Integral involving translates of $\{x\}$.

Question

We have two functions: $F(x)$ and $G(x)$

Suppose the improper integral has the value 1. $$\int_{1}^{\infty} F(x) G(x) \mathrm{d}x = 1$$

Can we find the value of the integral, where c is a constant? $$\int_{1}^{\infty} F(x+c) G(x) \mathrm{d}x$$

I am interested in generalized answers and not specific examples. Or, if any links/books' info are given where I can read similar problems, that would be highly appreciated.

EDIT: Due to the highly generalized nature of this question, I am giving the examples I am working upon.

$$F(x) = \{x\}$$ Where $\{x\}$ denotes the fractional part of x, and $$G(x) = \frac{\sin(x\log(x))}{x^{h+1}}$$ where constants are $0< h < 1$ and $c = 1/2$.

Answer 1

Do you know anything about $F$ and $G$? If not, then certainly no. Let $G(x) = 1$ if $x \leq 2$ and $0$ otherwise. Let $F(x) = 1$ if $x< 2$; we leave $F(x)$ undefined elsewhere for now. Then $\int_1^{\infty} F(x)G(x) = 1$.

Consider the function $h(c) = \int_1^\infty F(x+c)G(x)dx$ defined on $0 < c$. I claim that for any continuously differentiable function $H(c)$ such that $H(0) = 1$, I can find an extension to $F(x)$ (i.e. filling in the undefined spots) such that $h(c) = H(c)$.

Proof: Notice that by definition $h(c) = \int_{1+c}^{c+2} F(x) dx$. Then $h'(c) = F(c+2) - F(c+1)$. So start with $F(x) = 1$ for $1\leq x < 2$, then define $F(x) = H'(x-2) + H'(x - 3) + \ldots + H'(x - k) + F(x - k + 1)$, where $k$ is the integer where $1 \leq x - k + 1 < 2$. The by construction, $h'(c) = H'(c)$ and $h(0) = H(0)$, so the two functions agree.

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From the above we can conclude that just knowing $\int_1^\infty F(x) G(x) dx = 1$ tells us absolutely nothing about $\int_1^\infty F(x+c) G(x) dx$; it can be pretty much anything you want it to be.

Answer 2

Now that you have updated your question with the fact that $F(x) = \{x\}$ (the standard notation for the fractional part of $x$), there is quite a bit more that can be said. In particular, you can give an interpretation to $\int_1^{\infty} \{x + c\} G(x) dx$. Let $0 < c < 1$.

Then $$\{x + c\} = \begin{cases} \{x\} + c, &0 \leq \{x\} < 1 - c; \\ \{x + c\}= \{x\} + c - 1, &1-c \leq \{x\} < 1. \end{cases}$$

We have $$\begin{align} &\int_1^{\infty} \{x + c\} G(x) dx = \int_{1 \leq x < \infty, 0 \leq \{x\} < 1 - c} (\{x \} + c) G(x) dx + \int_{1 \leq x < \infty, 1-c \leq \{x\} < 1 } (\{x \} + c - 1) G(x) dx \\ &= \int_1^{\infty} \{x \} G(x) dx + c \int_{1 \leq x < \infty, 0 \leq \{x\} < 1 - c} G(x) dx + (c-1) \int_{1 \leq x < \infty, 1-c \leq \{x\} < 1 } G(x) dx \\ &= 1 + c \int_{1 \leq x < \infty, 0 \leq \{x\} < 1 - c} G(x) dx - (1-c) \int_{1 \leq x < \infty, 1-c \leq \{x\} < 1 } G(x) dx. \end{align}$$

The two remaining integrals constitute an average of sorts, weighted to account for the fact that they are being taken over different percentages of the interval $[1,\infty)$. The first integral gets weighted by $c$ but includes $1-c$ of the interval $[1,\infty)$, as it is being taken over the set $\cup_{i=1}^{\infty} [i,i+1-c)$. (Remember that $c$ is a fraction between $0$ and $1$.) The second integral gets weighted by $1-c$ but includes $c$ of the interval $[1,\infty)$, as it is being taken over the set $\cup_{i=2}^{\infty} [i-c,i)$. So $\int_1^{\infty} \{x + c\} G(x) dx$ just shifts the weights on the values of $G(x)$ in $\int_1^{\infty} \{x \} G(x) dx$ in the manner I just described. The resulting value for $\int_1^{\infty} \{x + c\} G(x) dx$ will be either greater or smaller than $1$, depending on whether the larger values of $G(x)$ over $[1, \infty)$ tend to clump just above each integer value of $x$ or just below.

Other than this, I think Willie Wong's answer still applies. In particular, you still can't get an exact answer for $\int_1^{\infty} \{x + c\} G(x) dx$ -- just an interpretation of it.

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You also asked for references for problems similar to yours. One such is the convolution of two functions $f$ and $g$, one form of which is

$$(f*g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t - \tau) d \tau.$$

Convolutions have lots of interesting properties and interpretations. See MathWorld's article on convolutions for more information.

Answer 3

In reasonably general situations (using, e.g, distributions if needed) the second integral can be differentiated with respect to $c$ and the derivatives will be $\int F^{(n)}(x)G(x) dx$. Just as a function is not determined by its value at a point, these derivatives are not determined by the value at $c=0$, so even the local behavior of perturbations of the original integral (the asymptotics of the second integral for small $c>0$) is not determined by the integral with $c=0$.

It's hard to think of examples where the second integral is determined, no matter how constrained $F$ and $G$ are. Consider the situation where $F(x)$ and $1/G(x)$ are polynomials of degrees 2 and 4 with positive coefficients. This is a very well behaved problem but the variation with respect to $c$ is undetermined by a single piece of information. If you require integer coefficients or a similar discretization then there is some hope of recovering $F$ and $G$ from the value of the first integral, but this is a Diophantine problem and no longer a question of real analysis.

Answer 4

If $c$ is negative, the second integral involves values $F(t)$ for $t<1$, which can be changed arbitrarily without affecting the first integral! But even if you restrict yourself to positive $c$, I guess there's nothing you can say in general without additional information.