Show that the Laplace transform of a $C^1$ function f equal to its derivative at 0. Let $g: R → R$ be a $C^1$ function. Show that
$\lim_{h\to \infty} h^2 \int_{t=0}^1 e^{-ht}[g(t)-g(0)]dt=g'(0)$.
My attempt: I divided the integration into two parts and use integrate by part and get $he^{-h}(g(0)-g(1)-h\int_{t=0}^1e^{-ht}g'(t)dt$.
I cannot see any connection between it and $g'(0)$.
 A: Make the change of variable $s=ht$. You get $h\int_0^{h} e^{-s} [g(\frac s h)-g(0)]ds=\int_0^{h} se^{-s} \frac {[g(\frac s h)-g(0)]} {s/h}ds$. Note that $\frac {[g(\frac s h)-g(0)]} {s/h} \to g'(0)$ boundedly since $\frac s h$ is restricted to $[0,1]$. [You can apply MVT and use the fact that $g'$ is bounded on $[0,1]$].  So an application of DCT gives the limit as $g'(0) \int_0^{\infty} se^{-s}ds=g'(0)$.
A: Hint: Insert $g(t)=g(0)+g'(t)t + t\epsilon(t)$ where $\epsilon(t)$ is bounded and goes to zero as $t\to 0$.
A: $$x^2 \int_0^1 e^{-tx} [f(t)-f(0)]\, dt \equiv I(x)
.$$
let $u=[f(t)-f(0)]\implies du=f'(t) dt$
let $dv=e^{-tx} dt \implies v=-\frac 1x e^{-tx}$
$$I (x) = -x  \bigg(     e^{-x}  [f(1)-f(0)]-  \int_0^1  e^{-tx}f'(t)  \, dt \bigg)  $$
But we know that the factor of $e^{-tx}$ will drive the first term to zero in the limit $x \to \infty$
SO
$$I (x)  \to x\int_0^1  e^{-tx} f'(t)  \, dt$$
let $u=  f'(t) \implies du=f''(t) dt$
let $dv=e^{-tx} dt\implies v=-\frac 1x e^{-tx}$
$$I (x)  \to - \bigg (e^{-x}f'(1) -f'(0)   -  \int_0^1  e^{-tx} f''(t)  \, dt \bigg )$$
$$ \to  f'(0)   +  \int_0^1  e^{-tx} f''(t)  \, dt $$
Now if you were to continue the same integration by parts substitutions, each iteration would gain one more inverse power of $x$ so the integral will tend to
$$I (x)  \to f'(0)+ \frac{   f''(0)}x + \frac{   f'''(0)}        {x^2 }+...$$
So
$$ \lim_{x\to \infty} I(x)=f'(0) $$
