Why make $A$ an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on $C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that.
Let $I_+ \subset C^\infty(\mathbb{R})$ be the ideal of functions that vanish on some neighborhood of $+\infty$ (for definiteness, say, on at least one interval of the form $[a,+\infty)$ with $a\in\mathbb{R}$). Let $A_+ = C^\infty(\mathbb{R})/I_+$. The quotient $A_+$ is an $\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at $+\infty$. Let $1_+$ be the image of the constant function $1\in C^\infty(\mathbb{R})$ under the quotient map. Define $I_-$, $A_-$ and $1_0$ 1_-$in the same way, by replacing$+\infty$with$-\infty$. Now, let$B=A_+\oplus A_-$and$N\subset B$be the linear subspace spanned by the element$1_+\oplus 1_-$. And finally let$A = B/N$, where we are just taking the quotient of linear spaces (the ring property of$B$ceases to be of importance). Any smooth function$f\in C^\infty(\mathbb{R})$has an indefinite integral$f_a(x) = \int_a^x f(y) dy$that is also in$C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image$[f_a]$of$f_a$in the linear space$A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say$f_a$and$f_b$) coincide. I think that letting$T(f) = [f_a]$will satisfy all the properties that you wanted of an "integration" map. 2 deleted 1 characters in body Why make$A$an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on$C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that. Let$I_+ \subset C^\infty(\mathbb{R})$be the ideal of functions that vanish on every some neighborhood of$+\infty$(for definiteness, say, on at least one interval of the form$[a,+\infty)$with$a\in\mathbb{R}$). Let$A_+ = C^\infty(\mathbb{R})/I_+$. The quotient$A_+$is an$\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at$+\infty$. Let$1_+$be the image of the constant function$1\in C^\infty(\mathbb{R})$under the quotient map. Define$I_-$,$A_-$and$1_0$in the same way, by replacing$+\infty$with$-\infty$. Now, let$B=A_+\oplus A_-$and$N\subset B$be the linear subspace spanned by the element$1_+\oplus 1_-$. And finally let$A = B/N$, where we are just taking the quotient of linear spaces (the ring property of$B$ceases to be of importance). Any smooth function$f\in C^\infty(\mathbb{R})$has an indefinite integral$f_a(x) = \int_a^x f(y) dy$that is also in$C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image$[f_a]$of$f_a$in the linear space$A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say$f_a$and$f_b$) coincide. I think that letting$T(f) = [f_a]$will satisfy all the properties that you wanted of an "integration" map. 1 Why make$A$an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on$C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that. Let$I_+ \subset C^\infty(\mathbb{R})$be the ideal of functions that vanish on every neighborhood of$+\infty$(for definiteness, say, on at least one interval of the form$[a,+\infty)$with$a\in\mathbb{R}$). Let$A_+ = C^\infty(\mathbb{R})/I_+$. The quotient$A_+$is an$\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at$+\infty$. Let$1_+$be the image of the constant function$1\in C^\infty(\mathbb{R})$under the quotient map. Define$I_-$,$A_-$and$1_0$in the same way, by replacing$+\infty$with$-\infty$. Now, let$B=A_+\oplus A_-$and$N\subset B$be the linear subspace spanned by the element$1_+\oplus 1_-$. And finally let$A = B/N$, where we are just taking the quotient of linear spaces (the ring property of$B$ceases to be of importance). Any smooth function$f\in C^\infty(\mathbb{R})$has an indefinite integral$f_a(x) = \int_a^x f(y) dy$that is also in$C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image$[f_a]$of$f_a$in the linear space$A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say$f_a$and$f_b$) coincide. I think that letting$T(f) = [f_a]\$ will satisfy all the properties that you wanted of an "integration" map.