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Question Number 193278 by Nimnim111118 last updated on 09/Jun/23

Please Help...!! ∫^( ∞) _( 0) x.e^(−x) .sinx.dx

$${Please}\:{Help}...!! \\ $$$$\:\:\:\:\underset{\:\:\:\:\mathrm{0}} {\int}^{\:\:\infty} {x}.{e}^{−{x}} .{sinx}.{dx}\: \\ $$$$ \\ $$

Answered by qaz last updated on 09/Jun/23

∫_0 ^∞ xe^(−x) sin xdx=−im∫_0 ^∞ xe^(−(1+i)x) dx =−im(1/((1+i)^2 ))=(1/2) −−−−− ∫_0 ^∞ xe^(−x) sin xdx=−DL{e^(−x) sin x}(s=0)=−D(1/((1+s)^2 +1))(s=0) =((2(1+s))/(((1+s)^2 +1)^2 ))(s=0)=(1/2)

$$\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} \mathrm{sin}\:{xdx}=−{im}\int_{\mathrm{0}} ^{\infty} {xe}^{−\left(\mathrm{1}+{i}\right){x}} {dx} \\ $$$$=−{im}\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$−−−−− \\ $$$$\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}} \mathrm{sin}\:{xdx}=−{D}\mathscr{L}\left\{{e}^{−{x}} \mathrm{sin}\:{x}\right\}\left({s}=\mathrm{0}\right)=−{D}\frac{\mathrm{1}}{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} +\mathrm{1}}\left({s}=\mathrm{0}\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}+{s}\right)}{\left(\left(\mathrm{1}+{s}\right)^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\left({s}=\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Nimnim111118 last updated on 10/Jun/23

Thank you sir. But, I dont clearly understand it.

$${Thank}\:{you}\:{sir}. \\ $$$${But},\:{I}\:{dont}\:{clearly}\:{understand}\:{it}. \\ $$

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