Question Number 156362 by MathSh last updated on 10/Oct/21
Answered by mindispower last updated on 10/Oct/21
$$\mid\Omega_{\mathrm{1}} +{i}\Omega_{\mathrm{2}} \mid^{\mathrm{2}} =\Omega_{\mathrm{1}} ^{\mathrm{2}} +\Omega_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Omega_{\mathrm{1}} +{i}\Omega_{\mathrm{2}} =\int_{−\infty} ^{\infty} {e}^{{i}\left({e}^{\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} \right)} {e}^{{x}−{sh}^{\mathrm{2}} \left({x}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{{i}\left({e}^{\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} \right)} {e}^{−{sh}^{\mathrm{2}} \left({x}\right)} \left({e}^{{x}} +{e}^{−{x}} \right){dx} \\ $$$${let}\:{u}={sh}\left({t}\right) \\ $$$$={du}={ch}\left({t}\right){dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{{i}\left(\mathrm{4}{sh}^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=\mathrm{2}{e}^{{i}} \int_{\mathrm{0}} ^{\infty} {e}^{\left(−\mathrm{1}+\mathrm{4}{i}\right){u}^{\mathrm{2}} } {du} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{{au}^{\mathrm{2}} } {du}={f}\left({a}\right),{Re}\left({a}\right)<\mathrm{0} \\ $$$${f}\left({a}\right)=\frac{\sqrt{\pi}}{\:\mathrm{2}\sqrt{{a}}} \\ $$$$\Omega_{\mathrm{1}} +{i}\Omega_{\mathrm{2}} =\frac{{e}^{{i}} }{\left(−\mathrm{1}+\mathrm{4}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\sqrt{\pi} \\ $$$$\mid\Omega_{\mathrm{1}} +{i}\Omega_{\mathrm{2}} \mid=\frac{\sqrt{\pi}}{\:\sqrt{\sqrt{\mathrm{17}}}}\Rightarrow\Omega_{\mathrm{1}} ^{\mathrm{2}} +\Omega_{\mathrm{2}} ^{\mathrm{2}} =\frac{\pi}{\:\sqrt{\mathrm{17}}} \\ $$$$ \\ $$$$\:\:\:\: \\ $$
Commented by MathSh last updated on 10/Oct/21
$$\mathrm{Perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mindispower last updated on 11/Oct/21
$${withe}\:{pleasur} \\ $$