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Question Number 147680 by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{find}\mathrm{by}\mathrm{residus}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2-\mathrm{x}+1)}^3}\mathrm{dx}$$

Commented by mathmax by abdo last updated on 23/Jul/21

$$\mathrm{the}\mathrm{Q}\mathrm{is}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2-\mathrm{x}+1)}^3}\mathrm{dx}$$

Commented by mathmax by abdo last updated on 23/Jul/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{\mathrm{x}}^2-\mathrm{x}+\mathrm{a}}\mathrm{dx}\mathrm{with}\mathrm{a}>\frac{1}{4}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2-\mathrm{x}+\mathrm{a})}^2}\Rightarrow {\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2({\mathrm{x}}^2-\mathrm{x}+\mathrm{a})\cos \left(2\mathrm{a}\right)}{{({\mathrm{x}}^2-\mathrm{x}+\mathrm{a})}^4}\mathrm{dx}$$$$=2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2-\mathrm{x}+\mathrm{a})}^3}\mathrm{dx}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2-\mathrm{x}+1)}^3}\mathrm{dx}=\frac{1}{2}{\mathrm{f}}^{{}^{″}}\left(1\right)$$$$\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{ix}}}{{\mathrm{x}}^2-\mathrm{x}+\mathrm{a}}\mathrm{dx}\right)\mathrm{let}\mathrm{\Lambda }\left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{{\mathrm{z}}^2-\mathrm{z}+\mathrm{a}}\mathrm{poles}?$$$$\mathrm{\Delta }=1-4\mathrm{a}<0\Rightarrow {\mathrm{z}}_1=\frac{1+\mathrm{i}\sqrt{4\mathrm{a}-1}}{2}\mathrm{and}{\mathrm{z}}_2=\frac{1-\mathrm{i}\sqrt{4\mathrm{a}-1}}{2}$$$$\mathrm{\Lambda }\left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}$$$${\int }_{\mathrm{R}}\mathrm{\Lambda }\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{\Lambda },{\mathrm{z}}_1)=\frac{{\mathrm{e}}^{{2\mathrm{iz}}_1}}{{\mathrm{z}}_1-{\mathrm{z}}_2}$$$$=\frac{{\mathrm{e}}^{2\mathrm{i}\left(\frac{1+\mathrm{i}\sqrt{4\mathrm{a}-1}}{2}\right)}}{\mathrm{i}\sqrt{4\mathrm{a}-1}}=\frac{{\mathrm{e}}^{\mathrm{i}-\sqrt{4\mathrm{a}-1}}}{\mathrm{i}\sqrt{4\mathrm{a}-1}}=\frac{{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}(\cos \left(1\right)+\mathrm{isin}\left(1\right))}{\mathrm{i}\sqrt{4\mathrm{a}-1}}\Rightarrow$$$${\int }_{\mathrm{R}}\mathrm{\Lambda }\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}(\cos \left(1\right)+\mathrm{isin}\left(1\right)}{\mathrm{i}\sqrt{4\mathrm{a}-1}}$$$$=\frac{2\pi }{\sqrt{4\mathrm{a}-1}}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}(\cos \left(1\right)+\mathrm{isin}\left(1\right))\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{2\pi \cos \left(1\right)}{\sqrt{4\mathrm{a}-1}}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=2\pi \cos \left(1\right){(4\mathrm{a}-1)}^{-\frac{1}{2}}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}$$$$\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=2\pi \cos \left(1\right)(-2\mathrm{a}{(4\mathrm{a}-1)}^{-\frac{3}{2}}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}$$$$+{(4\mathrm{a}-1)}^{-\frac{1}{2}}(-\frac{4}{2\sqrt{4\mathrm{a}-1}}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}})$$$$=2\pi \cos \left(1\right)\{-2\mathrm{a}{(4\mathrm{a}-1)}^{-\frac{3}{2}}-2{(4\mathrm{a}-1)}^{-\frac{3}{2}}\}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}$$$$=-4\pi \cos \left(1\right)\left\{(\mathrm{a}+1){(4\mathrm{a}-1)}^{-\frac{3}{2}}\right\}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}\Rightarrow$$$${\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)=-4\pi \cos \left(1\right)\{{(4\mathrm{a}-1)}^{-\frac{3}{2}}-6(\mathrm{a}+1){(4\mathrm{a}-1)}^{-\frac{5}{2}}){\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}$$$$-\frac{2}{\sqrt{4\mathrm{a}-1}}{\mathrm{e}}^{-\sqrt{4\mathrm{a}-1}}(\mathrm{a}+1){(4\mathrm{a}-1)}^{-\frac{3}{2}}\}\Rightarrow$$$${\mathrm{f}}^{{}^{″}}\left(1\right)=-4\pi \cos \left(1\right)\{(3^{-\frac{3}{2}}-12.3^{-\frac{5}{2}}){\mathrm{e}}^{-\sqrt{3}}-\frac{2}{\sqrt{3}}{\mathrm{e}}^{-\sqrt{3}}\left(2\right).3^{-\frac{3}{2}}\}$$$$\mathrm{and}\mathrm{\Psi }=\frac{1}{2}{\mathrm{f}}^{\left(2\right)}\left(1\right)$$$$$$

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