For some function $h \in \mathcal{S}(\mathbb{R})$ and $\theta, \eta \in \mathbb{R}$, we defining the Fourier transform as \begin{equation}\label{fourier} \tilde{h}^{\pm}(\theta) := \frac{1}{2 \pi} \int dp h(p)e^{\pm i \theta p}, \end{equation} the function $g(\theta) := -i/(\theta + i0^+)$ and having the symbol $\star$ denote the convolution, can we calculate the following \begin{align*} \int d\theta d\eta h(\theta)\frac{-i}{(\theta - \eta + i0^+)} h(\eta) & = \int d\theta h(\theta) (h \star g)(\theta) \\ & = 4 \pi^2 \int dp \widetilde{h}(p) \widetilde{h \star g}(p) \quad\quad \Big(\text{Plancherel's theorem}\Big)\\ & = 6 \pi^3 \int dp |\widetilde{h}(p)|^2 \widetilde{g}(p) \\ & = 4 \pi^2 \int dp |\widetilde{h}(p)|^2 \Theta(p) \quad\quad \Big(\text{Heaviside function}\Big)\\ & = 4 \pi^2 \int_0^\infty dp|\tilde{h}(p)|^2. \end{align*} I'm slightly confused about the prefactor $4 \pi$. From Plancherel's theorem, we would pick up a factor of $2\pi$, but that we are actually transforming a function $h$ and a convolution, so in the thirs line of the following computation, there have been 3 Fourier transforms performed, and so there should be a factor of $(2 \pi)^3$.
\begin{align} \int_{-\infty}^\infty |\tilde{h}(p)|^2 dp = \int_{-\infty}^\infty \tilde{h}(p)\overline{\tilde{h}(p)} dp & = \int_{-\infty}^\infty \Big[\frac{1}{2\pi} \int_{-\infty}^\infty h(\theta)e^{ip\theta} d\theta \Big]\Big[\frac{1}{2\pi} \int_{-\infty}^\infty \overline{h(\theta')}e^{-ip\theta'} d\theta' \Big] dp \\ & = \frac{1}{4 \pi^2} \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty h(\theta)\overline{h(\theta')} e^{i(\theta - \theta')p} d\theta d\theta' dp \\ & = \frac{1}{4 \pi^2} \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty h(\theta)\overline{h(\theta')} e^{i(\theta - \theta')p} dp d\theta d\theta' \\ & = \frac{2 \pi}{4 \pi^2} \int_{-\infty}^\infty\int_{-\infty}^\infty \delta(\theta - \theta') h(\theta) \overline{h(\theta')} d\theta d\theta' \\ & = \frac{1}{2\pi}\int_{-\infty}^\infty|h(\theta)|^2 d\theta \end{align} \begin{equation} \Rightarrow \int_{-\infty}^\infty|h(\theta)|^2 d\theta = 2 \pi \int_{-\infty}^\infty |\tilde{h}(p)|^2 dp. \end{equation} And we also pick up another factor of $\frac{1}{2\pi}$ from the Fourier transform of the function $g(\theta)$: \begin{equation} \tilde{g}(p) = \frac{1}{2 \pi} \Theta(p). \end{equation}
But I'm really not convinced about my argument. Am I making some critical or silly error here?