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I would like to calculate the following expectation: $$E \left[\sin(\pi U )\sum_{k=1}^\infty \frac {1}{(2k+U)(2k+1+U)}\right]$$ where $U$ is uniformly distributed on the interval $[0,1]$.

I want to show that it is equal to:

$ \int_{1}^{\infty}\frac{\sin(2\pi x)}{x}dx $

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  • $\begingroup$ It seemed like the equations disappeared in the last edit, so I rolled it back. If for some reason this is due to a but on my end (e.g. this), I apologize! $\endgroup$ – GeoMatt22 Apr 23 '17 at 5:10
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First, express the expectation $E$ as $$ E = \int_{0}^1 \sum_{k=1}^\infty \frac{\sin(\pi u) }{(2k +u)(2 k +1 +u)}\,\text{d}u. $$ Note that for $u$ between $0$ and $1$ $$ \frac{1}{(2k + u)(2 k +1 +u)} = \frac{1}{2k + u} - \frac{1}{2 k + 1 + u}. $$ Now multiplying by $\sin(\pi u)$ and integrating $$ \int_{0}^1 \frac{\sin(\pi u)}{(2k +u)(2 k +1 +u)}\,\text{d}u = \int_{0}^1 \frac{\sin(\pi u)}{ 2k +u} \, \text{d}u - \int_{0}^1 \frac{\sin(\pi u)}{ 2k + 1 +u} \, \text{d}u $$ At right hand side use the change of variable $z:= 2k + u$ in the first integral and $z:= 2k + 1 + u$ in the second $$ \int_{0}^1 \quad = \int_{2k }^{2k +1} \frac{\sin(\pi z)}{z} \, \text{d}z + \int_{2k+1}^{2k+2} \frac{\sin(\pi z)}{ z} \, \text{d}z = \int_{2k}^{2k+2} \frac{\sin(\pi z)}{ z} \, \text{d}z $$ because $\sin[\pi (z - 2k)] = \sin(\pi z)$ and $\sin[\pi (z - 2k -1)] = -\sin(\pi z)$. Summing these relations for $k = 1$ to $K$ and taking $K \to \infty$ we get $$ E = \int_{2}^{\infty} \frac{\sin(\pi z)}{ z} \, \text{d}z $$ which is the wanted result after a new change of variable $x:= z /2 $. Note that the integral is a semi-convergent (only) Riemann integral.

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