This question is from my assignments and I was unable to solve it and I think I can't solve it without any help. I have been following Tom M Apostol.
Assume f is continuous on $(0,+\infty)$ and let $F(z)= \int_{0}^{\infty} e^{-zt} f(t) dt$ , z=x+iy for x>c>0 . If s>c and a>0 prove that :
(a) F(s+a) = a$\int_{0}^{\infty} g(t) e^{-at} dt $ where $g(x)=\int_{0}^{\infty} e^{-st} f(t)dt$.
Attempt: LHS is $F(s+a) =\int_{0}^{\infty} e^{-(s+a)t} f(t) dt$ and RHS is a$\int_{0}^{\infty} e^{-at} \int_{0}^{t}e^{-sx} f(x) dx dt$ . I was thinking of using/ simplifying RHS to obtain LHS but I am unable to do so $e^{-at} \int_{0}^{t}e^{-sx} f(x) dx$. If there was $\infty$ instead of t then if would be F(s) but I am unable to understand what it would be in this case.
So, Can you please guide me in this question?
Suppose $$\int_{0}^{\infty} e^{-ut}|f(t)|dt<\infty$$ for real number $u>c>0$. Then $$\int_{0}^{\infty} dt\,e^{-(\mathbf{Re}(s)+a)x}|f(t)|<\infty$$ for $\mathbf{Re}(s)>c$.
Now $f$ satisfies the premise of Fubini's theorem to be applied. We switch the order of integration of the following double integral. \begin{align} a\int_{0}^{\infty} e^{-at} \int_{0}^{t}e^{-sx} f(x) dx dt&=\int_{0}^{\infty} dx\,e^{-sx}f(x)\int_x^\infty adt\,e^{-at} \\ &=\int_{0}^{\infty} dx\,e^{-sx}f(x)e^{-ax} \\ &=F(s+a). \end{align}