**Hint.** Recall that, from the definition of the Euler $\Gamma$ function, we have $$ \begin{align} \int_{0}^{\infty} e^{-bt} \, t^{-\alpha} \, dt = \frac{\Gamma(1-\alpha)}{b^{1-\alpha}}, \quad 0<\alpha<1, \Re b>0\. \tag1 \end{align} $$ Then put $b:=b_\epsilon:=\epsilon+i,\, \epsilon>0$, in $(1)$, let $\epsilon \to 0^+$ and take the real part to get $$ \begin{align} \int_{0}^{\infty} t^{-\alpha} \cos t \, dt & = \sin \left(\frac{\pi \alpha}{2}\right)\Gamma(1-\alpha), \quad 0<\alpha<1\. \tag2 \end{align} $$