Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
5  
I don't think you're allowed to switch the order of contour integration and complex differentiation without uniform convergence. –  anon Nov 18 '11 at 21:43
1  
@anon: Would there be something wrong with assuming that $f(t)$ is such that there is uniform convergence? –  becko Nov 19 '11 at 4:23

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
Thanks a lot. It was very helpful. –  becko Mar 21 '12 at 3:29
    
@becko: Glad to help! –  user26872 Mar 21 '12 at 3:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.