Question Number 219449 by Lukos last updated on 25/Apr/25
![L{sinx}=∫_0 ^∞ e^(−sx) sinx dx=∫_0 ^∞ e^(−sx) ((e^(ix) −e^(−ix) )/(2i))dx =(1/(2i))[∫_0 ^∞ e^(−(s−i)x) dx −∫_0 ^∞ e^(−(s+i)x) dx] =(1/(2i))[((−1)/(s−i))e^(−(s−i)x) +(1/(s+i))e^(−(s+i)x) ]_0 ^∞ =(1/(2i))[(1/(s−i))−(1/(s+i))]=(1/(2i))×((s+i−s+i)/((s−i)(s+i)))=(1/(2i))×((2i)/(s^2 +1))=(1/(s^2 +1))](https://www.tinkutara.com/question/Q219449.png)
$${L}\left\{{sinx}\right\}=\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} {sinx}\:{dx}=\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} \frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\int_{\mathrm{0}} ^{\infty} {e}^{−\left({s}−{i}\right){x}} {dx}\:\:−\int_{\mathrm{0}} ^{\infty} {e}^{−\left({s}+{i}\right){x}} {dx}\right]\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\frac{−\mathrm{1}}{{s}−{i}}{e}^{−\left({s}−{i}\right){x}} +\frac{\mathrm{1}}{{s}+{i}}{e}^{−\left({s}+{i}\right){x}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\frac{\mathrm{1}}{{s}−{i}}−\frac{\mathrm{1}}{{s}+{i}}\right]=\frac{\mathrm{1}}{\mathrm{2}{i}}×\frac{{s}+{i}−{s}+{i}}{\left({s}−{i}\right)\left({s}+{i}\right)}=\frac{\mathrm{1}}{\mathrm{2}{i}}×\frac{\mathrm{2}{i}}{{s}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$