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Question Number 26057 by abdo imad last updated on 18/Dec/17
find a integral form of L( e^(−x^2 ) ) L(f) means laplace transform of f .
$${find}\:{a}\:{integral}\:{form}\:{of}\:\:{L}\left(\:\:{e}^{−{x}^{\mathrm{2}} } \:\right) \\ $$$${L}\left({f}\right)\:{means}\:{laplace}\:{transform}\:{of}\:{f}\:. \\ $$
Commented by abdo imad last updated on 21/Dec/17
we define L(f(x)) =∫_0 ^∞ f(t) e^(−xt) dt with x≥0 with ∫_0 ^∞ /f(t)/dt convergent L(e^(−x^2 ) ) = ∫_0 ^∞ e^(−t^2 ) . e^(−xt) dt =∫_0 ^∞ e^(−( t^2 +xt)) dt =∫_0 ^∞ e^(−(t^2 +2(x/2)t+ (x^2 /4) −(x^2 /4))) dt = e^(x^2 /4) ∫_0 ^∞ e^(−(t+(x/2))^2 ) dt and by the changement α =t+(x/2) L( e^(−x^2 ) ) = e^(x^2 /4) ∫_(x/2) ^∝ e^(−α^2 ) dα = e^(x^2 /4) ( ∫_(x/2) ^0 e^(−α^2 ) dα + ∫_0 ^∝ e^(−α^2 ) dα ) but ∫_0 ^∝ e^(−α^2 ) dα=((√π)/2) ⇒ L( e^(−x^2 ) ) = e^(x^2 /4) ( ((√π)/2) − ∫_0 ^(x/2) e^(−α^2 ) dα )
$${we}\:{define}\:{L}\left({f}\left({x}\right)\right)\:=\int_{\mathrm{0}} ^{\infty} {f}\left({t}\right)\:{e}^{−{xt}} {dt}\:\:{with}\:\:{x}\geqslant\mathrm{0}\:{with}\:\int_{\mathrm{0}} ^{\infty} /{f}\left({t}\right)/{dt}\:{convergent} \\ $$$${L}\left({e}^{−{x}^{\mathrm{2}} } \right)\:\:=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } .\:{e}^{−{xt}} \:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\:{t}^{\mathrm{2}} +{xt}\right)} {dt}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({t}^{\mathrm{2}} +\mathrm{2}\frac{{x}}{\mathrm{2}}{t}+\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right)} {dt} \\ $$$$=\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({t}+\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} } {dt}\:\:\:{and}\:{by}\:{the}\:{changement}\:\:\alpha\:={t}+\frac{{x}}{\mathrm{2}} \\ $$$${L}\left(\:{e}^{−{x}^{\mathrm{2}} } \right)\:\:=\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\:\int_{\frac{{x}}{\mathrm{2}}} ^{\propto} \:\:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\:=\:\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\:\left(\:\:\int_{\frac{{x}}{\mathrm{2}}} ^{\mathrm{0}} \:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\:\:+\:\int_{\mathrm{0}} ^{\propto} \:\:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\:\:\right) \\ $$$${but}\:\:\int_{\mathrm{0}} ^{\propto} \:\:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha=\frac{\sqrt{\pi}}{\mathrm{2}}\:\:\:\: \\ $$$$\Rightarrow\:\:\:{L}\left(\:{e}^{−{x}^{\mathrm{2}} } \right)\:\:=\:\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\:\left(\:\:\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:\:−\:\:\int_{\mathrm{0}} ^{\frac{{x}}{\mathrm{2}}} \:\:{e}^{−\alpha^{\mathrm{2}} } \:{d}\alpha\:\:\:\right) \\ $$$$ \\ $$

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