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.
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.