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Question Number 38468 by maxmathsup by imad last updated on 25/Jun/18

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(2x\right)sh\left(3x\right)}{4+x^2}dx$$

Commented by math khazana by abdo last updated on 27/Jun/18

$$I=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sh\left(3x\right)e^{ix}}{4+x^2}dx\right)let$$$$A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sh\left(3x\right)e^{ix}}{4+x^2}dx$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{(e^{3x}-e^{-3x})e^{ix}}{x^2+4}dx$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{(3+i)x}-e^{(-3+i)x}}{x^2+4}dx$$$$let\phi \left(z\right)=\frac{e^{(3+i)z}-e^{(-3+i)z}}{z^2+4}thepolesof\phi are$$$$2iand-2i$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,2i)but$$$$\phi \left(z\right)=\frac{e^{(3+i)z}-e^{(-3+i)z}}{(z-2i)(z+2i)}\Rightarrow$$$$Res(\phi ,2i)=\frac{e^{(3+i)\left(2i\right)}-e^{(-3+i)\left(2i\right)}}{4i}$$$$=\frac{e^{-2+6i}-e^{-2-6i}}{4i}=\frac{e^{-2}}{4i}2isin\left(6\right)=\frac{e^{-2}}{2}sin\left(6\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-2}}{2}sin\left(6\right)$$$$=i\pi e^{-2}sin\left(6\right)\Rightarrow A=i\frac{\pi }{2}{e}^{-2}sin\left(6\right)but$$$$I=Im\left(A\right)\Rightarrow I=\frac{\pi }{2e^2}sin\left(6\right).$$

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