Question Number 48043 by maxmathsup by imad last updated on 18/Nov/18 | ||
$${let}\:{f}\left(\alpha\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\alpha{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}\:{dx} \\ $$$$\left.\mathrm{1}\right){calculate}\:{f}\left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{2}{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx}\:. \\ $$ | ||
Commented by maxmathsup by imad last updated on 20/Nov/18 | ||
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left(\alpha\right)={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\alpha{x}^{\mathrm{3}} } }{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx}\right)\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\alpha{z}^{\mathrm{3}} } }{{z}^{\mathrm{2}} +\mathrm{8}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{e}^{{i}\alpha{z}^{\mathrm{3}} } }{{z}^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{2}}{i}\right)^{\mathrm{2}} }\:=\frac{{e}^{{i}\alpha{z}^{\mathrm{3}} } }{\left({z}−\mathrm{2}\sqrt{\mathrm{2}}{i}\right)\left({z}+\mathrm{2}\sqrt{\mathrm{2}}{i}\right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\mathrm{2}\sqrt{\mathrm{2}}{i}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\:{but} \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}\sqrt{\mathrm{2}}} \:\:\:\:\left({z}−\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\varphi\left({z}\right)\:=\frac{{e}^{{i}\alpha\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } }{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:=\frac{{e}^{−{i}\alpha\left(\mathrm{16}{i}\sqrt{\mathrm{2}}\right)} }{\mathrm{4}{i}\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} }{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:.\frac{{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} }{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:=\frac{\pi\:{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} }{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} . \\ $$$$\left.\mathrm{2}\right)\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx}\:={f}\left(\mathrm{2}\right)\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:{e}^{\mathrm{32}\sqrt{\mathrm{2}}} . \\ $$ | ||