I have to show that $\int_{\lambda}^{\infty} (z-\lambda)^{\delta} \, \frac{{e}^{\frac{-z^2}{2}}}{\sqrt{2\pi}} dz = \frac{2^{-1-\frac{\delta}{2}} \, e^{\frac{-{\lambda}^2}{2}} }{\sqrt{\pi}} \, \Gamma(1+\delta) \, U \Big(\frac{1+\delta}{2},\frac{1}{2},\frac{{\lambda}^2}{2} \Big)$. This is at least, what Mathematica gives.
I have $\lambda,\delta \geq 0$. I have already reached the following:
$\int_{\lambda}^{\infty} (z-\lambda)^{\delta} \, \frac{{e}^{\frac{-z^2}{2}}}{\sqrt{2\pi}} dz = \int_{\lambda}^{\infty} \sum_{k=0}^{\infty} \binom{\delta}{k} (-\lambda)^{\delta-k} z^k \frac{{e}^{\frac{-z^2}{2}}}{\sqrt{2\pi}} dz = \sum_{k=0}^{\infty} \binom{\delta}{k} (-\lambda)^{\delta-k} \int_{\lambda}^{\infty} z^k \frac{{e}^{\frac{-z^2}{2}}}{\sqrt{2\pi}} dz $.
If $u=\frac{z^2}{2}$, the summation of definite integrals become
$\sum_{k=0}^{\infty} \binom{\delta}{k} (-\lambda)^{\delta-k} \int_{\frac{\lambda^2}{2}}^{\infty} (2u)^{\frac{k-1}{2}} \frac{e^{-u}}{\sqrt{2\pi}} du = \sum_{k=0}^{\infty} \binom{\delta}{k} (-\lambda)^{\delta-k} \frac{2^\frac{k}{2}}{\sqrt{\pi}} \int_{\frac{\lambda^2}{2}}^{\infty} u^{\frac{k+1}{2}-1} e^{-u} \, du $
By the definition of the incomplete gamma function (and using gamma functions for the binomial coefficient), I am stucked at
$\sum_{k=0}^{\infty} \binom{\delta}{k} (-\lambda)^{\delta-k} \frac{2^\frac{k}{2}}{\sqrt{\pi}} \, \Gamma \Big( \frac{k+1}{2}, \frac{\lambda^2}{2} \Big) = \sum_{k=0}^{\infty} \frac{\Gamma(\delta+1)}{\Gamma(\delta-k+1) \, \Gamma(k+1)} \, (-\lambda)^{\delta-k} \, \frac{2^\frac{k}{2}}{\sqrt{\pi}} \, \Gamma \Big( \frac{k+1}{2}, \frac{\lambda^2}{2} \Big) $.
I need to arrive at the result involving the Tricomi confluent hypergeometric function. I would really appreciate your help. Thank you so much!