# how do we prove that a sum of two periods is still a period?

Kontsevich and Zagier define periods as the values of absolutely convergent integrals $$\int_\sigma f$$ where $$f$$ is a rational function with rational coefficients and $$\sigma$$ is a semi-algebraic subset of $$\mathbb{R}^n$$. How do we prove that the sum of two such numbers is still of this form? I've tried a few things but they don't seem to work...

Let $$\alpha$$ and $$\beta$$ be two periods corresponding respectively to two absolutely convergent integrals $$\int_\sigma f(x)dx$$ and $$\int_\tau g(y)dy$$, where $$f$$ (resp. $$g$$) is a rational function on $$\Bbb Q$$ with $$r$$ (resp. $$s$$) variables and $$\sigma$$ (resp. $$\tau$$) is a semi-algebraic subset of $$\Bbb R^r$$ (resp. $$\Bbb R^s$$).
Setting $$\omega:=\sigma\times\left\lbrace0\right\rbrace\times(0,1)^s\coprod(0,1)^r\times\left\lbrace1\right\rbrace\times\tau$$, one immediately gets that $$\alpha+\beta=\int_\omega \left[(1-t)f(x)+tg(y)\right]dxdydt$$which is again an absolutely convergent integral, so that $$\alpha+\beta$$ is a period.
• The "$\int_\sigma f$" being used as a definition of a period is presumably an ordinary volume integral - no surface integrals allowed. $\omega$ has zero volume, so the integral is zero. – Dap Apr 11 at 6:48