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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 $n$). –  Gunnar Magnusson Jun 28 '11 at 20:21
    
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. –  SteveH Jun 28 '11 at 20:47
    
@Gunnar: exponentiation. I'm going to look into it, thnx –  JBSnorro Jun 28 '11 at 21:08
    
@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 0 –  JBSnorro Jun 28 '11 at 21:41
    
$f$ 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 $k,r,R$. –  SteveH Jun 28 '11 at 22:15
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1 Answer 1

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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

enter image description here

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? –  JBSnorro Jun 29 '11 at 15:02
    
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. –  Andrew Jun 29 '11 at 18:56
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