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I've been stuck on a particular integral I encountered. I don't need an exact solution, I doubt it even exists.

$f(x)=\frac{e^{-i (r+R-k) x} \left(i-2 e^{i (r+R) x} r x-R x+e^{2 i r x} (R x-i)\right)}{ x^3}$

I'm tasked to find$\int_{-\infty}^\infty{f^n(x) dx}$ for very large integer n and $0 < r < R$

Any suggestions on how to do so? Thanks

EDIT: ok, I've made some progress:

The following Laurent series gives f(x): $f(x)=\sum _{m=0}^{\infty } a_mx^m$ with $a_m=\frac{(i (r-R))^{2+m} R-R (-i (r+R))^{2+m}}{r (2+m)!}-\frac{i (i (r-R))^{3+m}-i (-i (r+R))^{3+m}}{r (3+m)!};$

which is related to the contour (a circle at any non-zero distance from $x = 0$) integral via $\oint_C f(x) = 2 \pi i a_{-1} =0$ when there is only one singular point.

But this was all for $n=1$, and I don't know how $\oint_C f(x)$ relates to $\int_{-\infty}^\infty{f(x) dx}$, let alone when $n\neq1$.

On $f^n(x)$, I didn't find an explicit expression for the corresponding coefficients, but did find that all coefficients are $0$ for $m < 0$. I don't know what that implies for $\int_{-\infty}^\infty{f^n(x) dx}$, please elaborate.

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    If this were my problem, I would look into computing it numerically. The function decreases fairly rapidly and so you may be able to get some reasonable estimates. If you explain where the function comes from, maybe more people can help you.2011-06-29

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Let $k=0$. Then Fourier transform of $f$ is $ g(y)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-i x y} f(y)\,dy= $ $ \frac{1}{2} \sqrt{\frac{\pi }{2}} \left((r-R-t)^2 (-\text{sgn}(r-R-t))+(r+R+t)^2 \text{sgn}(-r-R-t)-\right. $ $ 2 R (r+R+t) \text{sgn}(-r-R-t)+2 R (-r+R+t) \text{sgn}(r-R-t)+4 r t \text{sgn}(t)\Big). $ It is non-negative on $\mathbb R$ and $\mathrm{supp}\, g =[-R-r,0]\,$. For example, $r=1$, $R=2$ gives

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If $a=\int_{-\infty}^\infty g(y)\,dy\ $ then $p(y)=g(y)/a\,$ is a probability density function of some continuous random variable $\xi$. Convolution of $g$'s corresponds to $f^n$. By the central limit theorem for large $n$ its graphics will be a bell-shaped curve (with support on $[-n(R+r),0]$). For $g*g*g$ from previous example:

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Multiplying the $f$ by $e^{ i k x}$ means a shift for $g$ on $k$. Also integral of $f^n$ on $\mathbb R$ is equal to its Fourier transform at the origin. So for fixed $r$ and $R$ the result as a function of $k$ is positive and bell-shaped curve on $[0,n(r+R)]$ and zero otherwise. If denote $m$ and $d$ expectation and dispersion of $\xi$, it has to be sort of $\frac{a^n e^{-\frac{n (k+m)^2}{2 d}}}{\sqrt{2 \pi d n}}.$ But would it give the exact asymptotic is not clear since the central limit theorem is directly applicable for $|k(n)+m|\le c/\sqrt n$, where $c$ is a constant.

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    Yes, $a$, $m$ and $d$ are functions of $r$ and $R$. In the previous answer I made a miscalculation. The integral will be not equal to zero for $k\in(0,R+r)$. May be not, but that has to be proven.2011-06-29