Expectation of a sum of squares of normal variables with different variances. Let $Z_1 \sim N(0, 1), Z_2\sim N(0, \sigma^2) $, what is the expectation of $Y=\sqrt{Z_1^2 + Z_2^2}$? Is there a generalization of chi distribution based on normal variables with different variance?
Edit:
$Z_1$ and $Z_2$ are independent.
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\begin{align}\color{#66f}{\large{\mathbb E}\bracks{Y}}&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{\exp\pars{-Z_{1}^{2}/2} \over \root{2\pi}}\,
{\exp\pars{-Z_{2}^{2}/\pars{2\sigma^{2}}} \over \root{2\pi}\sigma}\,
\root{Z_{1}^{2} + Z_{2}^{2}}\,\dd Z_{1}\,\dd Z_{2}
\\[5mm]&={1 \over 2\pi\sigma}\int_{0}^{2\pi}\int_{0}^{\infty}
\exp\pars{-\,\half\,r^{2}\bracks{%
\sin^{2}\pars{\phi} + {\cos^{2}\pars{\phi} \over \sigma^{2}}}}r^{2}\,\dd r\,\dd\phi
\\[5mm]&={1 \over 2\pi\sigma}\int_{0}^{2\pi}{2^{3/2} \over \bracks{%
\sin^{2}\pars{\phi} + \cos^{2}\pars{\phi}/ \sigma^{2}}^{3/2}}
\int_{0}^{\infty}\exp\pars{-\,\half\,r^{2}}r^{2}\,\dd r\,\dd\phi
\\[5mm]&={\root{2} \over \pi\sigma}\int_{0}^{2\pi}{\dd\phi \over \bracks{%
\sin^{2}\pars{\phi} + \cos^{2}\pars{\phi}/ \sigma^{2}}^{3/2}}\
\underbrace{\int_{0}^{\infty}\exp\pars{-\,\half\,r^{2}}r^{2}\,\dd r}
_{\ds{=\ \color{#c00000}{\root{\pi \over 2}}}}
\\[5mm]&={1 \over \root{\pi}\sigma}\ \underbrace{%
\int_{0}^{2\pi}{\dd\phi \over \bracks{%
\sin^{2}\pars{\phi} + \cos^{2}\pars{\phi}/ \sigma^{2}}^{3/2}}}
_{\ds{\color{#c00000}{2\root{\sigma}\bracks{\,{\rm E}\pars{1 - \sigma} + \root{\sigma}\,{\rm E}\pars{1 - {1 \over \sigma}}}}}}
\end{align}
where
$\ds{{\rm E}\pars{m} \equiv
     \int_{0}^{\pi/2}\root{1 - m\sin^{2}\pars{\phi}}\,\dd\phi}$ is the Complete Elliptic Integral.

Then
  $$\color{#66f}{\large{\mathbb E}\bracks{y}
=
{2 \over \root{\pi}}\bracks{{{\rm E}\pars{1 - \sigma} \over \root{\sigma}}
+{\rm E}\pars{1 - {1 \over \sigma}}}}
$$

