Question Number 38728 by maxmathsup by imad last updated on 28/Jun/18
$${find}\:{L}\:\left(\:\frac{{e}^{−\frac{{x}}{{a}}} }{{a}}\right)\:\:{with}\:{a}\neq\mathrm{0}\:\:{and}\:{L}\:{laplace}\:{transfom}. \\ $$
Commented by abdo mathsup 649 cc last updated on 29/Jun/18
![L((1/a) e^(−(x/a)) )= ∫_0 ^∞ (1/a)e^(−(t/a)) e^(−xt) dt =(1/a)∫_0 ^∞ e^((−x−(1/a))t) dt =(1/a) [ (1/(−x−(1/a))) e^((−x−(1/a))t) ]_0 ^(+∞) =(1/a) (1/(x +(1/a))) = (1/(ax+1)) .](https://www.tinkutara.com/question/Q38777.png)
$${L}\left(\frac{\mathrm{1}}{{a}}\:{e}^{−\frac{{x}}{{a}}} \right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{a}}{e}^{−\frac{{t}}{{a}}} \:{e}^{−{xt}} {dt} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{\left(−{x}−\frac{\mathrm{1}}{{a}}\right){t}} {dt} \\ $$$$=\frac{\mathrm{1}}{{a}}\:\:\left[\:\frac{\mathrm{1}}{−{x}−\frac{\mathrm{1}}{{a}}}\:{e}^{\left(−{x}−\frac{\mathrm{1}}{{a}}\right){t}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{{a}}\:\frac{\mathrm{1}}{{x}\:+\frac{\mathrm{1}}{{a}}}\:=\:\frac{\mathrm{1}}{{ax}+\mathrm{1}}\:. \\ $$