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Question Number 118307 by bramlexs22 last updated on 16/Oct/20

$${\underset{0}{\int }}^{\mathrm{\infty }}\frac{x^2-2}{x^4+x^2+1}dx$$

Commented by Lordose last updated on 16/Oct/20

$$\mathrm{see}\mathrm{qst}118030$$

Answered by TANMAY PANACEA last updated on 16/Oct/20

$$\int \frac{x^2-2}{x^4+x^2+1}dx$$$$\int \frac{dx}{x^2+1+\frac{1}{x^2}}-\int \frac{\frac{2}{x^2}}{x^2+1+\frac{1}{x^2}}dx$$$$\frac{1}{2}\int \frac{1+\frac{1}{x^2}+1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}-\int \frac{1+\frac{1}{x^2}-(1-\frac{1}{x^2})}{x^2+\frac{1}{x^2}+1}$$$$\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+3}+\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-1}-\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+3}+\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-1}$$$$\frac{3}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-1}\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+3}$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{p}\mathit{l}\mathit{s}\mathit{u}\mathit{s}\mathit{e}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}$$$$\int \frac{dy}{y^2-a^2}and\int \frac{dy}{y^2+a^2}andputlimitpls$$$$\mathit{i}\mathit{p}\mathit{r}\mathit{e}\mathit{f}\mathit{e}\mathit{r}\mathit{n}\mathit{o}\mathit{t}\mathit{t}\mathit{o}\mathit{r}\mathit{e}\mathit{m}\mathit{e}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}$$$$\mathit{r}\mathit{a}\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{i}\mathit{l}\mathit{i}\mathit{k}\mathit{e}\mathit{h}\mathit{o}\mathit{w}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}\mathit{i}\mathit{s}\mathit{d}\mathit{e}\mathit{r}\mathit{i}\mathit{v}\mathit{e}\mathit{d}$$

Answered by Bird last updated on 17/Oct/20

$$letI={\int }_0^{\mathrm{\infty }}\frac{x^2-2}{x^4+x^2+1}dx\Rightarrow$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2-2}{x^4+x^2+1}dxweconsider$$$$\phi \left(z\right)=\frac{z^2-2}{z^4+z^2+1}polesof\phi ?$$$$z^4+z^2+1=0\Rightarrow u^2+u+1=0(u=z^2)$$$$\mathrm{\Delta }=-3\Rightarrow u_1=\frac{-1+i\sqrt{3}}{2}=e^{\frac{i2\pi }{3}}$$$$u_2=e^{-\frac{i2\pi }{3}}$$$$\Rightarrow \phi \left(z\right)=\frac{z^2-2}{(z^2-e^{\frac{i2\pi }{3}})(z^2-e^{-\frac{i2\pi }{3}})}$$$$=\frac{z^2-2}{(z-e^{\frac{i\pi }{3}})(z+e^{\frac{i\pi }{3}})(z-e^{-\frac{i\pi }{3}})(z+e^{-\frac{i\pi }{3}})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \{Res(\phi ,e^{\frac{i\pi }{3}})+Res(\phi ,-e^{-\frac{i\pi }{3}})\}$$$$Res(\phi ,e^{\frac{i\pi }{3}})=\frac{e^{\frac{i2\pi }{3}}-2}{2e^{\frac{i\pi }{3}}\left(2isin\left(\frac{2\pi }{3}\right)\right)}$$$$=\frac{e^{-\frac{i\pi }{3}}(e^{\frac{i2\pi }{3}}-2)}{4i\left(\frac{\sqrt{3}}{2}\right)}=\frac{e^{\frac{i\pi }{3}}-2e^{-\frac{i\pi }{3}}}{2i\sqrt{3}}$$$$Res(\phi ,-e^{-\frac{i\pi }{3}})=\frac{e^{-\frac{i2\pi }{3}}-2}{-2e^{-\frac{i\pi }{3}}(-2isin\left(\frac{2\pi }{3}\right))}$$$$=\frac{e^{-\frac{i\pi }{3}}-2e^{\frac{i\pi }{3}}}{2i\sqrt{3}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{2i\pi }{2i\sqrt{3}}\{e^{\frac{i\pi }{3}}-2e^{-\frac{i\pi }{3}}$$$$+e^{-\frac{i\pi }{3}}-2e^{\frac{i\pi }{3}}\}$$$$=\frac{\pi }{\sqrt{3}}\{2cos\left(\frac{\pi }{3}\right)-4cos\left(\frac{\pi }{3}\right))$$$$=\frac{\pi }{\sqrt{3}}\{-2\left(\frac{1}{2}\right)[=-\frac{\pi }{\sqrt{3}}=2I\Rightarrow$$$$I=-\frac{\pi }{2\sqrt{3}}$$

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