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

We have 2 features:. $F(x)$ and also $G(x)$

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

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

I want generalised solutions and also general instances. Or, if any kind of links/books' details are offered where I can read comparable troubles, that would certainly be very valued.

EDIT: Due to the very generalised nature of this inquiry, I am offering the instances I am functioning upon.

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

0
2019-05-18 21:55:05
Source Share

If $c$ is adverse, the 2nd indispensable entails values $F(t)$ for $t<1$, which can be transformed randomly without influencing the first indispensable! Yet also if you limit on your own to favorable $c$, I presume there is absolutely nothing you can claim as a whole without added details.

0
2019-05-21 07:36:10
Source

Do you recognize anything concerning $F$ and also $G$? Otherwise, after that absolutely no. Allow $G(x) = 1$ if $x \leq 2$ and also $0$ or else. Allow $F(x) = 1$ if $x< 2$ ; we leave $F(x)$ undefined in other places in the meantime. After that $\int_1^{\infty} F(x)G(x) = 1$.

Take into consideration the function $h(c) = \int_1^\infty F(x+c)G(x)dx$ specified on $0 < c$. I assert that for any kind of continually differentiable function $H(c)$ such that $H(0) = 1$, I can locate an expansion to $F(x)$ (i.e. completing the undefined places) such that $h(c) = H(c)$.

Evidence: Notice that necessarily $h(c) = \int_{1+c}^{c+2} F(x) dx$. After that $h'(c) = F(c+2) - F(c+1)$. So start with $F(x) = 1$ for $1\leq x < 2$, after that specify $F(x) = H'(x-2) + H'(x - 3) + \ldots + H'(x - k) + F(x - k + 1)$, where $k$ is the integer where $1 \leq x - k + 1 < 2$. The by building and construction, $h'(c) = H'(c)$ and also $h(0) = H(0)$, so both features concur.

From the over we can end that feeling in one's bones $\int_1^\infty F(x) G(x) dx = 1$ informs us definitely nothing concerning $\int_1^\infty F(x+c) G(x) dx$ ; it can be virtually anything you desire it to be.

0
2019-05-21 07:34:09
Source

In sensibly basic scenarios (making use of, e.g, circulations if required) the 2nd indispensable can be set apart relative to $c$ and also the by-products will certainly be $\int F^{(n)}(x)G(x) dx$. Equally as a function is not established by its value at a factor, these by-products are not established by the value at $c=0$, so also the neighborhood actions of perturbations of the initial indispensable (the asymptotics of the 2nd indispensable for tiny $c>0$) is not established by the indispensable with $c=0$.

It is tough to consider instances where the 2nd indispensable is established, despite just how constricted $F$ and also $G$ are. Take into consideration the scenario where $F(x)$ and also $1/G(x)$ are polynomials of levels 2 and also 4 with favorable coefficients. This is a quite possibly acted trouble yet the variant relative to $c$ is obscure by a solitary item of details. If you call for integer coefficients or a comparable discretization after that there is some hope of recouping $F$ and also $G$ from the value of the first indispensable, yet this is a Diophantine trouble and also no more an inquiry of actual evaluation.

0
2019-05-21 07:32:32
Source

Now that you have actually upgraded your inquiry with the reality that $F(x) = \{x\}$ (the criterion symbols for the fractional component of $x$), there is a fair bit extra that can be claimed. Specifically, you can offer an analysis to $\int_1^{\infty} \{x + c\} G(x) dx$. Allow $0 < c < 1$.

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

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 2 continuing to be integrals comprise approximately types, weighted to make up the reality that they are being taken control of various percents of the interval $[1,\infty)$. The first indispensable obtains weighted by $c$ yet consists of $1-c$ of the interval $[1,\infty)$, as it is being taken control of the set $\cup_{i=1}^{\infty} [i,i+1-c)$. (Remember that $c$ is a portion in between $0$ and also $1$.) The 2nd indispensable obtains weighted by $1-c$ yet consists of $c$ of the interval $[1,\infty)$, as it is being taken control of the set $\cup_{i=2}^{\infty} [i-c,i)$. So $\int_1^{\infty} \{x + c\} G(x) dx$ simply changes the weights on the values of $G(x)$ in $\int_1^{\infty} \{x \} G(x) dx$ in the fashion I simply defined. The resulting value for $\int_1^{\infty} \{x + c\} G(x) dx$ will certainly be either better or smaller sized than $1$, relying on whether the bigger values of $G(x)$ over $[1, \infty)$ often tend to glob simply over each integer value of $x$ or simply below.

Apart from this, I assume Willie Wong is solution still uses. Specifically, you still can not get a specific solution for $\int_1^{\infty} \{x + c\} G(x) dx$ - - simply an analysis of it.

You additionally requested for referrals for troubles comparable to your own. One such is the convolution of 2 features $f$ and also $g$, one kind of which is

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

Convolutions have great deals of intriguing buildings and also analyses. See MathWorld is write-up on convolutions for additional information.

0
2019-05-21 05:22:44
Source