I am reading a paper about fractional differential equation. One of the piece said as follow:

By applying the Parseval–Plancherel theorem we may show: \begin{equation} \int_0^\infty v(t)B_\alpha v(t)dt=\frac{1}{2\pi}\int_0^\infty y^{-\alpha}\cos(\pi\alpha)|\hat{v}(iy)|^2dy>0 \end{equation} for $-1<\alpha<1$ where $\hat{v}$ denotes the Laplace transform of $v$ and \begin{equation} B_\alpha v(t)=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}v(s)ds \end{equation} for $t>0$.

I have known about Parseval theorem and Plancherel theorem. They are very important theorems in Fourier analysis. But I can not find the corresponding theorem about Laplace transform. I am not familiar with integral transform field. Could anyone help me to deduct the equality and introduce the theorem to me? Also, hope one can recommend some references including the introduction of this theorem.

Thanks in advance.

  • $\begingroup$ You need to use $\langle f,g \ast h \rangle = \langle \hat{f},\hat{g} \hat{h} \rangle$ and the Laplace/Fourier transform of $t^{\alpha-1} 1_{t > 0}$ and the convolution theorem $\endgroup$ – reuns Sep 9 '17 at 17:40

Complex Laplace transforms can be written as Fourier transforms. Let $$\hat f(z)=∫_{0}^{∞}dt\exp[izt]f(t),\;\mathrm{Im}z>0.$$

Put $z=ω+iδ$, $ω$ real and $δ>0$. Then ($θ(t)$ is the Heaviside step function) $$\hat f(ω+iδ) = ∫_{-∞}^{+∞}dt\exp[iωt]θ(t)\exp[-δt]f(t),$$
$$θ(t)\exp[-δt]f(t) = \frac{1}{2\pi}∫_{-∞}^{+∞}dω\exp[-iωt]\hat f(ω+iδ).$$

Maybe this is useful.

  • $\begingroup$ Thank you for your hint. In fact, I have try this method. But I can deal with the integral kernel $e^{-\delta t}$ when doing the Fourier transforms. What is more, I do not know how is the $\hat{v}(iy)$ and $\cos(\pi\alpha)$ appearance. (Maybe the $\cos(\pi\alpha)$ is given by the $e^{\pi \alpha i}$ but I do not know how it is appearance). $\endgroup$ – Lion Feb 9 '14 at 15:10
  • $\begingroup$ I have prove it by using the transform that $s=i\omega$ in Fourier transform. $\endgroup$ – Lion Feb 13 '14 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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