Is the following proposition true?
Proposition. For any $a,k\in \mathbb{R}$, \begin{equation} \int_a^{\infty} dx e^{ikx} = 2\pi \delta(k). \end{equation} (End)
I think it is true based on the following argument. First, let me change the integration variable from $x$ to $x'$ by \begin{equation} x' = x-\lambda, \end{equation} with $\lambda>0$. Then, \begin{equation} \int_a^{\infty} dx e^{ikx} =\int_{a-\lambda}^{\infty} dx' e^{ik(x'+\lambda)} = e^{ik\lambda} \int_{a-\lambda}^{\infty} dx' e^{ikx'}. \end{equation} Now, by letting $\lambda\rightarrow +\infty$, the factors in the last expression approach, as distribution of $k$, \begin{equation} e^{ik\lambda} \rightarrow \begin{cases} 0 & \text{ for } k\not=0\\ 1 & \text{ for } k =0 \end{cases} , \end{equation} and \begin{equation} \int_{a-\lambda}^{\infty} dx' e^{ikx'} \rightarrow \int_{-\infty}^{\infty} dx' e^{ikx'} =2\pi\delta(k). \end{equation} Hence, \begin{equation} e^{ik\lambda}\int_{a-\lambda}^{\infty} dx' e^{ikx'} \rightarrow 2\pi\delta(k). \end{equation} By this, I think the proposition is true.
However, I don't see this representation in the books or web pages. (The lower bound $a$ of the integral is always written as $-\infty$.) I guess it may be wrong, and I would like someone point out mistakes in my argument.
Some of the mistakes:
First, the function $e^{ikx}$ is not integrable, hence one should be cautious with with these kind of formal manipulations. The formula with $a=−\infty$ is true only if you know that the integral in this case is not a true integral but a shortcut for the Fourier transform in the sense that distributions.
The convergence of $e^{ikx}$ as a distribution of $k$ is also false: it converges to $0$. There is no meaning to the "value when $k=0$". The proof of this fact follows easily from the Riemann-Lebesgue lemma which implies in particular that for any $\varphi\in C^\infty_c$, $$ \langle e^{ikx},\varphi\rangle = \int_{\Bbb R} e^{ikx}\,\varphi(x)\,\mathrm d x \underset{k\to\infty}\to 0. $$
The limit of a product is not the product of limits for distributions.
The right solution:
The good way to interpret the formula "$\int_a^\infty e^{-ikx}\,\mathrm d x$" is as the Fourier transform in the sense of distributions of the function $\mathbf{1}_{x>a}$, that is defined as the distribution such that for any test function $\varphi\in C^\infty_c$ $$ \langle \mathcal{F}(\mathbf{1}_{x>a}),\varphi\rangle = \langle \mathbf{1}_{x>a},\widehat{\varphi}\rangle = \int_a^\infty \widehat{\varphi}(k)\,\mathrm d k. $$ Now, $\mathbf{1}_{x>a} = H(x-a)$ where $H$ is the Heaviside function. Using the well-known formula of the Fourier transform of the Heaviside function, $$ \mathcal{F}(\mathbf{1}_{x>a}) = e^{-ika} \mathcal{F}(H) = \pi\,e^{-ika} \left(\delta_0 + \mathrm{vp}(\tfrac{1}{i\pi k})\right), $$ where $\mathrm{vp}$ denotes the Cauchy principal value. In your case, you have $k$ instead of $-k$ in the integral, so you get $$ ``\int_a^\infty e^{ikx}\,\mathrm d x" = \mathcal{F}(\mathbf{1}_{x>a})(-k) = \pi\,e^{ika} \left(\delta_0 - \mathrm{vp}(\tfrac{1}{i\pi k})\right). $$