You wish to (or will) show that $$ \int_z^\infty {x\over z} {1\over \sqrt{2\pi} }e^{-x^2/2}\,dx={\phi(z)\over z}, $$ where $\phi(z)={1\over\sqrt{2\pi}} e^{-z^2/2}$.
Hint:
Let $u={x^2\over 2}$. Then $du=x\,dx$, when $x=z$, $u=z^2/2$, and when $x=\infty$, $u= \infty$.
Solution:
Substitution then gives $$ \int_z^\infty {x\over z} {1\over \sqrt{2\pi} }e^{-x^2/2}\,dx =\int_{z^2/2}^{\infty}{1\over z\sqrt{2\pi}} e^{-u}\,du ={-1\over z\sqrt{2\pi}} e^{-u}\Biggl|_{z^2/2}^\infty ={e^{-z^2/2}\over z\sqrt{2\pi}}={\phi(z)\over z}. $$