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Let $m$ and $n$ are two integers such that ,

  1. If $m \neq n$ then $$\int _a^b {\dfrac{f_n(x)}{f_m(x)} \ dx}=0$$
  2. If $m=n$ then $$\int _a^b {\dfrac{f_n(x)}{f_n(x)} \ dx}=\int _a^b 1 \ dx=b-a$$

I am looking for general properties of such functions ( such as how to find their differential equation, generating functions, polynomials, etc. )

For now I have found one example which is

$$ f_n(x)= e^{2\pi inx}\ \rm{for} \ a=0, b=1 .$$

How can we find the general properties of such functions or other functions that satisfy the above property ? Thank you for answers.

EDIT: We can find a function expression via using $f_n(x)$.

$$g(x)= \cdots +a_{-2}f_{-2}(x)+a_{-1}f_{-1}(x)+a_0f_0(x)+a_1f_1(x)+a_2f_2(x)+\cdots $$

$$ \int _a^b {\dfrac{g(x)}{f_m(x)} \ dx} =a_m (b-a)$$

or

$$g(x)= \cdots + \frac{b_{-2}}{f_{-2}(x)}+ \frac{b_{-1}}{f_{-1}(x)}+ \frac{b_{0}}{f_{0}(x)}+\frac {b_{1}}{f_{1}(x)}+\frac {b_{2}}{f_{2}(x)}+\cdots$$

$$ \int _a^b g(x)f_m(x) dx=b_m (b-a)$$

Thus

$$\frac{a_m}{b_m}=\frac{\int _a^b {\dfrac{g(x)}{f_m(x)} \ dx} }{\int _a^b g(x)f_m(x) dx}$$

Maybe it can be helpful further analysis to find such functions.

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    If $f_n$ is such a family, then $gf_n$ also works, where $g$ is any non-vanishing function on $(a,b)$.2012-04-19
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    But is the second condition necessary ? . I mean if $m = n$ then every definite integral yields the same thing. Is there some need to again specify it as a condition ? .2012-07-09
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    But are you sure that $\large \int _a^b {\dfrac{f_n(x)}{f_m(x)} \ dx}=0$ when $\large f_n(x)=\large e^{2\pi inx} \ \rm{for} \ a=0, b=1 $ ? . I dont see how $ \int _0^1 \large e^{2\pi i (n-m ) x} \ dx = 0 $2012-07-09
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    @Iyengar: m=n is not condition. It is just info about the result if $m=n$. if $m$ is not equal to $n$ , it is really zero if you solve the integral $\frac{e^{2\pi i(n-m)1}}{2\pi i(n-m)}-\frac{1}{2\pi i(n-m)}=0$2012-07-09
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    @Mathlover : Oh ok, but I didn't try, I would appreciate if you can specify how , if you are free. I got this $\dfrac{ e^{2 \pi i (n-m)} -1}{2 \pi i (n-m)} $ .2012-07-09
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    if m is not equal to n , $n-m$ will be negative or positive integer that different from zero. And we know $e^{2\pi ik} =\cos 2\pi k + i \sin 2\pi k $. if you put any integer for $k$ you will get $e^{2\pi ik} =\cos 2\pi k + i \sin 2\pi k =1$.2012-07-09
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    @Mathlover : Good implication. Thank you for detailed explanation . I feel ashamed that I am not able to think behind that.2012-07-09
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    you are welcome . I am very much happy if you got the point.2012-07-09
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    @Mathlover : Did you try periodic functions like $\sin (2 \pi x)$ or $\cos (2 \pi x)$ ?. I think they too admit to $0$. You must see that both functions should be equal after plugging the limits of the integral. P.S. : I beg you to use @<> tag when typing a comment. Your comments are not indicated on my side, and then I have to trace your question and find the comments repeatedly.2012-07-09
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    @Iyengar: Yes. but I got very hard integral expressions when I try. Maybe someone can help to find their results if zero or not for $m \neq n$ .2012-07-09
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    @Mathlover : Your edit looks good !!, may be it can lead to some seminal result.2012-07-09

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