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Question Number 143575 by Mathspace last updated on 15/Jun/21

find L(e^(−(√x)) )

$${find}\:{L}\left({e}^{−\sqrt{{x}}} \right) \\ $$

Answered by Dwaipayan Shikari last updated on 15/Jun/21

L(e^(−(√x)) )=∫_0 ^∞ e^(−sx) e^(−(√x)) dx =2∫_0 ^∞ te^(−st^2 −t) dt =2∫_0 ^∞ te^(−s(t+(1/( 2s)))^2 +(1/(4s))) dt =2e^(1/(4s)) ∫_(1/(2s)) ^∞ (z−(1/(2s)))e^(−sz^2 ) dz z^2 =u =e^(1/(4s)) ∫_(1/( 4s^2 )) ^∞ e^(−su) du−(e^(1/(4s)) /s)∫_(1/(2s)) ^∞ e^(−sz^2 ) dz z=(m/( (√s))) =(1/s)e^(1/(4s)) (e^(−(1/(4s))) )−(e^(1/(4s)) /s)∫_(1/(2s(√s))) ^∞ e^(−m^2 ) dm =(1/s)−(e^(1/(4s)) /s)(((√π)/2)−((√π)/2)erf((1/(2s(√s)))))

$$\mathscr{L}\left({e}^{−\sqrt{{x}}} \right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} {e}^{−\sqrt{{x}}} {dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {te}^{−{st}^{\mathrm{2}} −{t}} {dt}\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {te}^{−{s}\left({t}+\frac{\mathrm{1}}{\:\mathrm{2}{s}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}{s}}} {dt}\: \\ $$$$=\mathrm{2}{e}^{\frac{\mathrm{1}}{\mathrm{4}{s}}} \int_{\frac{\mathrm{1}}{\mathrm{2}{s}}} ^{\infty} \left({z}−\frac{\mathrm{1}}{\mathrm{2}{s}}\right){e}^{−{sz}^{\mathrm{2}} } {dz}\:\:\:\:\:\:\:{z}^{\mathrm{2}} ={u} \\ $$$$={e}^{\frac{\mathrm{1}}{\mathrm{4}{s}}} \int_{\frac{\mathrm{1}}{\:\mathrm{4}{s}^{\mathrm{2}} }} ^{\infty} {e}^{−{su}} {du}−\frac{{e}^{\frac{\mathrm{1}}{\mathrm{4}{s}}} }{{s}}\int_{\frac{\mathrm{1}}{\mathrm{2}{s}}} ^{\infty} {e}^{−{sz}^{\mathrm{2}} } {dz}\:\:\:\:{z}=\frac{{m}}{\:\sqrt{{s}}} \\ $$$$=\frac{\mathrm{1}}{{s}}{e}^{\frac{\mathrm{1}}{\mathrm{4}{s}}} \left({e}^{−\frac{\mathrm{1}}{\mathrm{4}{s}}} \right)−\frac{{e}^{\frac{\mathrm{1}}{\mathrm{4}{s}}} }{{s}}\int_{\frac{\mathrm{1}}{\mathrm{2}{s}\sqrt{{s}}}} ^{\infty} {e}^{−{m}^{\mathrm{2}} } {dm} \\ $$$$=\frac{\mathrm{1}}{{s}}−\frac{{e}^{\frac{\mathrm{1}}{\mathrm{4}{s}}} }{{s}}\left(\frac{\sqrt{\pi}}{\mathrm{2}}−\frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left(\frac{\mathrm{1}}{\mathrm{2}{s}\sqrt{{s}}}\right)\right) \\ $$

Commented by mathmax by abdo last updated on 16/Jun/21

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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