Question Number 36853 by tawa tawa last updated on 06/Jun/18
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{laplace}\:\mathrm{of}\:\:\:\:\:\mathrm{L}\left\{\frac{\mathrm{e}^{−\mathrm{at}} \:−\:\mathrm{e}^{−\mathrm{bt}} }{\mathrm{t}}\right\} \\ $$
Commented by prof Abdo imad last updated on 06/Jun/18
![L{ ((e^(−ax) −e^(−bx) )/x)} =∫_0 ^∞ f(t)e^(−xt) dt ∫_0 ^∞ ((e^(−at) −e^(−bt) )/t) e^(−xt) dt =ϕ(x) so if a>0 ,b>0 we have ϕ^′ (x) = ∫_0 ^∞ (∂/∂x){ ((e^(−at) −e^(−bt) )/t) e^(−xt) }dt =−∫_0 ^∞ ( e^(−at) −e^(−bt) )e^(−xt) dt =∫_0 ^∞ (e^(−(b+x)t) −e^(−(a+x)t) )e^(−xt) dt = ∫_0 ^∞ e^(−(b+x)t) dt − ∫_0 ^∞ e^(−(a+x)t) dt =((−1)/(b+x))[ e^(−(b+x)t) ]_(t=0) ^∞ +(1/(a+x))[ e^(−(a+x)t) ]_(t=0) ^∞ = (1/(b+x)) −(1/(a+x)) ⇒ϕ(x)= ln∣b+x∣−ln∣a+x∣ +c =ln∣ ((b+x)/(a+x))∣ +c but c=lim_(x→+∞) {ϕ(x)−ln(((b+x)/(a+x)))} ∃m>0 / ∣ϕ(x)∣<m ∫_0 ^∞ e^(−xt) d t =(m/x) →0(x→+∞) ⇒c =0 and ϕ(x)=ln (((b+x)/(a+x))) .](https://www.tinkutara.com/question/Q36889.png)
$${L}\left\{\:\frac{{e}^{−{ax}} \:−{e}^{−{bx}} }{{x}}\right\}\:=\int_{\mathrm{0}} ^{\infty} \:\:{f}\left({t}\right){e}^{−{xt}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{at}} \:−{e}^{−{bt}} }{{t}}\:{e}^{−{xt}} {dt}\:=\varphi\left({x}\right) \\ $$$${so}\:{if}\:{a}>\mathrm{0}\:,{b}>\mathrm{0}\:\:\:{we}\:{have} \\ $$$$\varphi^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\partial}{\partial{x}}\left\{\:\:\frac{{e}^{−{at}} \:−{e}^{−{bt}} }{{t}}\:{e}^{−{xt}} \right\}{dt} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\left(\:{e}^{−{at}} \:−{e}^{−{bt}} \right){e}^{−{xt}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\left({e}^{−\left({b}+{x}\right){t}} \:−{e}^{−\left({a}+{x}\right){t}} \right){e}^{−{xt}} {dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({b}+{x}\right){t}} {dt}\:−\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({a}+{x}\right){t}} {dt} \\ $$$$=\frac{−\mathrm{1}}{{b}+{x}}\left[\:{e}^{−\left({b}+{x}\right){t}} \right]_{{t}=\mathrm{0}} ^{\infty} \:\:\:\:+\frac{\mathrm{1}}{{a}+{x}}\left[\:{e}^{−\left({a}+{x}\right){t}} \right]_{{t}=\mathrm{0}} ^{\infty} \\ $$$$=\:\frac{\mathrm{1}}{{b}+{x}}\:−\frac{\mathrm{1}}{{a}+{x}}\:\Rightarrow\varphi\left({x}\right)=\:{ln}\mid{b}+{x}\mid−{ln}\mid{a}+{x}\mid\:+{c} \\ $$$$={ln}\mid\:\frac{{b}+{x}}{{a}+{x}}\mid\:+{c}\:{but}\:{c}={lim}_{{x}\rightarrow+\infty} \left\{\varphi\left({x}\right)−{ln}\left(\frac{{b}+{x}}{{a}+{x}}\right)\right\} \\ $$$$\exists{m}>\mathrm{0}\:/\:\mid\varphi\left({x}\right)\mid<{m}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} {d}\:{t}\:=\frac{{m}}{{x}}\:\rightarrow\mathrm{0}\left({x}\rightarrow+\infty\right) \\ $$$$\Rightarrow{c}\:=\mathrm{0}\:{and}\:\varphi\left({x}\right)={ln}\:\left(\frac{{b}+{x}}{{a}+{x}}\right)\:. \\ $$
Commented by tawa tawa last updated on 06/Jun/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$