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?

• I don't think you're allowed to switch the order of contour integration and complex differentiation without uniform convergence.
– anon
Commented Nov 18, 2011 at 21:43
• @anon: Would there be something wrong with assuming that $f(t)$ is such that there is uniform convergence?
– a06e
Commented Nov 19, 2011 at 4:23

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.