Why use a particular regularization for $\int_0^\infty \mathrm{d}x\,e^{i p x}$? There are many badly defined integrals in physics.
I want to discuss one of them which I see very often. 
$$\int_0^\infty \mathrm{d}x\,e^{i p x}$$
I have seen this integral in many physical problems. Many people seem to think it is a well defined integral, and is calculated as follows:
We will use regularization (we introduce a small real parameter $\varepsilon$ and after calculation set $\varepsilon = 0$.
$$I_0=\int_0^\infty \mathrm{d}x\,e^{i p x}e^{ -\varepsilon x}=\frac{1}{\varepsilon-i p}=\frac{i}{p}$$
But I can obtain an arbitrary value for this integral!
I will use regularization too, but I will use another parametrization:
$$I(\alpha)=\int_0^\infty \mathrm{d}x\,e^{i p x}=\int_0^\infty dx \left(1+\alpha\frac{\varepsilon \sin px}{p}\right)e^{i p x}e^{ -\varepsilon x}$$
where $\varepsilon$ is a regularization parameter and $\alpha$ is an arbitrary value
using $\int_0^\infty \mathrm{d}x\,\sin{(\alpha x)} e^{ -\beta x}=\frac{\alpha}{\alpha^2+\beta^2}$ 
After a not-so-difficult calculation I obtain that
$I(\alpha)=\frac{i}{p}\left(1+\frac{\alpha}{2}\right)$.
This integral I have often seen in intermediate calculation. But usually people do not take into account of this problem, and just use $I_0$. I don't understand why?
I know only one example when I can explain why we should use $I_0$. In field theory when we calculate $U(-\infty,0)$, where $U$ is an evolution operator, It is proportional to $\int^0_{-\infty} \mathrm{d}t\,e^{ -iE t}$. It is necessary for the Weizsaecker-Williams approximation in QED, or the DGLAP equation in QCD, because in axiomatic QFT we set $T\to \infty(1-i\varepsilon)$.
My question is: Why, in calculation of the integral $\int_0^\infty \mathrm{d}x\,e^{i p x}$, do people use $I_0$? Why people use $e^{ -\varepsilon x}$ regularization function? In my point of view this regularization no better and no worse than another.
 A: When you introduce an auxiliary variable, such as a regularization parameter, at the end of the calculation you have to take the limit that sets the expression back to the original one. If you introduce multiple auxiliary variables, you have to do this for all of them. Otherwise you're just doing a different integral. In this case specifically,
$$\begin{align}
I(\alpha)
&=\int_0^\infty \mathrm{d}x\,e^{i p x} \\
&=\lim_{\alpha\to 0}\lim_{\varepsilon\to 0}\int_0^\infty dx \left(1+\alpha\frac{\varepsilon \sin px}{p}\right)e^{i p x}e^{ -\varepsilon x} \\
&= \lim_{\alpha\to 0} I(\alpha) \\
&= I(0) = I_0
\end{align}$$
Without the $\lim_{\alpha\to 0}$, you're not calculating $\int e^{ipx}$, you're calculating something else. 
Sure, if you put the $\varepsilon$ limit inside the integral then it looks like the value of $\alpha$ doesn't matter, but you can't actually make that conclusion, because limits don't commute with integrals in general. And clearly, in this case, the value of $\alpha$ does matter.
A: Something that fixes $\frac{i}{p}$ uniquely is that, independently of regulator, it is the constant term of the asymptotic expansion of
$$\int_0^b \mathrm{e}^{ipx} \,\mathrm{d}x = \frac{i - \mathrm{e}^{ipb}}{p}$$
Moreover, consider applying your regulator with $\alpha \neq 0$ to a case where the integral does converge, such as $p=i$. Does it still yield the correct answer? Naively, one might think the second term in the integrand doesn't matter because $\varepsilon \rightarrow 0$ so $\frac{\varepsilon \sin p x}{p} \rightarrow 0$. The same would be true in this convergent scenario. But it clearly does matter because $\alpha \neq 0$ yields the wrong answer. Thus you must let $\alpha \rightarrow 0$ at the end, or your method will be inconsistent with convergent integrals.
