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Question Number 96658 by mathmax by abdo last updated on 03/Jun/20
determine L(e^(−x^2 −x) ) with L laplace transform
$$\mathrm{determine}\:\mathrm{L}\left(\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{x}} \right)\:\:\:\mathrm{with}\:\mathrm{L}\:\mathrm{laplace}\:\mathrm{transform} \\ $$
Answered by mathmax by abdo last updated on 04/Jun/20
L(e^(−x^2 −x) ) =∫_0 ^∞ e^(−t^2 −t) e^(−xt) dt =∫_0 ^∞ e^(−t^2 −(x+1)t) dt =∫_0 ^∞ e^(−(t^2 +2t×((x+1)/2) +(((x+1)^2 )/4)−(((x+1)^2 )/4))) =e^(((x+1)^2 )/4) ∫_0 ^∞ e^(−{t+((x+1)/2)}^2 ) dt =_(t+((x+1)/2)=u) e^(((x+1)^2 )/4) ∫_((x+1)/2) ^∞ e^(−u^2 ) du =e^(((x+1)^2 )/4) { ∫_((x+1)/2) ^0 e^(−u^2 ) du +∫_0 ^∞ e^(−u^2 ) du} =e^(((x+1)^2 )/4) {(π/2)−∫_0 ^((x+1)/2) e^(−u^2 ) du} ⇒ L(e^(−x^2 −x) ) =(π/2)e^(((x+1)^2 )/4) −e^(((x+1)^2 )/4) ∫_0 ^((x+1)/2) e^(−u^2 ) du
$$\mathrm{L}\left(\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{x}} \right)\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} −\mathrm{t}} \:\mathrm{e}^{−\mathrm{xt}} \:\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} −\left(\mathrm{x}+\mathrm{1}\right)\mathrm{t}} \:\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}×\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}\:+\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\right)} \:=\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\left\{\mathrm{t}+\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}\right\}^{\mathrm{2}} } \mathrm{dt} \\ $$$$=_{\mathrm{t}+\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}=\mathrm{u}} \:\:\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \:\int_{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\:=\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \:\left\{\:\int_{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}} ^{\mathrm{0}} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\:+\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\right\} \\ $$$$=\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \left\{\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\right\}\:\Rightarrow \\ $$$$\mathrm{L}\left(\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{x}} \right)\:=\frac{\pi}{\mathrm{2}}\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \:−\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\: \\ $$
Commented by mathmax by abdo last updated on 04/Jun/20
sorry L(e^(−x^2 −x) ) = e^(((x+1)^2 )/2) {((√π)/2) −∫_0 ^((x+1)/2) e^(−u^2 ) du} ⇒ L(e^(−x^2 −x) ) =((√π)/2) e^(((x+1)^2 )/4) −e^(((x+1)^2 )/4) ∫_0 ^((x+1)/2) e^(−u^2 ) du
$$\mathrm{sorry}\:\mathrm{L}\left(\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{x}} \right)\:=\:\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}} \left\{\frac{\sqrt{\pi}}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\right\}\:\Rightarrow \\ $$$$\mathrm{L}\left(\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{x}} \right)\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \:−\mathrm{e}^{\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \:\mathrm{du} \\ $$

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