Another way of proving of smoothness is to use theorem 9.8 [Rudin, Real and complex analysis]
Let $h_\lambda(x)=\sqrt{\frac{2}{\pi}} \frac{\lambda}{\lambda^2+x^2}$, for $\lambda>0, x\in \mathbb{R}$, and let $H_\lambda(t)=e^{-\lambda |t|}$.
Then $H_\lambda (t)=\frac{1}{\sqrt{2\pi}} \int_\mathbb{R} h_\lambda(x) e^{-itx}dx $ $h_\lambda(x)=\frac{1}{\sqrt{2\pi}} \int_\mathbb{R} H_\lambda (t) e^{itx} dt$ ( the function $H_\lambda$ is the Fourier transform of $h_{\lambda}$, and the function
$h_{\lambda}$ is the inverse Fourier transform of $H_\lambda $).
Theorem. Let a function$f: \mathbb{R} \rightarrow \mathbb{R}$ be integrable. Then $(f*h_\lambda)(x)=\frac{1}{2\pi}\int_\mathbb{R} H_\lambda (t) {\cal F} (f)(t) e^{ixt} dt, $ where ${\cal F} (f)(t)=\frac{1}{\sqrt{2 \pi}} \int_R f(x) e^{-itx} dx$ - is the Fourier transform of $f$.
(This Theorem follows immediately by Fubini theorem and fact that $h_\lambda$ is the inverse Fourier transform of function $H_\lambda $).
The function under integral in the formula for convolution $f*h_\lambda$ is continuous with respect to $x,t$ and has derivatives of all order with respect to $x$, which (since ${\cal F}(f)$ is bounded) are bounded by integrable functions $t \mapsto C_k t^k e^{-\lambda |t|}$, where $C_k$, for $k=0,1,2...$, are constants. Hence $(f*h_\lambda)^{(k)}(x)=\frac{1}{2 \pi} \int_R H(\lambda t) {\cal F} (f)(t) (it)^k e^{ixt} dt$ and consequently the convolution is smooth.
In the same way, taking $h_\lambda(x)=\frac{1}{\lambda} e^{-\frac{x^2}{2\lambda^2}}$, and $H_\lambda(t)=e^{-\frac{\lambda^2 t^2}{2}}$, for $\lambda >0$, $x \in \mathbb{R}$, may show that $f* h_\lambda $ is smooth.