In mathematics, an absolutely integrable function is a function whose absolute value is integrable, meaning that the integral of the absolute value over the whole domain is finite.

For a real-valued function, since

$${\displaystyle \int |f(x)|\,dx=\int f^{+}(x)\,dx+\int f^{-}(x)\,dx}$$

$${\displaystyle f^{+}(x)=\max(f(x),0),\ \ \ f^{-}(x)=\max(-f(x),0),}$$

both ${\textstyle \int f^{+}(x)\,dx}$ and ${\textstyle \int f^{-}(x)\,dx}$ must be finite. In Lebesgue integration, this is exactly the requirement for any measurable function f to be considered integrable, with the integral then equaling ${\textstyle \int f^{+}(x)\,dx-\int f^{-}(x)\,dx}$, so that in fact "absolutely integrable" means the same thing as "Lebesgue integrable" for measurable functions.

The same thing goes for a complex-valued function. Let us define

$${\displaystyle f^{+}(x)=\max(\Re f(x),0)}$$

$${\displaystyle f^{-}(x)=\max(-\Re f(x),0)}$$

$${\displaystyle f^{+i}(x)=\max(\Im f(x),0)}$$

$${\displaystyle f^{-i}(x)=\max(-\Im f(x),0)}$$

${\displaystyle \Re f(x)}$

${\displaystyle \Im f(x)}$

${\displaystyle f(x)}$

$${\displaystyle |f(x)|\leq f^{+}(x)+f^{-}(x)+f^{+i}(x)+f^{-i}(x)\leq {\sqrt {2}}\,|f(x)|}$$

$${\displaystyle \int |f(x)|\,dx\leq \int f^{+}(x)\,dx+\int f^{-}(x)\,dx+\int f^{+i}(x)\,dx+\int f^{-i}(x)\,dx\leq {\sqrt {2}}\int |f(x)|\,dx}$$