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Question Number 93632 by abdomathmax last updated on 14/May/20

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2-3}{{(x^2-x+1)}^2}dx$$

Commented by abdomathmax last updated on 15/May/20

$$letI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2-3}{{(x^2-x+1)}^2}dx\Rightarrow I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2-3}{{({(x-\frac{1}{2})}^2+\frac{3}{4})}^2}dx$$$${=}_{x-\frac{1}{2}=\frac{\sqrt{3}}{2}t}\frac{16}{9}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{(\frac{1}{2}+\frac{\sqrt{3}}{2}t)}^2-3}{{(t^2+1)}^2}\times \frac{\sqrt{3}}{2}dt$$$$=\frac{8\sqrt{3}}{9}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{(1+\sqrt{3}t)}^2-12}{4{(t^2+1)}^2}dt$$$$=\frac{2\sqrt{3}}{9}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1+2\sqrt{3}t+3t^2-12}{{(t^2+1)}^2}dtlet\phi \left(z\right)=\frac{3z^2+2\sqrt{3}z-11}{{(z^2+1)}^2}$$$$polesof\phi ?$$$$\phi \left(z\right)=\frac{3z^2+2\sqrt{3}z-11}{{(z-i)}^2{(z+i)}^2}sothepolesare\stackrel{-}{+}i\left(doubles\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(2-1)!}{\left\{{(z-i)}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$$$={lim}_{z\to i}{\left(\frac{3z^2+2\sqrt{3}z-11}{{(z+i)}^2}\right)}^{\left(1\right)}$$$$={lim}_{z\to i}\frac{(6z+2\sqrt{3}){(z+i)}^2-2(z+i)(3z^2+2\sqrt{3}z-11)}{{(z+i)}^4}$$$$={lim}_{z\to i}\frac{(6z+2\sqrt{3})(z+i)-2(3z^2+2\sqrt{3}z-11)}{{(z+i)}^3}$$$$=\frac{(6i+2\sqrt{3})\left(2i\right)-2(-3+2\sqrt{3}i-11)}{{\left(2i\right)}^3}$$$$=\frac{-12+4\sqrt{3}i+6-4\sqrt{3}i+22}{-8i}=\frac{22-6}{-8i}=\frac{16}{-8i}=\frac{-2}{i}$$$$=2i\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \left(2i\right)=-4\pi \Rightarrow$$$$I=\frac{2\sqrt{3}}{9}(-4\pi )=\frac{-8\pi \sqrt{3}}{9}$$$$★I=\frac{-8\pi \sqrt{3}}{9}★$$$$$$$$$$$$$$

Answered by Ar Brandon last updated on 14/May/20

$$\frac{{\mathrm{x}}^2-3}{({\mathrm{x}}^2-\mathrm{x}+1)}\equiv \frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2-\mathrm{x}+1}+\frac{\mathrm{cx}+\mathrm{d}}{({\mathrm{x}}^2-\mathrm{x}+1)}$$$$=\frac{1}{{\mathrm{x}}^2-\mathrm{x}+1}+\frac{\mathrm{x}-4}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}$$$$\Rightarrow \mathrm{K}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2-3}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{\mathrm{x}}^2-\mathrm{x}+1}\mathrm{dx}+{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{x}-4}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}$$$$\Rightarrow \mathrm{K}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{(\mathrm{x}-\frac{1}{2})}^2+\frac{3}{4}}\mathrm{dx}+\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2\mathrm{x}-1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}-\frac{7}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}$$$$\Rightarrow \mathrm{K}=\frac{2\sqrt{3}}{3}\arctan \left[\frac{2\sqrt{3}}{3}(\mathrm{x}-\frac{1}{2})\right]-\frac{1}{2({\mathrm{x}}^2-\mathrm{x}+1)}-\frac{7}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}$$$$\mathrm{J}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{({\mathrm{x}}^2+1)-({\mathrm{x}}^2-1)}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}$$$$2\mathrm{J}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2+1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2-1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}$$$$\Rightarrow 2\mathrm{J}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2+1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1-\frac{1}{{\mathrm{x}}^2}}{{(\mathrm{x}-1+\frac{1}{\mathrm{x}})}^2}\mathrm{dx}$$$$\Rightarrow 2\mathrm{J}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2+1}{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}\mathrm{dx}+\frac{\mathrm{x}}{{\mathrm{x}}^2-\mathrm{x}+1}$$

