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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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    Is $f^n$ exponentation or iteration of the function $f$? If its the first then you can probably work out an exact solution using the residue formula of complex analysis; via that formula your integrand can be replaced by a finite sum (the number of terms may depend on $n$).2011-06-28
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    A contour integral could be pretty tricky to set up because there appears to be terms of the form $e^{iaz}$ with $a$ positive and negative appearing.2011-06-28
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    @Gunnar: exponentiation. I'm going to look into it, thnx2011-06-28
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    @Gunnar: I tried. The only singular point in f^n(x) is at x = 0, but I find a residue of 0 there.... Which would imply that its integral is 02011-06-28
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    $f$ has only a removable singularity at $x=0$. Also, What contour did you use? One cannot draw any conclusions without proving convergence of the contour integral to the integral on $\mathbb R$. I would be surprised if the integral is $0$ for arbitrary $k,r,R$.2011-06-28
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    @Steve, I used as contour $$|x|=1$$2011-06-28
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    JBSnorro, the integral over $|z|=1$ is going to be $0$ because $f$ is an entire function (removable singularity at $z=0$). The contour must in the limit contain the real line for the contour integral to relate to the integral on $\mathbb R$.2011-06-28
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    I didn't think about this deeptly, but does this integral even converge? For $n=1$ the first term doesn't I think, as it behaves as $1/x^3$ around zero.2011-06-28
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    @Steve, I've outlined my attempts in an edit of my post. Can you plz specify on how to proceed ? Thank you2011-06-28
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    JBSnorro, you expansion is incorrect; the function has no pole at $x=0$, at least by my calculation. Also, since the function is entire (other than a removable singularity), trying to use contour integration is pretty hopeless.2011-06-29
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    @Steve, Yes, I had just figured that out, as I was editing my post when you commented. Thank you. Any other suggestions how I could acquire the solution?2011-06-29
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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:

enter image description here

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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    Thank you so much. This is very helpful. I suppose your $m$ and $d$ are functions of $r$ and $R$, since it are the mean and deviation of the function $g(y)$? And in my case k is a constant, and n the (large) variable. Does that change anything?2011-06-29
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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