Question

I already asked a similar question http://mathoverflow.net/questions/48742/is-there-natural-integration-constant-closed, but it seems it was not understood properly, so I am trying now to formulate it differently.

Is there an operator R[f] that plays in integral world the same role as Ramanujan sum plays in the world of series?

It is known that Ramanujan sum plays the role of natural integration constant for discrete integration, that is if F(x) is the discrete integral of f(x), usually postulated that

$$F(0)=\sum^{\Re}f(x)$$

This yields the functions which have some very useful properties, for example, Bernoulli polynomials (which are the results of discrete integration of power function), the Hurwitz Zeta function (which is generalization of Bernoulli polynomials), etc. Discrete integrals which normalized with Ramanujan sum (or equal method) called "balanced" (as opposed to some functions normalized without it, for example to be zero in zero such as Harmonic numbers).

One of the properties of balanced functions is that $$\int_0^1 F(x)\ dx=0$$

To outline the properties of such operator in integral world we should admit that such operator should be symmetric against zero (unlike the Ramanujan sum).

It is also highly desirable (if possible) that R[exp(x)]=1 thus making integral of exponent itself exponent and thus invariant against differintegral operator. This would allow to provide integration constants for trigonometric functions that would work in agreement with known expressions for differintegral:

$$D^q \sin x=\sin \left(x+\frac{q\pi}{2}\right)$$

P.S. See my own answer below.

Answer 1

I will try to answer this question myself.

My aim was to find the most natural and universal constant of integration which would allow to define an operator of "natural integration", unambiguously selecting one distinguished integral function for any given function.

So let's designate as $F(x)=\int_N f(x) dx$ the natural integral which we are trying to define and $R[f]=F(0)$ is the value of the natural integral in zero, the constant of integration we are trying to find.

First of all we know that operation of integration is symmetrical over choice of direction. This means that we should require that $R[f(x)]=-R[f(-x)]$. This follows from the fact that $(F(x))'|_0=-(f(-x))'|_0$ for any function F. It allows us to directly define the constant of integration for even functions: its natural integral should be odd function and $R[f]=0$ if $F[0]$ is defined.

Since all functions can be represented as a sum of even and odd function, we only have to define $R[f]$ for odd functions.

This is a more difficult task.

But we can spot one more property of natural integral. If we want it to be independent of any particular point on the real axis, any shift in the given function should lead to a corresponding shift in the integral without any other change. This means that if a function can be made even by shifting it along real axis, we can find its natural integral by applying the rule for even functions.

I.e. for continuous f, if $f(x_0-x)=f(x_0+x)$ for any $x$, then $R[f]=\int_{x_0}^0 f(t)dt$.

A function can have more than one axis of symmetry though, but if they are more than one, the function is periodic, and $R[f]$ is still unique.

This method allows us to find natural integrals for sine, cosine, hyperbolic sine and cosine as well as exponent as in the following table:

$\int_N \sin x dx = -\cos x$

$\int_N \cos x dx = \sin x$

$\int_N \sinh x dx = \cosh x$

$\int_N \cosh x dx = \sinh x$

$\int_N \exp x dx = \exp x$

But how can we define the natural integrals for other functions?

To cover all analytic functions we have to define the natural integral on polynomials.

First of all we spot that natural integral of hyperbolic sine has 1 in zero. This means that natural integration it term by term adds a sequence that sums up to 1. The most simple sequence of this kind is 1/2+1/4+1/8+1/16+... . Since integrating exponent gives the same result, it is logical that the terms which stay in odd position after integration contribute nothing. Similarly as integrating minus sine gives the same result in zero, we can conclude that all terms that stay on even positions but do not divide by 4 also contribute nothing.

Simplifying all said above and accounting for a factorial which exists in each term we can obtain a simple formula:

$$R[f]=\sum _{k=1}^{\infty} \frac{f^{(4k-1)}(0)}{2^k}$$

and thus the general formula for natural integral:

$$F(x)=\sum _{k=1}^{\infty} \frac{f^{(4k-1)}(0)}{2^k}+\int_0^x f(t) dt$$

This formula again confirm the listed above results, derived by another method, but also adds the following list for polynomials:

$\int_N 0 dx = 0$

$\int_N 1 dx = x$

$\int_N x dx =\frac{x^2}{2}$

$\int_N x^2 dx = \frac{x^3}3$

$\int_N x^3 dx = 3+\frac{x^4}4$

$\int_N x^4 dx = \frac{x^5}5$

$\int_N x^5 dx = \frac{x^6}6$

$\int_N x^6 dx = \frac{x^7}7$

$\int_N x^7 dx = 1260 + \frac{x^8}8$

$\int_N x^n dx =\frac{x^{n + 1}}{n + 1}+\begin{cases} & \frac{n!}{2^{\frac{n + 1}4}}, & \mbox{if n+1 divides by 4,} \ 0, & \mbox{otherwise} \end{cases}$