Show that the Laplace transform of a $C^1$ function f equal to its derivative at 0.

Question

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 $g'(0)$.

Answer 1

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 $g'(0) \int_0^{\infty} se^{-s}ds=g'(0)$.

Answer 2

Hint: Insert $g(t)=g(0)+g'(t)t + t\epsilon(t)$ where $\epsilon(t)$ is bounded and goes to zero as $t\to 0$.

Answer 3

$$x^2 \int_0^1 e^{-tx} [f(t)-f(0)]\, dt \equiv I(x) .$$ 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 $$ \lim_{x\to \infty} I(x)=f'(0) $$