Definite integral with a complex number in Euler form Well... I spent an hour trying to figure out how to go from lhs to rhs:
$$\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\left( \int _{ k }^{ +\infty  } e^{ -iux }dx \right) du=\frac { 1 }{ 2 } +\frac { 1 }{ \pi  } \int _{ 0 }^{ +\infty  } \Re \left[ \frac { \phi _{ T }(u)e^{ -iuk } }{ iu }  \right] du$$
The $\phi$ being a characteristic function.
What I get is :
$$\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\frac { 1 }{ u } \left[ \sin(ux) \right] _{ k }^{ +\infty  }du+\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\frac { i }{ u } \left[ \cos(ux) \right] _{ k }^{ +\infty  }du$$ 
which seems to me will go nowhere since I don't see a way to solve the improper integrals with the cos and sin...
Would appreciate a hint, thanks.
 A: For the internal integral you get:
$$
\int _{ k }^{ +\infty  } e^{ -iux }dx=\biggr[\frac{e^{-iux}}{-iu}\biggr]_{k}^{\infty},
$$
whatever that means... So you have 
$$
\begin{eqnarray}
\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\left( \int _{ k }^{ +\infty  } e^{ -iux }dx \right) du&=&\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\biggr[\frac{e^{-iux}}{-iu}\biggr]_{k}^{\infty} du\\
&=&\lim_{z\to \infty}\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\biggr[\frac{e^{-iuz}}{-iu}-\frac{e^{-iuk}}{-iu}\biggr] du\\
&=&\lim_{z\to \infty}\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\frac{e^{-iuz}}{-iu} du\\
&&-\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du
\end{eqnarray}
$$
Let's assume $
\lim_{z\to \infty}\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\frac{e^{-iuz}}{-iu} du =\frac12
$ for the moment.
Now split the remaining integral in 2 parts $\int_{-\infty}^0 \dots du$ and $\int_0^{\infty} \dots du$ to get:
$$
\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ +\infty  } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du=
\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ 0  } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du+
\frac { 1 }{ 2\pi  } \int _{ 0  }^{ +\infty  } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du.
$$
Now substitute $u=-u'$, use $\phi_T(u)=\phi_T(-u)^*$ to get
$$
\frac { 1 }{ 2\pi  } \int _{ -\infty  }^{ 0  } \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du=
\frac { 1 }{ 2\pi  } \int _{ 0 }^{ \infty   } \phi _{ T }(-u')\frac{e^{iu'k}}{iu'} du'=
\frac { 1 }{ 2\pi  } \int _{ 0 }^{ \infty   } \left(\phi _{ T }(u')\frac{e^{-iu'k}}{-iu'}\right)^* du'
$$
Write $u$ for $u'$ again since it doesn't matter and combine the integrants
$$\frac { 1 }{ 2\pi  } \int _{ 0 }^{ \infty   } \left(\phi _{ T }(u)\frac{e^{-iuk}}{-iu}\right)^* + \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du=
\frac { 1 }{ 2\pi  } \int _{ 0 }^{ +\infty  }2 \Re \left[ \frac { \phi _{ T }(u)e^{ -iuk } }{ iu }  \right] du$$
