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
$\frac { 1 }{ 2\pi } \int _{ 0 }^{ \infty } \left(\phi _{ T }(u)\frac{e^{-iuk}}{-iu}\right)^* + \phi _{ T }(u)\frac{e^{-iuk}}{-iu} du= \frac { 1 }{ 2\pi } \int _{ 0 }^{ +\infty }2 \Re \left[ \frac { \phi _{ T }(u)e^{ -iuk } }{ iu } \right] du$