In an attempt to find two integrable functions $f$ and $g$, say $\in L^1(\mathbb{R}$), whose product is not in $L^1(\mathbb{R})$, I thought about finding a more general example :
Let $f(x) = \frac{1}{x} $ and $g(x) = x$, we have that
\begin{equation} \int_{-\infty}^{\infty} f(x)g(x) dx = \int_{-\infty}^{\infty}1 dx = \infty \end{equation}
But is it correct to say the following (with a similar reasoning for $g(x)$) ?
\begin{equation} \int_{-\infty}^{\infty}\frac{1}{x} dx = \int_{0}^{\infty} \frac{1}{x}+\frac{-1}{x} dx = \int_{0}^{\infty} 0 dx = 0 + C \end{equation} We'd have found an example of two integrable functions whose product is not integrable, but I wonder if the reasoning is correct since $\int_{0}^{\infty} 1/x dx$ is not finite. This would be similar to say $\infty - \infty = 0$ , which is not always true.
Also, any idea on the initial problem, i.e finding $f,g\in L^1(\mathbb{R})$ s.t. $fg \not \in L^1(\mathbb{R})$ ?
The Hölder inequality proves that $$ \|fg\|_1 \leq \|f\|_2 \|g\|_2 $$ so such a $f,g$ must be searched among non-$L^2(\mathbb{R})$ functions. We can consider $$ f(x)= \begin{cases} \frac{1}{\sqrt{|x|}} & \text{if } x \in [-1,1]\\ x^{-2} & \text{otherwise} \end{cases} $$ This is $L^1(\mathbb{R})$ as $\int_{-1}^1 f(x) dx = 4$ and $\int_1^{+\infty} f(x) dx = 1 $. But $$ f(x)^2= \begin{cases} \frac{1}{|x|} & \text{if } x \in [-1,1]\\ x^{-4} & \text{otherwise} \end{cases} $$ is not $L^1(\mathbb{R})$ as $|x|^{-1}$ is not integrable in any neighborhood of the origin. Therefore $\|f\|_1 < \infty$ but $\|f^2\|_1 = \infty$.