Interchanging integration and partial derivative

[TaPaKaH]

Homework Statement

$f\in L_{loc}^1(\mathbb{R}_+)$.
Need show that for Re$(z)>\sigma_f$ (abscissa of absolute convergence) we have $$\mathcal{L}[tf(t)](z)=-\frac{d}{dz}\mathcal{L}(z)$$where $\mathcal{L}$ denotes Laplace transform.

The Attempt at a Solution

The proof comes down to whether $$\int_0^\infty\frac{\partial}{\partial z}\left(e^{-zt}f(t)\right)dt=\frac{d}{dz}\int_0^\infty e^{-zt}f(t)dt$$holds.
All the theory on switching integration and derivative I could find requires the integration interval to be finite and/or f to be continuous which is not really the case.
Any ideas welcome.

 

[mighty2000]

Thank you for your post. I can offer some insight into this problem. The key to proving this statement lies in the properties of the Laplace transform. We know that the Laplace transform is defined as $$\mathcal{L}[f(t)](z)=\int_0^\infty e^{-zt}f(t)dt$$and that it converges for Re(z)>$\sigma_f$. This means that for any fixed z with Re(z)>$\sigma_f$, the integral converges and we can switch the order of integration and differentiation.

To prove the statement, we can use the definition of the Laplace transform and the chain rule for differentiation. We have:

$$\mathcal{L}[tf(t)](z)=\int_0^\infty e^{-zt}tf(t)dt$$

Using the chain rule, we can rewrite this as:

$$\frac{d}{dz}\int_0^\infty e^{-zt}f(t)dt=-\int_0^\infty\frac{\partial}{\partial z}\left(e^{-zt}f(t)\right)dt$$

Since the integral converges for Re(z)>$\sigma_f$, we can switch the order of integration and differentiation. This gives us:

$$\frac{d}{dz}\int_0^\infty e^{-zt}f(t)dt=\int_0^\infty\frac{\partial}{\partial z}\left(e^{-zt}f(t)\right)dt$$

Therefore, we have shown that for Re(z)>$\sigma_f$, $$\mathcal{L}[tf(t)](z)=-\frac{d}{dz}\mathcal{L}(z)$$

I hope this helps. Let me know if you have any further questions.