Inmathematics, anabsolutely integrable function is afunction whoseabsolute value isintegrable, meaning that the integral of the absolute value over the wholedomain is finite.

For areal-valued function, since$${\displaystyle \int |f(x)|dx=\int f^{+}(x)dx+\int f^{-}(x)dx}$$where$${\displaystyle f^{+}(x)=max(f(x),0),f^{-}(x)=max(-f(x),0),}$$

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

The same thing goes for acomplex-valued function. Let us define$${\displaystyle f^{+}(x)=max(\mathrm{\Re }f(x),0)}$$$${\displaystyle f^{-}(x)=max(-\mathrm{\Re }f(x),0)}$$$${\displaystyle f^{+i}(x)=max(\mathrm{\Im }f(x),0)}$$$${\displaystyle f^{-i}(x)=max(-\mathrm{\Im }f(x),0)}$$where${\displaystyle \mathrm{\Re }f(x)}$ and${\displaystyle \mathrm{\Im }f(x)}$ are thereal and imaginary parts of${\displaystyle f(x)}$. Then$${\displaystyle |f(x)|\le f^{+}(x)+f^{-}(x)+f^{+i}(x)+f^{-i}(x)\le \sqrt{2}|f(x)|}$$so$${\displaystyle \int |f(x)|dx\le \int f^{+}(x)dx+\int f^{-}(x)dx+\int f^{+i}(x)dx+\int f^{-i}(x)dx\le \sqrt{2}\int |f(x)|dx}$$This shows that the sum of the four integrals (in the middle) is finite if and only if the integral of the absolute value is finite, and the function is Lebesgue integrable only if all the four integrals are finite. So having a finite integral of the absolute value is equivalent to the conditions for the function to be "Lebesgue integrable".

• "Absolutely integrable function – Encyclopedia of Mathematics". Retrieved9 October 2015.

• Tao, Terence,Analysis 2, 3rd ed., Texts and Readings in Mathematics, Hindustan Book Agency, New Delhi.

• Integral calculus

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