# Is there integral analog of Ramanujan sum?

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.

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Are you asking if there is a canonical constant attached to the antiderivative of an arbitrary integrable function? I'm pretty sure the answer is "no", but it might become "yes" if you restrict your functions to some suitably nice family such as trigonometric sums of a certain form. –  S. Carnahan Dec 20 '10 at 9:02
The question is, can it be defined an operator which for any integral gives a constant which is analogous to the Ramanujan sum of series? A desirable consequence of this is that the integrals, normalized with such constant express certain properties that are not observed in integrals, normalized in other way, such as the above mentioned expression with differintegral. –  Anixx Dec 20 '10 at 9:26
Anixx: I'm sorry, but I am still a little bit confused by your exposition. Can you more clearly separate (perhaps visually) the motivation of the question and your suggestions of possible nice properties for the object you are looking for? (For example, should "balanced" be a requirement? It it not clear whether the paragraph beginning "One of the properties" is concerning the Ramanujan sum or the operator that you want.) And what do you mean by "symmetric against zero"? Thanks. –  Willie Wong Dec 20 '10 at 12:14
No, "balanced" is only property of discrete integrals normalized with Ranujan sum (as you can see, it is not even symmetric against zero). If we are talking about integrals, there may look for analog of "balanced" functions. Regarding "symmetric against zero" it means that R[f(x)]=-R[f(-x)]. This follows from the fact that $(f(x))'|_0=-(f(-x))'|_0$. –  Anixx Dec 20 '10 at 12:43
See my own answer below. –  Anixx Apr 2 '11 at 4:32
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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:

$\int_N 0 dx = 0$

$\int_N 1 dx = x$

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

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

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

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

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

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

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

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

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Sorry, I don't follow the point of this. Doesn't it seem strange to you that some monomials are getting weird constant terms and others aren't? –  David Hansen Apr 2 '11 at 18:09
The constant terms play the same role here as Bernoulli numbers in discrete integrating monomials. Since the general natural formula for discrete integrating monomials is $$\sum_x x^n=\frac{B_{n+1}(x)}{n+1}$$ The corresponding table for discrete integrals would be $$\sum_x 1 = -\frac12 + x$$ $$\sum_x x = \frac{1}{12}-\frac{x}{2}+\frac{x^2}{2}$$ $$\sum_x x^2 = \frac{x}{6}-\frac{x^2}{2}+\frac{x^3}{3}$$ $$\sum_x x^3 = -\frac{1}{120}+\frac{x^2}{4}-\frac{x^3}{2}+\frac{x^4}{4}$$ $$\sum_x x^4 = -\frac{x}{30}+\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}$$ –  Anixx Apr 2 '11 at 21:34
As you can see, here also the constant term not always present. The constant terms here are just Bernoulli numbers devided by (n+1). –  Anixx Apr 2 '11 at 21:34
If $f=x(1-x)$, your definitions above yield both $R[f]=-1/12$ and (if you want natural integration to be a linear operator, which I assume you do) $R[f]=0$. –  David Hansen Apr 3 '11 at 4:07
Good point. The formula $F(x)=\sum _{k=1}^{\infty} \frac{f^{(4k-1)}(0)}{2^k}+\int_0^x f(t) dt$ gives a linear operator anyway, but this as you pointed, contradicts the preposition of shift invariance. So, the preposition is not correct (does not satisfy linearity) so we should derive the values for trigonometric functions from other considerations (for example, simple differintegral formulas). –  Anixx Apr 3 '11 at 8:27