Fourier transform of a $L^1 \cap L^{\infty} $ function is in $L^1$

Question

I am looking to prove that if $f$ is in $L^1(\mathbb{R}^d) \cap L^{\infty}(\mathbb{R}^d)$ and is such that $\hat{f} \geq 0$, then $\hat{f} \in L^1(\mathbb{R}^d)$. Does it suffice to show that the expression \begin{equation} \int_{\mathbb{R}^d} e^{-\frac{|x|^2}{2k}}\hat{f}(x) dx \end{equation} converges when $k \rightarrow \infty$ to $\int_{\mathbb{R}^d} \hat{f}(x) dx$, (which it does, by the monotone convergence theorem) and then conclude by observing that the first integrand is in $L^1(\mathbb{R}^d)$ for a fixed $k \in \mathbb{N}$?