Fourier transform, Heaviside Step function

Question

I was solving this exercise: Given the function $$ P(x) = \Theta(x)e^{-x}$$ where $\Theta(x)$ is the Heaviside function. I need to find the expression of: $$ f(x) = \int \int P(x_1)P(x_2)\delta(x-x_1-x_2)dx_1dx_2$$ First of all, I recognized that the double integral is the convolution of $P(x)$ with itself, $f(x) = P(x) \ast P(x)$. Therefore I used the Fourier transform, applying the well-known property: $$\mathcal{F}[f\ast g](k) = \sqrt{2\pi}\hat{f}(k)\hat{g}(k) $$ which, in our case, reduces to: $$\mathcal{F}[f](k) = \sqrt{2\pi}(\hat{P}(k))^2 $$ Now some questions started arising: our function can be written as: $$f(x) = \left\{ \begin{array} $e^{x}, x >0 \\ 0, x<0 \end{array} \right. $$ But the Fourier transform's domain of integration is $]-\infty,+\infty[$ So how do I make sure that the integral of the fourier transform behaves well, namely do I need to consider an approximated form like the following: $$ \lim_{\epsilon \to 0^+} \left[ \int_{-\infty}^{-\epsilon} P(x) + \int_{+\epsilon}^{\infty} P(x) \right]$$ Now what I did is restrain the domain of integration to $[0,\infty]$ and calculated: $$\hat{P}(k) = \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty}e^{-x(ik+1)}dx = ... = \frac{1}{\sqrt{2\pi}}\frac{1}{ik+1} $$ So I thought that maybe I could get away with this: $$ \hat{P}(k) = \left\{ \begin{align*} \frac{1}{\sqrt{2\pi}}\frac{1}{ik+1}, x>0 \\ 0, x <0 \end{align*} \right.$$ But this doesn't make much sense. I then checked the solution and found out that my professor doesn't make any domain restrictions, in fact, he only says that: $$ \hat{P}(k) = \frac{1}{\sqrt{2\pi}}\frac{1}{ik+1} \tag{1}$$ I thought this was clearly wrong since it doesn't take into consideration that the step function presents a jump discontinuity at $x=0$ and just accounts for when $x>0$. I tried finishing the exercise by adopting $(1)$ and what I found is this: $$ \hat{f}(k) = \frac{1}{\sqrt{2\pi}}\frac{1}{(ik+1)^2}$$ By using the antitransform I finally get: $$ f(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{ikx}}{(ik+1)^2}dk = \frac{1}{2\pi}I$$
Now I solved this integral as I usually do, namely the residue theorem: I consider the trivial complex extension of the integrand: $$f(z) = \frac{e^{izx}}{(iz-i^2)^2}$$ which has a second-order pole at $z = i$. I then considered the two cases $x < 0, x> 0$. When $x$ is positive I can choose a standard contour(a semi-circumference of radius $R$) that encloses the pole and reduces to $I$ on the real axis when we take the limit $R \to +\infty$. Doing all the calculations (which I'll leave out) I obtain: $$ I = 2\pi i Res(f,z=i) = 2\pi x e^{-x} $$ When $x$ is negative, I instead choose the same contour but in the $\Im(z)<0$ semi-plane. But this contour does not enclose any pole, therefore, from the residue theorem, we get: $$I = 0, x<0$$ Therefore I wrote the solution in the following way: $$ f(x) = \Theta(x)x e^{-x}$$ I checked the solution and the expression I just obtained is indeed correct. So my professor was right all along I suppose? The transform of $P(x)$ is unique and represented by $(1)$. And somehow everything turns out smoothly when solving the complex integral. It is as though the Fourier transform of $P(x)$ adapts to our initial piecewise smooth function, allowing us not to make any distinction for $x \ge 0$ and for $ x<0$.
I might not have the necessary mathematical background to understand a complete and thorough answer, but I'd like to get a deeper insight into this anyway. Any help or answer will be obviously well-received.

Answer 1

Firstly, note that $\hat{P}(k)$ can't depend on values of $x$, because it is a function of $k$ only; indeed, the Fourier transform maps the variable $x$ to a conjugated variable $k$, so that $x$ disappears in the transformation $P(x) \mapsto \hat{P}(k)$ because of the integration with respect to $x$.

Secondly, when calculating the integral $$ \int_\mathbb{R} \theta(x)e^{-x}e^{-ikx} \,\mathrm{d}x = \int_0^\infty e^{-(1+ik)x} \,\mathrm{d}x = \left.-\frac{e^{-(1+ik)x}}{1+ik}\right|_{x=0}^{x=\infty} = \frac{1}{1+ik}, $$ $x$ stays positive, since the domain of integration is reduced to $\mathbb{R}_+$ because of the Heaviside function, that is why you must ignore the case $x < 0$.

Finally, let's recall that an integrand vanishing at some points on the domain of integration doesn't lead necessarily to a vanishing integral $-$ another good example being the characteristic function $-$, because the integral depends on all the points of the domain of integration. In consequence, you can't make an argument such as $f(x) = 0$ for some $x$ $\Longrightarrow$ $\int f = 0$; the latter case would mean that $f \equiv 0$ on the whole domain of integration.