Laplace transform and differentiation Let $F(s)$ be the Laplace transform of $f(t)$:
$$F\left(s\right)=\int_{0}^{\infty}e^{-st}f\left(t\right)dt$$
It then follows that $f(t)$ can be recovered from $F(s)$ by the inverse Laplace transform:
$$f\left(t\right)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F\left(s\right)ds$$
From the Laplace transform formula one can prove by integration by parts that the Laplace transform of the derivative $f'(t)$ is given by $sF(s)-f(0)$; so that, applying the inverse Laplace formula again:
$$f'\left(t\right)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}\left[sF\left(s\right)-f\left(0\right)\right]ds$$
However, if one differentiates with respect to $t$ the inverse Laplace transform formula giving $f(t)$ from $F(s)$, one obtains:
$$f'\left(t\right)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}sF\left(s\right)ds$$
and from this it seems that the Laplace transform of $f'(t)$ is just $sF(s)$. Since these two results are inconsistent, I think I'm missing something here. Can someone help me?
 A: Consider the problematic part of the integral
$$- f(0) \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}  ds \ e^{st}$$
The curve $\gamma$ is to the right of all the singularities of the argument.
But there are no such singularities, so the integral is zero.
Here we have assumed that $t>0$ so that as we push the contour to the left the integral is suppressed. 
Let us be slightly more careful.
By definition, the Laplace transform of $f(x)$ is
$$F(s) = \int_0^\infty d x \ e^{-s x} f(x).$$
From this definition it can be shown that the inverse transform is 
$$\Theta(x)f(x) = \frac{1}{2\pi i} \int_C ds \ e^{s x} F(s),$$
where $C$ is the appropriate contour,
and where $\Theta(x)$ is the Heaviside step function. 
(Your problem originates from neglecting this factor in several places.)
But this implies
$$\Theta(x)f'(x) + \Theta'(x) f(x) = \frac{1}{2\pi i} \int_C ds \ e^{s x} s F(s).$$
Notice that
$$\begin{eqnarray}
\Theta'(x) f(x) &= & \delta(x)f(0) \\
&=& \frac{1}{2\pi} \int_{-\infty}^\infty d t e^{i x t} f(0) \\
&=& \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} d s e^{s x} f(0) \\
&=& \frac{1}{2\pi i} \int_C d s e^{s x} f(0).
\end{eqnarray}$$
Here we have used the usual integral representation of the Dirac delta function, done the change of variables $t=-is$, and pushed the contour to agree with $C$ (which is allowed since there are no singularities).
Thus we find 
$$\Theta(x)f'(x) = \frac{1}{2\pi i} \int_C ds \ e^{s x} [s F(s)-f(0)].$$
as required. 
