I'm trying to solve the following Volterra integral of the first kind for $w(x)$.
\begin{equation} P(x)=\int_{-\infty}^{\infty}[d_1(x-x_0)+d_2(x)]w(x_0)dx_0 \end{equation}
My attempt so far: I'm aware that I can break this integral up into a convolution integral and a simpler integral. If I define the constant $C = \int_{-\infty}^{\infty}w(x_0)dx_0$ then
\begin{equation} P(x)=\int_{-\infty}^{\infty}d_1(x-x_0)w(x_0)dx_0 + d_2(x)\int_{-\infty}^{\infty}w(x_0)dx_0 \\ P(x)=\int_{-\infty}^{\infty}d_1(x-x_0)w(x_0)dx_0 + Cd_2(x) \end{equation} If I Fourier transform both sides then: \begin{equation} F[P(x)-Cd_2(x)]=F[d_1(x)]\times F[w(x)] \\ w(x) = F^{-1}\frac{F[P(x)-Cd_2(x)]}{F[d_1(x)]} \end{equation}
But this is not an ideal format because $C$ is a depends on $w(x)$ so I guess I'd have to solve this iteratively somehow. Perhaps there are better ways? An approximate solution is fine.
Just follow what you have, and assume the Fourier transformations are well-defined. Note your $C = \hat{w}(0)$.
$\hat{P}(\zeta) - \hat{w}(0) \hat{d_2}(\zeta) = \hat{d_1}(\zeta) \hat{w}(\zeta)$,
You can first solve $\hat{w}(0)$ by setting $\zeta = 0$, which means
$$\hat{w}(0) = \frac{\hat{P}(0)}{\hat{d_1}(0) + \hat{d_2}(0)}$$
Then you can solve the rest of $\hat{w}(\zeta) = \frac{1}{\hat{d_1}(\zeta)}(\hat{P}(\zeta) - \hat{w}(0) \hat{d_2}(\zeta))$.
Note: if $\hat{P}(0) = \hat{d_1}(0) + \hat{d_2}(0) = 0$, then you can set $\hat{w}(0)$ as an arbitrary number.