Expected value of $\sin(\pi U )\sum_{k=1}^\infty \frac {1}{(2k+U)(2k+1+U)}$ 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 $
 A: 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.
