TITLE: Integral involving translates of $\{x\}$.
QUESTION [6 upvotes]: 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 $c = 1/2$.

REPLY [2 votes]: 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.

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.