I'm not sure if the following result is correct, but I also have no clue where I could have possibly made a mistake.
\begin{align} \int_0^\infty J_0(x)\,dx &= \frac{1}{2}\int_{-\infty}^\infty J_0(x)\,dx \tag1\\ &=\frac{1}{4\pi}\int_{-\infty}^\infty \int_{-\pi}^\pi e^{ix\cos\phi} \, d\phi \, dx \tag2\\ &=\frac{1}{4\pi}\int_{-\pi}^\pi \int_{-\infty}^\infty e^{ix\cos\phi} \, dx \, d\phi \tag3\\ &=\frac{1}{2}\int_{-\pi}^\pi \delta(\cos\phi)\,d\phi \tag4\\ &=\frac{1}{2}\int_{-\pi}^\pi \left[\sum_{n=-\infty}^\infty \delta\left(\phi-\left(n+\frac{1}{2}\right)\pi\right)\right] \, d\phi \tag5\\ &=\frac{1}{2}\cdot 2 = 1 \tag6 \end{align}
In $(1)$, I used the fact that $J_0(x)$ is even.
In $(2)$, I used the integral representation of the Bessel Function of the First Kind.
In $(3)$, I used Fubini's Theorem to interchange orders of integration.
In $(4)$, I used the complex definition of the Delta Function.
In $(5)$, I used the fact that $\delta(f(x)) = \sum\frac{\delta(x-x_i)}{|f'(x_i)|}$, where the $x_i$ are the roots of the function $f(x)$.
In $(6)$, I used that only two delta functions contribute to the integral - the ones with singularities at $\pm\frac{\pi}{2}$, and the integral over each of these singularities is $1$.
However, the reason that I am doubting my result is because the Bessel Function $J_0(x)$ looks somewhat like $\frac{\sin(x)}{\sqrt{x}}$ for large $x$ and actually encloses a larger area with the $x$-axis close to $0$ (as $J_0(0)= 1$, whereas $\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\sqrt{x}}=0$). Therefore, I would expect that
$$\int_0^\infty J_0(x) \, dx \geq \int_0^\infty \frac{\sin(x)}{\sqrt{x}} = \sqrt{\frac{\pi}{2}}\approx 1.25$$
would hold.
Is my intuition wrong here, and if so, which part of my intuition is off? Or have I made a computational mistake somewhere?