Question Number 173068 by mnjuly1970 last updated on 06/Jul/22
Answered by Mathspace last updated on 06/Jul/22
![Φ=∫_0 ^∞ (∫_0 ^∞ e^(−x) sin(x+y)dx)e^(−y) dy but ∫_0 ^∞ e^(−x) sin(x+y)dx =Im(∫_0 ^∞ e^(−x+i(x+y)) dx)and ∫_0 ^∞ e^(−x+i(x+y)) dx =e^(iy) ∫_0 ^∞ e^((−1+i)x) dx =e^(iy) [(1/(−1+i))e^((−1+i)x) ]_0 ^∞ =e^(iy) (−(1/(−1+i)))=(e^(ih) /(1−i)) =(((1+i)e^(iy) )/2)=(((1+i)(cosy+isiny))/2) =(1/2){cosy+isiny+icosy−siny} ⇒Im(...)=(1/2)(cosy+siny)⇒ Φ=∫_0 ^∞ (1/2)(cosy+siny)e^(−y) dy 2Φ=Re(∫_0 ^∞ e^(−y+iy) dy)+Im(∫_0 ^∞ e^(−y+iy) dy) or ∫_0 ^∞ e^((−1+i)y) dy =[(1/(−1+i))e^((−1+i)y) ]_0 ^∞ =(1/(1−i)) =((1+i)/2) ⇒Re(....)=(1/2) and Im(...)=(1/2) ⇒ 2Φ=1 ⇒Φ=(1/2)](https://www.tinkutara.com/question/Q173073.png)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {sin}\left({x}+{y}\right){dx}\right){e}^{−{y}} {dy} \\ $$$${but}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {sin}\left({x}+{y}\right){dx} \\ $$$$={Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{x}+{i}\left({x}+{y}\right)} {dx}\right){and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}+{i}\left({x}+{y}\right)} {dx} \\ $$$$={e}^{{iy}} \int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−\mathrm{1}+{i}\right){x}} {dx} \\ $$$$={e}^{{iy}} \left[\frac{\mathrm{1}}{−\mathrm{1}+{i}}{e}^{\left(−\mathrm{1}+{i}\right){x}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$={e}^{{iy}} \left(−\frac{\mathrm{1}}{−\mathrm{1}+{i}}\right)=\frac{{e}^{{ih}} }{\mathrm{1}−{i}} \\ $$$$=\frac{\left(\mathrm{1}+{i}\right){e}^{{iy}} }{\mathrm{2}}=\frac{\left(\mathrm{1}+{i}\right)\left({cosy}+{isiny}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cosy}+{isiny}+{icosy}−{siny}\right\} \\ $$$$\Rightarrow{Im}\left(…\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({cosy}+{siny}\right)\Rightarrow \\ $$$$\Phi=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}}\left({cosy}+{siny}\right){e}^{−{y}} {dy} \\ $$$$\mathrm{2}\Phi={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{y}+{iy}} {dy}\right)+{Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{y}+{iy}} {dy}\right) \\ $$$${or}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{\left(−\mathrm{1}+{i}\right){y}} {dy} \\ $$$$=\left[\frac{\mathrm{1}}{−\mathrm{1}+{i}}{e}^{\left(−\mathrm{1}+{i}\right){y}} \right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{\mathrm{1}−{i}} \\ $$$$=\frac{\mathrm{1}+{i}}{\mathrm{2}}\:\Rightarrow{Re}\left(….\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:{Im}\left(…\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{2}\Phi=\mathrm{1}\:\Rightarrow\Phi=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 06/Jul/22
$$\mathrm{bravo}\:\mathrm{sir}.. \\ $$
Commented by Mathspace last updated on 06/Jul/22
$${you}\:{are}\:{welcome} \\ $$
Answered by Eulerian last updated on 07/Jul/22
$$\: \\ $$$$\:\Omega\:=\:\Im\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\mathrm{e}^{−\left(\mathrm{x}+\mathrm{y}\right)} \:\mathrm{e}^{\mathrm{i}\left(\mathrm{x}+\mathrm{y}\right)} \:\mathrm{dxdy} \\ $$$$\:\:\:\:\:\:=\:\Im\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\mathrm{e}^{−\left(\mathrm{1}−\mathrm{i}\right)\mathrm{x}} \:\mathrm{e}^{−\left(\mathrm{1}−\mathrm{i}\right)\mathrm{y}} \:\mathrm{dxdy} \\ $$$$\:\:\:\:\:\:=\:\Im\:\left(\int_{\mathrm{0}} ^{\:\infty} \mathrm{e}^{−\left(\mathrm{1}−\mathrm{i}\right)\mathrm{y}} \:\mathrm{dy}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\Im\:\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{i}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$