What is the correct cosine-like integral representation of Dirac delta?

Question

exploring the integral representations of the Dirac delta I found this in terms of an integral of cosine function (from wolfram's database, https://functions.wolfram.com/GeneralizedFunctions/DiracDelta/07/01/01/ ) $$\delta(x)=\frac{1}{\pi}\int_{-\infty}^{\infty}\cos{xt}dt $$ but in Wikipedia there is $$\delta(x-\alpha)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} d p \cos (p x-p \alpha)$$ that reduces to $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\cos{xt}dt$$ So, What is the correct expression?

I tried to show the second one $$\begin{aligned} \delta(x) &=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixt}dt\\ &=\frac{1}{2}\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixt}dt+\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixt}dt \right)\\ &=\frac{1}{2}\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ixt}dt+\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ixt}dt \right)\\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}cos(xt)dt \end{aligned}$$ where I used the usual formula $\frac{e^{iz}+e^{-iz}}{2}=cos(z)$.

Greetings and thank you.

Answer 1

Since $\delta(x)$ is real, it is easy to show that the Wikipedia formula is the correct one by taking the real part of the $e^{ixt}$ formula.

\begin{align} \delta(x) = \mathrm{Re}(\delta(x)) &= \frac{1}{2\pi}\mathrm{Re}\left(\int_{-\infty}^{\infty} e^{ixt}dt\right)\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty} \cos(xt)dt. \end{align}

The wolfram formula looks wrong to me. Maybe it was intended to be a one-sided integral?