Question Number 38197 by maxmathsup by imad last updated on 22/Jun/18
$${find}\:{a}\:{simple}\:{form}\:{of}\:{L}\left({e}^{−\sqrt{{x}}} \right)\:\:{L}\:{is}\:{laplace}\:{transform} \\ $$
Commented by math khazana by abdo last updated on 25/Jun/18
![L(e^(−(√x)) )= ∫_0 ^∞ f(t)e^(−xt) dt=∫_0 ^∞ e^(−(√t) −xt) dt changement (√t)=u give L(e^(−(√x)) )= ∫_0 ^∞ e^(−u−xu^2 ) 2u du =−(1/x) ∫_0 ^∞ −2xu e^(−xu^2 ) e^(−u) du −(1/x){[ e^(−xu^2 ) e^(−u) ]_0 ^(+∞) +∫_0 ^∞ e^(−u) e^(−xu^2 ) } =(1/x) −(1/x) ∫_0 ^∞ e^(−(xu^2 +u)) du but ∫_0 ^∞ e^(−(xu^2 +u)) du=∫_0 ^∞ e^(−{((√x)u)^2 +2 (((√x)u)/(√x)) + (1/x) −(1/x)}) du = ∫_0 ^∞ e^(−{ ((√x)u +(1/(√x)))^2 } +(1/x)) du =e^(1/x) ∫_0 ^∞ e^(−{ ((√x)u+(1/(√x)))^2 }) du =e^(1/x) ∫_(1/(√x)) ^(+∞) e^(−t^2 ) (dt/(√x)) ( chang.(√x)u +(1/(√x)) =t) =(e^(1/x) /(√x)) ∫_(1/(√x)) ^(+∞) e^(−t^2 ) dt ⇒ L(e^(−(√x)) ) =(1/x) −(e^(1/(√x)) /(x(√x))) ∫_(1/(√x)) ^(+∞) e^(−t^2 ) dt .](Q38438.png)
$${L}\left({e}^{−\sqrt{{x}}} \right)=\:\int_{\mathrm{0}} ^{\infty} \:\:{f}\left({t}\right){e}^{−{xt}} {dt}=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\sqrt{{t}}\:\:−{xt}} \:{dt} \\ $$$${changement}\:\sqrt{{t}}={u}\:{give} \\ $$$${L}\left({e}^{−\sqrt{{x}}} \right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{e}^{−{u}−{xu}^{\mathrm{2}} } \:\mathrm{2}{u}\:{du} \\ $$$$=−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:−\mathrm{2}{xu}\:{e}^{−{xu}^{\mathrm{2}} } \:\:{e}^{−{u}} {du}\: \\ $$$$−\frac{\mathrm{1}}{{x}}\left\{\left[\:{e}^{−{xu}^{\mathrm{2}} } \:{e}^{−{u}} \right]_{\mathrm{0}} ^{+\infty} \:\:+\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}} {e}^{−{xu}^{\mathrm{2}} } \right\} \\ $$$$=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({xu}^{\mathrm{2}} \:+{u}\right)} {du}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left({xu}^{\mathrm{2}} \:+{u}\right)} {du}=\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left\{\left(\sqrt{{x}}{u}\right)^{\mathrm{2}} \:+\mathrm{2}\:\frac{\sqrt{{x}}{u}}{\sqrt{{x}}}\:\:+\:\frac{\mathrm{1}}{{x}}\:\:−\frac{\mathrm{1}}{{x}}\right\}} {du} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left\{\:\left(\sqrt{{x}}{u}\:+\frac{\mathrm{1}}{\sqrt{{x}}}\right)^{\mathrm{2}} \:\right\}\:+\frac{\mathrm{1}}{{x}}} \:\:{du} \\ $$$$={e}^{\frac{\mathrm{1}}{{x}}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left\{\:\left(\sqrt{{x}}{u}+\frac{\mathrm{1}}{\sqrt{{x}}}\right)^{\mathrm{2}} \right\}} {du} \\ $$$$={e}^{\frac{\mathrm{1}}{{x}}} \:\:\:\int_{\frac{\mathrm{1}}{\sqrt{{x}}}} ^{+\infty} \:\:\:\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{dt}}{\sqrt{{x}}}\:\:\left(\:\:{chang}.\sqrt{{x}}{u}\:+\frac{\mathrm{1}}{\sqrt{{x}}}\:={t}\right) \\ $$$$=\frac{{e}^{\frac{\mathrm{1}}{{x}}} }{\sqrt{{x}}}\:\int_{\frac{\mathrm{1}}{\sqrt{{x}}}} ^{+\infty} \:\:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:\Rightarrow \\ $$$${L}\left({e}^{−\sqrt{{x}}} \right)\:=\frac{\mathrm{1}}{{x}}\:−\frac{{e}^{\frac{\mathrm{1}}{\sqrt{{x}}}} }{{x}\sqrt{{x}}}\:\:\:\int_{\frac{\mathrm{1}}{\sqrt{{x}}}} ^{+\infty} \:\:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:. \\ $$$$ \\ $$