Prove sum of 2 periods is a period Periods form a ring but I couldn't find any proof in the literature. This unaccepted answer at MathOverflow
 https://mathoverflow.net/questions/326977/how-do-we-prove-that-a-sum-of-two-periods-is-still-a-period
claims to have answered the question. But it is vague and seems to me like cheating.

Q: Please, give an easy to understand proof of the fact that sum of 2 periods is a period by 'elementary' means without using any fancy notations (like the ones used it that MO link) and without heavyweight terminology so that even a XVIII-th century mathematician or a high school student with only basic knowledge of Riemann integral can understand it? Thanks.

I tried to understand the 'proof' in the link or find a proof myself but so far couldn't.
 A: Consider periods $r_1,r_2$ defined by absolutely convergent integrals
$$r_i=\int_{\mathbb R^{n_i}}\frac{f_i(x)}{g_i(x)}\;[\phi_i(x)]\;d^{n_i}x\tag{1}$$
where $f_i$ and $g_i$ are $n_i$-variable polynomials with rational coefficients, $\phi_i$ is an $n_i$-variable quantifier-free formula in the language of ordered fields, and where $[\phi_i(x)]=1$ if $\phi_i(x)$ holds, and $[\phi_i(x)]=0$ otherwise.
We want to write $r_1+r_2$ as an integral of the same form.
Without loss of generality, $n_1\leq n_2.$
But we can easily express $r_1$ as a period using $n_2$ variables instead of $n_1$: 
$$r_1=\int_{\mathbb R^{n_2}}
\frac{f_1(x_1,\dots,x_{n_1})}{g_1(x_1,\dots,x_{n_1})}
\; [\phi_1(x)\wedge (0<x_{n_1+1}<1)\wedge\dots\wedge (0<x_{n_2}<1)]\;d^{n_2}x$$
so we can assume $n_1=n_2.$ Let $n=n_1=n_2$ from now on.
Define $(n+1)$-variable formulas $\psi_i$ by
$$\psi_i(x_1,\dots,x_n,y)=\phi_i(x)\wedge (g_i(x)^2y^4<f_i(x)^2)\wedge(f_i(x)g_i(x)y>0).$$
I claim that
$$r_i=\int_{\mathbb R^{n+1}}2y\;[\psi_i(x,y)]\;d^nx\;dy\tag{2}$$
The condition $(g_i(x)^2y^4<f_i(x)^2)\wedge(f_i(x)g_i(x)y>0)$ holds for $y$ in the interval $(0,\sqrt{f_i(x)/g_i(x)})$ if $f_i(x)/g_i(x)>0,$ and in the interval $(\sqrt{-f_i(x)/g_i(x)},0)$ if $f_i(x)/g_i(x)<0.$ If $f_i(x)=0,$ the condition does not hold for any $y.$ The integral $\int 2y [\psi_i(x_1,\dots,x_n,y)]dy$ in each case is exactly $f_i(x)/g_i(x).$ So integrating out $y$ in (2) gives the original integral (1).
Hence
$$r_1+r_2=\int_{\mathbb R^{n+2}}2y\;[(\psi_i(x,y)\wedge(1<z<2))\vee(\psi_i(x,y)\wedge(2<z<3))]\;d^nx\;dy\;dz.$$
