Proving that $\widehat{g*f}\in L^1(\mathbb{R})$ with $f(t) = \frac{1}{\sqrt{1 + t^4}}$ and $g(t) = e^{-|t|}$ The question is as follows:


  Verify whether the Fourier of the  convolution: $f * g$ belongs to $L^1(\mathbb{R})$? Where  $$f(t) = \frac{1}{\sqrt{1 + t^4}}~~and ~~~~g(t) = e^{-|t|}$$


$\textbf{Efforts:}$
As we know $L^1(\mathbb{R}) $ is closed under convolution. So is it enough to show that $f(t)$ and $g(t)$ are in $L^1(\mathbb{R})$ for to see that their convolution is also in $L^1(\mathbb{R})$?
Can you please let me know what conditions has to satisfy for the convolution be in $L^1(\mathbb{R})$?
Indeed, see here
\begin{split}
\widehat{g}(x)= \int_{-\infty}^{\infty}e^{-|t|}e^{-ix t}\,dt
&amp;=&amp;\int_{-\infty}^{0}e^{t}e^{-ix t}\,dt+\int_{0}^{\infty}e^{-t}e^{-ix t}\,dt\\
 &amp;=&amp;\left[ \frac{e^{(1-ix)t}}{1-ix} \right]_{-\infty}^0-\left[\frac{e^{-(1+ix)t}}{1+ix} \right]_{0}^{\infty}\\
&amp;=&amp;\frac{1}{1-ix}+\frac{1}{1+ix}\\
&amp;=&amp;\color{blue}{\frac{2}{x^2+1}\in  L^1(\Bbb R)}.
\end{split}

Therefore,  we have  that $\widehat{g}\in L^1$ and we know that  $\widehat{f} \in L^\infty$ (Since $f\in L^1$ ) hence $$\color{red}{\widehat{f*g}=\widehat{f}\displaystyle\widehat{g}  \in L^1}$$

Proof of the claim above
$$|\widehat{f}(a)|=\left|\int f(x)e^{-ixa}\, dx\right| \le \int \left|f(x)\, \right|dx =\|f\|_1 $$

that is $$\|\widehat{f}\|_{\infty}\le \|f\|_1 $$
$f, g , f*g $ are non negative and since $f,g\in L^1\cap L^{\infty}$ we may apply Fubini's theorem to

$$ \int_{-\infty}^{+\infty}(f*g)(t)\,dt = \int_{-\infty}^{+\infty}\int_{-\infty}^{\infty}f(x)\,g(t-x)\,dx\,dt $$
leading to $\|f*g \|_1 = \|f\|_1\cdot \|g\|_1 = \frac{1}{\sqrt{\pi}}\,\Gamma\left(\frac{1}{4}\right)^2.$