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Question Number 156362 by MathSh last updated on 10/Oct/21

Answered by mindispower last updated on 10/Oct/21

$$\mid {\mathrm{\Omega }}_1+i{\mathrm{\Omega }}_2{\mid }^2={\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_2^2$$$${\mathrm{\Omega }}_1+i{\mathrm{\Omega }}_2={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{i(e^{2x}+e^{-2x})}{e}^{x-{sh}^2\left(x\right)}$$$$={\int }_0^{\mathrm{\infty }}{e}^{i(e^{2x}+e^{-2x})}{e}^{-{sh}^2\left(x\right)}(e^x+e^{-x})dx$$$$letu=sh\left(t\right)$$$$=du=ch\left(t\right)dt$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{i(4{sh}^2\left(x\right)+1)}{e}^{-u^2}du$$$$=2e^i{\int }_0^{\mathrm{\infty }}{e}^{(-1+4i)u^2}du$$$${\int }_0^{\mathrm{\infty }}{e}^{{au}^2}du=f\left(a\right),Re\left(a\right)<0$$$$f\left(a\right)=\frac{\sqrt{\pi }}{2\sqrt{a}}$$$${\mathrm{\Omega }}_1+i{\mathrm{\Omega }}_2=\frac{e^i}{{(-1+4i)}^{\frac{1}{2}}}\sqrt{\pi }$$$$\mid {\mathrm{\Omega }}_1+i{\mathrm{\Omega }}_2\mid =\frac{\sqrt{\pi }}{\sqrt{\sqrt{17}}}\Rightarrow {\mathrm{\Omega }}_1^2+{\mathrm{\Omega }}_2^2=\frac{\pi }{\sqrt{17}}$$$$$$$$$$

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

$$withepleasur$$

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