# Integral of the inverse function over the real numbers

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

$$$$\int_{-\infty}^{\infty} f(x)g(x) dx = \int_{-\infty}^{\infty}1 dx = \infty$$$$

But is it correct to say the following (with a similar reasoning for $$g(x)$$) ?

$$$$\int_{-\infty}^{\infty}\frac{1}{x} dx = \int_{0}^{\infty} \frac{1}{x}+\frac{-1}{x} dx = \int_{0}^{\infty} 0 dx = 0 + C$$$$ We'd have found an example of two integrable functions whose product is not integrable, but I wonder if the reasoning is correct since $$\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. $$fg \not \in L^1(\mathbb{R})$$ ?

• check the definition of $L^1(X)$ function, you're missing an absolute value in the integrand Commented Jan 22 at 15:06
• Yes I know, sorry if I have been unclear. That's why I put "more general example" Commented Jan 22 at 17:20
• In general you say that a function $f : X \to \mathbb{C}$ is integrable if $\int_X |f| d \mu$ exists and is finite. This because one wants to avoid problems with things like $\infty - \infty$, so $L^1(X)$ is precisely the vector space of integrable functions. Commented Jan 23 at 8:42

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 $$\|f^2\|_1 = \infty$$.