Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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 )

$${we}\:{define}\:{L}\left({f}\left({x}\right)\right)\:=\int_{\mathrm{0}} ^{\infty} {f}\left({t}\right)\:{e}^{−{xt}} {dt}\:\:{with}\:\:{x}\geqslant\mathrm{0}\:{with}\:\int_{\mathrm{0}} ^{\infty} /{f}\left({t}\right)/{dt}\:{convergent} \\ $$$${L}\left({e}^{−{x}^{\mathrm{2}} } \right)\:\:=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } .\:{e}^{−{xt}} \:{dt}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\:{t}^{\mathrm{2}} +{xt}\right)} {dt}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({t}^{\mathrm{2}} +\mathrm{2}\frac{{x}}{\mathrm{2}}{t}+\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right)} {dt} \\ $$$$=\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({t}+\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} } {dt}\:\:\:{and}\:{by}\:{the}\:{changement}\:\:\alpha\:={t}+\frac{{x}}{\mathrm{2}} \\ $$$${L}\left(\:{e}^{−{x}^{\mathrm{2}} } \right)\:\:=\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\:\int_{\frac{{x}}{\mathrm{2}}} ^{\propto} \:\:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\:=\:\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\:\left(\:\:\int_{\frac{{x}}{\mathrm{2}}} ^{\mathrm{0}} \:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\:\:+\:\int_{\mathrm{0}} ^{\propto} \:\:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha\:\:\right) \\ $$$${but}\:\:\int_{\mathrm{0}} ^{\propto} \:\:{e}^{−\alpha^{\mathrm{2}} } {d}\alpha=\frac{\sqrt{\pi}}{\mathrm{2}}\:\:\:\: \\ $$$$\Rightarrow\:\:\:{L}\left(\:{e}^{−{x}^{\mathrm{2}} } \right)\:\:=\:\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:\:\left(\:\:\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:\:−\:\:\int_{\mathrm{0}} ^{\frac{{x}}{\mathrm{2}}} \:\:{e}^{−\alpha^{\mathrm{2}} } \:{d}\alpha\:\:\:\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com