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How do you work out $\int e^{itx} \sin^2(t) / t^2 dt$?

Given the integral is with respect to $t$, I'm thinking this is about the inversion formula, but the exponent would have to be $-itx$. (In that case, the result would be the density for the sum of two independent uniformly distributed random variables on $[-1, 1]$.)

The other thing is that, if you swap $x$ and $t$, the integral might be a characteristic function, but I don't think $\sin^2(x) / x^2$ can be a density (it integrates to greater than 1).

johnsmith
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  • If you mean the definite integral $\int_{\mathbb{R}} e^{itx} \sin^2(t) / t^2 dt$, the following posts can help you: https://math.stackexchange.com/questions/253230/a-fourier-transform-using-contour-integral?noredirect=1, https://math.stackexchange.com/questions/3079815/calculate-int-limits-infty-infty-frac-sin2x-coswxx2dx-using?noredirect=1. – Gonçalo Jan 26 '24 at 23:15
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    $\int e^{itx} \sin^2(t) / t^2 dt$ is same as $\int e^{-itx} \sin^2(t) / t^2 dt$ since the imaginary part is $0$. – Kavi Rama Murthy Jan 26 '24 at 23:16
  • @Gonçalo: the OP ask for a method based on Fourier transform. Your link does not offer clear answers. – Mittens Jan 26 '24 at 23:21
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    According to WolframAlpha the inetgral has a solution.

    If you now how to solve this equation with $exp(-itx)$ and say the answer is $I(t)$ then integrating with the $exp(+itx)$ kernel will result in the answer $I^{\ast}(t)$, i.e., the complex conjugated output

     – Dennis Marx Jan 26 '24 at 23:21
  • @Mittens, those answers may not be clear for you, but are clear for others. Let the OP decide whether they are useful or not. – Gonçalo Jan 26 '24 at 23:33

1 Answers1

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The Fourier transform of the law of the symmetric uniform distribution on $[-1,1]$ is given by $$\frac12\int^1_{-1}e^{-itx}\,dx=\frac{\sin t}{t}$$

The Fourier transform of the distribution of the sum of two i.i.d symmetric uniform random variables $X_1,X_2$ on $[-1,1]$ is (I leave details to the OP) $$\phi_{X_1+X_2}(t)=E[e^{it(X_1+X_2)}]=\frac{\sin^2}{t^2}$$

Since $\phi_{X_1+X_2}\in L_1(\mathbb{R})$ (I leave details to the OP), the inversion formula states that $$\frac{1}{2\pi}\int^\infty_{-\infty}e^{-itx}\frac{\sin^2}{t^2}\,dt=f(x)$$ where $f$ is the density of the law of $X_1+X_2$, that is (I leave details to the OP) \begin{align} f(x)&=\frac14\big(\mathbb{1}_{(-1,1)}*\mathbb{1}_{(-1,1)}\big)(x)=\frac{1}{4}\int^1_{-1}\mathbb{1}_{(-1,1)}(x-y)\,dy\\ &=\frac14(2-|x|)_+ \end{align}

Mittens
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  • For $t \leq 0$, $F_{X_1 + X_2}(t) = \frac{1}{8}t^2 + \frac{1}{2}t + \frac{1}{2}$. For $t \in (0, 2]$, $F_{X_1 + X_2}(t) = \frac{-1}{8}t^2 + \frac{1}{2}t + \frac{1}{2}$. For $t > 2$, $F_{X_1 + X_2}(t) = 1$. So for $x \leq 0$, $f(x) = \frac{1}{4}x + \frac{1}{2}$. For $x \in (0, 2]$, $f(x) = \frac{-1}{4}x + \frac{1}{2}$. For $x > 2$, $f(x) = 0$. I think this is a less advanced way of working out $f$ (I couldn't follow indicators and the $+$ subscript). – johnsmith Jan 27 '24 at 23:47
  • @johnsmith: $x_+=\max(x,0)$; in a similar way, $x_-=-\min(0,x)=\max(0,-x)$. The function $t\mapsto\mathbb{1}{A}(t)$ has value $1$ if $t\in A$ and $0$ otherwise. Recall that if $X$ and $Y$ are random variables whose distributions have densities $f$ and $g$ respectively, then the distribution of $X+Y$ also has density $h$ and it given by the convolution of the densities $f$, and $g$, i.e. $h(x)=\int^\infty{-\infty}f(x-y) g(y),dy$. – Mittens Jan 27 '24 at 23:55