Looking for set of functions with special properties wrt integration. Good day,
I look for set of analytic (if possible elementary) functions with special properties.
Let $(a,b)$ be an interval $a<b$ and $x_n$ a sequence with 
1) $x_{n+1}<x_n$ and
2) $x_n \to a$ as $n\to\infty$
The sequence can be chosen any way one wants, some special way is OK for me. The set of functions $f_n(x)$ should satisfy:
$$\int_a^{x_m}f_n(x) dx = \delta_{m,n}$$
It is easy to do with non-analytic functions, e.g.
$f_n(x) = 1/(x_n-x_{n-1})$ for $x \in(x_{n-1},x_n)$,
$f_n(x) = -1/(x_{n+1}-x_{n})$ for $x \in(x_{n},x_{n+1})$
and zero elsewhere.
Thank you.
EDIT: Motivation
One often approximates a function $f$ by approximation $A_f$ by imposing an infinite number of constraints. Examples:
1) Taylor/Neumann series and Pade match derivatives:
$$\frac{d^nf}{dx^n}=\frac{d^n A_f}{dx^n},~n=1,2,3,\dots$$
2) Fourier (like) series match integrals
$$ \int sin(nx)f(x)dx= \int sin(nx) A_f(x)dx,$$
$$ \int cos(nx)f(x)dx= \int cos(nx) A_f(x)dx$$
$$n=1,2,3,\dots$$
and more examples could be found. Often, in combination with further restriction (smoothness, analyticity) the approximation is valid.
I try to combine integral and differential approach hoping for expansion
$$A_g(x)= \sum c_n f_n(x)$$
$$c_n = \int_a^{x_n} g(x)dx$$
I have no proof this expansion has to work, well it certainly works in some trivial cases ($g(x)=0$ or $g(x)=\sum_k c_k f_k(x)$). I hope the class of function which can be approximated is much larger...
 A: Well, it seems I have found such functions, but approximation is not working. Here is logical method of construction.
1) Start by finding the anti-derivative (integral) $F$ of $f$ (i.e. $F'=f$) on the interval $(0,1)$. A good starting point is
$$F(x)=sin(\pi/x)$$
and take its crossing with $y=0$ zero as $x_n$, i.e.
$$x_n = 1/n.$$
2) Because this is anti-derivative, you want the value to be defined at $x=0$, the lower limit of the interval. Thus you make the modification
$$F(x)=x ~sin(\pi/x).$$
This does not change its crossing points with zero, and can be evaluated at upper ($x=x_n$) and lower ($x=0$) boundary of each integral (at least as limit)
$$ \int_0^{x_n} f(x)dx = [F(x)]^{x_n}_0 = 0-0 = 0$$
3) So far so good. Now you need $F$ to get at one crossing not the value 0 but 1! To do this you define
$$F_n(x)=\frac{x~sin(\pi/x)}{(-1)^{(n+1)}n\pi(x-1/n)}.$$
With this construction all zeros are kept except the zero at $x=1/n=x_n$ where the value (well, the limit) is 1.
The set of function I looked for then can be written as:
$$f_n(x)= \frac{d}{dx}\left[ \frac{x~sin(\pi/x)}{(-1)^{(n+1)}n\pi(x-1/n)} \right]$$
which, I do not write explicitly because of length but can easily be done.
Conclusion: Trying the approximation I see it fails. It gets the integral correct thanks to wild oscillations, but even a large number of terms does not force the approximant to approximate the function.
