I have been solving problems related to Laplace transformation involving the initial/final value theorem but got stuck at a step. The question is as follows:
Given $L{[\int^{\infty}_{t}\frac{e^{-x}}{x}}dx](s)=sf(s)$,for s>0. Then find $f(s)$ using the final value theorem.
Here are the steps which followed:
$L[f(t)]=\int^{\infty}_{0}f(t)e^{-st}dt$ where $f(t)=\int^{\infty}_{t}\frac{e^{-x}}{x}dx$
$L[f(t)]=\int^{\infty}_{0}[\int^{\infty}_{t}\frac{e^{-x}}{x}dx]e^{-st}dt$
Using the change of order of integral, we can write:
$L[f(t)]=\int^{\infty}_{0}[\int^{x}_{0} e^{-st}dx]\frac{e^{-x}}{x}dt$
$L[f(t)]=\frac{-1}{s}\int^{\infty}_{0}[e^{-sx}-1]\frac{e^{-x}}{x}dx$
$L[f(t)]=\frac{-1}{s}\int^{\infty}_{0}[\frac{e^{-x(s+1)}}{x}-\frac{e^{-x}}{x}]dx$
I am not able to perform the Laplace transformation on the above last expression as it is not integrable. The final value theorem states the if we know Laplace transform$f(s)$ of any function $f(t)$ then we can find $f(\infty)$ by solving, $$f(\infty)=\lim_{s\to 0} s.f(s)$$ But I have no idea how I can use this theorem to find $f(s)$.
Am I solving it wrongly? Should I approach in a different way? What can I do to find $f(s)$?
Please clarify my doubts.
Thank you!
Best Answer
Your approch leads to the final answer. After your last step, apply Proof of Frullani's theorem, $$\int_0^{+\infty} {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}} {x}}\,dx = f\left( 0 \right) {\ln\left( \frac{b} {a}\right)} $$ with $f(x)=e^{-x}$, $a=s+1$, $b=1$. It follows that $$L\left[\int^{\infty}_{t}\frac{e^{-x}}{x}\,dx\right](s)=\frac{\ln(s+1)}{s}.$$ However I don't know how to use the Final Value Theorem in order to get the same answer. Maybe you can take a look at this answer Proving of Integral $\int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}dx = \ln\left(\frac{a}{b}\right)$ where the Frullani's theorem is proved by using the Laplace transform.