My question is:
Are there any two functions $f(x)$ and $g(x)$ Which hold that $\int_{a}^{\infty} f(x)dx$ and$\int_{a}^{\infty} g(x)dx$ converge conditionally but $\int_{a}^{\infty} f(x)g(x)dx$ absolutely converges?
I know that if $\int_{a}^{\infty} f(x)dx$ and$\int_{a}^{\infty} g(x)dx$ converge conditionally then $\int_{a}^{\infty} |f(x)|dx $ and $\int_{a}^{\infty} |g(x)|dx$ diverges.
And if $\int_{a}^{\infty} f(x)g(x)dx$ absolutely converges then $\int_{a}^{\infty} |f(x)g(x)|dx$ also converges.
Thanks.
\begin{align} g(x)&=\frac{\sin x}{\sqrt[4]{x}}\mathbb{1}_{(0,1]}(x) +\frac{\sin x}{x}\mathbb{1}_{[1,\infty)}(x)\\ f(x)&=\mathbb{1}_{(0,1]}(x) +\frac{\sin x}{x}\mathbb{1}_{[1,\infty)}(x) \end{align} Then $$(fg)(x)=\frac{\sin x}{\sqrt[4]{x}}\mathbb{1}_{(0,1]}(x) + \frac{\sin^2x}{x^2}\mathbb{1}_{[1,\infty)}(x)$$ is in integrable (in the sense of Lebesgue) and so absolutely integrable. Indeed, both $f$ and $g$ are conditionally integrable in the sense that $\lim_{a\rightarrow\infty}\int^a_0g$ and $\lim_{a\rightarrow\infty}\int^a_0 f$ exists, but $\int^\infty_a|f|=\infty=\int^\infty_0|g|$. We have that $$\int^\infty_0|fg|=\int^1_0|fg|+\int^\infty_1|fg|\leq \int^1_0\frac{dx}{x^{1/4}} +\int^\infty_1\frac{dx}{x^2}<\infty $$