Question Number 31107 by abdo imad last updated on 02/Mar/18
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }\:\:{with}\:{n}>\mathrm{1}. \\ $$
Commented by abdo imad last updated on 06/Mar/18
$${let}\:{put}\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{{n}} }\:\:{with}\:{n}\geqslant\mathrm{2}\:{we}\:{have} \\ $$$${x}^{\mathrm{2}} \:+{x}+\mathrm{1}={x}^{\mathrm{2}} \:+\mathrm{2}{x}\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\:{the}\:{ch}. \\ $$$${x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}\:\Rightarrow\:\:{I}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)^{{n}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$$$=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }=\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }\:{but} \\ $$$${we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }=\pi\:\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }\:\Rightarrow \\ $$
Commented by abdo imad last updated on 06/Mar/18
$$\Rightarrow\:\:{A}_{{n}} =\pi\sqrt{\mathrm{3}}\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }\:. \\ $$