Answered by Ar Brandon last updated on 14/May/20

$$\frac{x^2-3}{{(x^2-x+1)}^2}\equiv \frac{\mathrm{a}x+\mathrm{b}}{x^2-x+1}+\frac{\mathrm{c}x+\mathrm{d}}{{(x^2-x+1)}^2}$$$$=\frac{1}{x^2-x+1}+\frac{x-4}{{(x^2-x+1)}^2}$$$$\Rightarrow \mathrm{K}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2-3}{{(x^2-x+1)}^2}dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{x^2-x+1}\mathrm{d}x+{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x-4}{{(x^2-x+1)}^2}\mathrm{d}x$$$$\Rightarrow \mathrm{K}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{(x-\frac{1}{2})}^2+\frac{3}{4}}\mathrm{d}x+\mathrm{L}$$$$\Rightarrow \mathrm{K}=\frac{2\sqrt{3}}{3}{\left[\arctan \frac{2\sqrt{3}}{3}(x-\frac{1}{2})\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+\mathrm{L}$$$$\Rightarrow \mathrm{K}=\frac{2\sqrt{3}}{3}\pi +\mathrm{L}$$$${\mathrm{L}}_{\mathrm{n}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{A}x+\mathrm{B}}{{(x^2+\mathrm{p}x+\mathrm{q})}^{\mathrm{n}}}\mathrm{d}x=\frac{\mathrm{A}}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2x+\mathrm{p}}{{(x^2+\mathrm{p}x+\mathrm{q})}^{\mathrm{n}}}\mathrm{d}x+(\mathrm{B}-\frac{\mathrm{Ap}}{2}){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{(x^2+\mathrm{p}x+\mathrm{q})}^{\mathrm{n}}}\mathrm{d}x$$$$\Rightarrow {\mathrm{L}}_{\mathrm{n}}=-{\left[\frac{\mathrm{A}}{2(\mathrm{n}-1){(x^2+\mathrm{p}x+\mathrm{q})}^{\mathrm{n}-1}}\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+(\mathrm{B}-\frac{\mathrm{Ap}}{2}){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{[{(x+\frac{\mathrm{p}}{2})}^2+{\left(\sqrt{\mathrm{q}-\frac{{\mathrm{p}}^2}{4}}\right)}^2]}^{\mathrm{n}}}\mathrm{d}x$$$$\mathrm{A}=1,\mathrm{B}=-4,\mathrm{p}=-1,\mathrm{q}=1,\mathrm{n}=2$$$$\Rightarrow \mathrm{L}=-\frac{7}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{[{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2]}^2}\mathrm{d}x$$$${\mathrm{J}}_{\mathrm{n}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{(x^2+{\mathrm{u}}^2)}^{\mathrm{n}}}\mathrm{d}x$$$$\Rightarrow {\mathrm{J}}_{\mathrm{n}-1}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{(x^2+{\mathrm{u}}^2)}^{\mathrm{n}-1}}\mathrm{d}x={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+{\mathrm{u}}^2}{{(x^2+{\mathrm{u}}^2)}^{\mathrm{n}}}\mathrm{d}x$$$$\Rightarrow {\mathrm{J}}_{\mathrm{n}-1}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}x\cdot \frac{2x}{{(x^2+{\mathrm{u}}^2)}^{\mathrm{n}}}\mathrm{d}x+{\mathrm{u}}^2{\mathrm{J}}_{\mathrm{n}}$$$$\Rightarrow {2\mathrm{J}}_{\mathrm{n}-1}={[x\int \frac{2x}{{(x^2+{\mathrm{u}}^2)}^{\mathrm{n}}}\mathrm{d}x]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\int \frac{2x}{x^2+{\mathrm{u}}^2}\mathrm{d}x\mathrm{d}x+{2\mathrm{u}}^2{\mathrm{J}}_{\mathrm{n}}$$$$\Rightarrow {2\mathrm{J}}_{\mathrm{n}-1}=-{\left[\frac{x}{(\mathrm{n}-1){(x^2+{\mathrm{u}}^2)}^{\mathrm{n}-1}}\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+\frac{1}{\mathrm{n}-1}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{(x^2+{\mathrm{u}}^2)}^{\mathrm{n}-1}}\mathrm{d}x+{2\mathrm{u}}^2{\mathrm{J}}_{\mathrm{n}}$$$$\Rightarrow {2\mathrm{J}}_{\mathrm{n}-1}=\frac{1}{\mathrm{n}-1}{\mathrm{J}}_{\mathrm{n}-1}+{2\mathrm{u}}^2{\mathrm{J}}_{\mathrm{n}}$$$$\Rightarrow {\mathrm{J}}_{\mathrm{n}}=\frac{2\mathrm{n}-3}{{2\mathrm{u}}^2(\mathrm{n}-1)}{\mathrm{J}}_{\mathrm{n}-1}$$$$\mathrm{u}=\frac{\sqrt{3}}{2},\mathrm{n}=2$$$$\Rightarrow -\frac{2}{7}\mathrm{L}=\frac{2}{3}{\mathrm{J}}_1$$$${\mathrm{J}}_1={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\mathrm{d}x$$$${\mathrm{J}}_1=\frac{2\sqrt{3}}{3}{\left[\arctan \frac{2\sqrt{3}}{3}(x-\frac{1}{2})\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{2\sqrt{3}}{3}\pi$$$$\Rightarrow \mathrm{L}=-\frac{14\sqrt{3}}{9}\pi$$$$\Rightarrow \mathrm{K}=\frac{2\sqrt{3}}{3}\pi -\frac{14\sqrt{3}}{9}\pi$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2-3}{{(x^2-x+1)}^2}dx=-\frac{8\sqrt{3}}{9}\pi$$

Commented by mathmax by abdo last updated on 14/May/20

$$thankyousirforthishardwork.$$

Commented by Ar Brandon last updated on 15/May/20

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