A formula for delta function in quantum mechanics I met a formula for delta function in a QM book ( not in English). The formula is used in the scattering theory. Its form is 
$$\lim_{\alpha\rightarrow \infty}\exp[i\alpha x]=2i\delta(x), ~~(x\geq 0).$$ Before, I never saw this form. Is this formula correct? Where can I find relevant references? 
Supplementary note:
The above formula may be wrong. In a problem set I find a proof for a similar formula:
$$\lim_{\alpha\rightarrow \infty}\alpha\exp[i\alpha x]=2i\delta(x), ~~(x\geq 0).$$
This proof refers to the book "Quantum collision theory" by C.J.Joachain, Chapter 3.
Now I paste the proof here.  
Assuuming $f(x)$ is a slowly-varying function of $x$. We have
$$\int_{-a}^af(x)\delta(x)dx=f(0),~~(a>0)$$ and
$$\int_{0}^af(x)\delta(x)dx=f(0)/2,~~(a>0).\tag{1}$$
Then we consider an integral
$$I=\int_0^a\alpha f(x)e^{i\alpha x}dx,~~(\alpha\rightarrow \infty).$$
For $x>0$, the factor $e^{i\alpha x}$, which oscillates rapidly, has no contribution on the integral. The value of the integral mainly comes from the $x\approx 0$, thus we can take $f(x) \sim f(0)$ and obtain
$$I=f(0)\int_0^a\alpha e^{i\alpha x}dx=-if(0)[e^{i\alpha a}-1].$$
When $\alpha \rightarrow \infty$, the value of $e^{i\alpha a}$ is indefinite. We can take its local average value, namely zero. Therefore,
$$I=\int_0^a\alpha f(x)e^{i\alpha x}dx=if(0), ~~\alpha \rightarrow \infty.$$
Comparing this result with Eq.(1), we can obtain
$$\alpha e^{i\alpha x}=2i\delta(x), ~~\alpha\rightarrow \infty.$$
 A: The Fourier transform of a smooth test function $f(x)$ defined on the whole real line will be tend to zero as $\alpha\to \infty$. If, however, it is non-zero only for $x\ge0$ and has a jump at the origin from $0$ on the negative axis to $f(0)$, and is thereafter smooth ,then it will go to zero as $-f(0)/i\alpha$. So have you missed out a factor of $\alpha$ in your formula? 
Something like 
$$
\lim_{\alpha\to \infty} \alpha e^{i\alpha x}= i \delta(x)?
$$
I don't know about the "$2$" though....
A: First of all, it is the book's author's responsibility to prove the formula or give a reference. Second, I suspect the formula is wrong. My reasoning (which is not conclusive) is as follows.
As you know, delta-function is not a function, but a distribution, and for a test function $f(x)$ we have $\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$. Let us assume (and this may be controversial) that $\int_0^\infty f(x)\delta(x)dx=\frac{1}{2}f(0)$. Then, assuming the formula is correct, we obtain $$\int_0^\infty f(x)\frac{1}{2i}\lim_{\alpha\to\infty}\exp(i\alpha x)dx=\frac{1}{2}f(0).$$Let us try $f(x)=\exp(-s x)$, where $s>0$. We obtain:$$\int_0^\infty \exp(-s x)\frac{1}{2i}\lim_{\alpha\to\infty}\exp(i\alpha x)dx=\frac{1}{2i}\lim_{\alpha\to\infty}\frac{-1}{i\alpha-s}=\frac{1}{2}.$$This does not look correct.  
EDIT (3/16/2019): @mike stone's suggestion may make the formula correct: if we consider the following initial formula: $$\lim_{\alpha\to\infty}\alpha\exp(i\alpha x)=2i\delta(x)$$ for $x\geq 0$, my calculation seems to give a correct result. 
