^{\frac{1}{\mathrm{s}}}\left(\mathrm{we}\mathrm{suppise}\mathrm{s}\mathrm{real}\right)$$$$\Rightarrow {\mathrm{A}}_{\mathrm{s}}=\frac{1}{\mathrm{s}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{{\mathrm{t}}^2+\mathrm{t}+1}\mathrm{dt}\mathrm{roots}\mathrm{of}{\mathrm{t}}^2+\mathrm{t}+1=0\to \mathrm{\Delta }=-3\Rightarrow$$$${\mathrm{t}}_1=\frac{-1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}\mathrm{and}{\mathrm{t}}_2=\frac{-1-\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}}\Rightarrow$$$${\mathrm{A}}_{\mathrm{s}}=\frac{1}{\mathrm{s}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{(\mathrm{t}-{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}})(\mathrm{t}-{\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}})}\mathrm{dt}=\frac{1}{\mathrm{s}}{\int }_0^{\mathrm{\infty }}(\frac{1}{\mathrm{t}-{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}}-\frac{1}{\mathrm{t}-{\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}}}){\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}\mathrm{dt}$$

Commented by mathmax by abdo last updated on 19/Dec/20

$${\mathrm{A}}_{\mathrm{s}}=\frac{1}{\mathrm{s}\left(2\mathrm{isin}\left(\frac{\pi }{3}\right)\right)}\{{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{\mathrm{t}-{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}}\mathrm{dt}-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{\mathrm{t}-{\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}}}\mathrm{dt}\}$$$${\mathrm{A}}_{\mathrm{s}}=\frac{1}{\mathrm{is}\sqrt{3}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{\mathrm{t}+{\mathrm{e}}^{\mathrm{i}(\pi +\frac{2\pi }{3})}}\mathrm{dt}-\frac{1}{\mathrm{is}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{\mathrm{t}+{\mathrm{e}}^{\mathrm{i}(\pi -\frac{2\pi }{3})}}\mathrm{dt}\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{\mathrm{t}+{\mathrm{e}}^{\mathrm{i}\left(\frac{5\pi }{3}\right)}}\mathrm{dt}{=}_{\mathrm{t}={\mathrm{e}}^{\mathrm{i}\frac{5\pi }{3}}\mathrm{z}}{\int }_0^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{\frac{\mathrm{i}5\pi }{3}}\right)}^{\frac{1}{\mathrm{s}}-1}{\mathrm{z}}^{\frac{1}{\mathrm{s}}-1}}{{\mathrm{e}}^{\frac{\mathrm{i}5\pi }{3}}(1+\mathrm{z})}{\mathrm{e}}^{\frac{\mathrm{i}5\pi }{3}}\mathrm{dz}$$$$={\mathrm{e}}^{\frac{5\mathrm{i}\pi }{3}(\frac{1}{\mathrm{s}}-1)}\times \frac{\pi }{\sin \left(\frac{\pi }{\mathrm{s}}\right)}\mathrm{also}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{s}}-1}}{\mathrm{t}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}}\mathrm{dt}{=}_{\mathrm{t}={\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\mathrm{z}}{\int }_0^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\right)}^{\frac{1}{\mathrm{s}}-1}{\mathrm{z}}^{\frac{1}{\mathrm{s}}-1}}{{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}(1+\mathrm{z})}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}}\mathrm{dz}$$$$={\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}(\frac{1}{\mathrm{s}}-1)}\times \frac{\pi }{\sin \left(\frac{\pi }{\mathrm{s}}\right)}\Rightarrow$$$${\mathrm{A}}_{\mathrm{s}}=\frac{1}{\mathrm{is}\sqrt{3}}\times \frac{\pi }{\sin \left(\frac{\pi }{\mathrm{s}}\right)}({\mathrm{e}}^{\frac{5\mathrm{i}\pi }{3}(\frac{1}{\mathrm{s}}-1)}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{3}(\frac{1}{\mathrm{s}}-1)})\mathrm{rest}\mathrm{to}\mathrm{simplify}{\mathrm{A}}_{\mathrm{s}}$$$$...\mathrm{be}\mathrm{continued}...$$

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