How to calculate the following integral $I=\int_{0}^{1}\frac{1}{\sqrt{y}}e^{-(x^2+y^2)/2y}dy$? 
I need to calculate the following integration $$I=\int_{0}^{1}\frac{1}{\sqrt{y}}e^{-(x^2+y^2)/2y}dy$$

I am trying to find some substitution but failed so far. I don't see if there any way to simplify this further. So how to do it? Can anyone give me a hint or a solution? Thanks.
Note 
this integral came up when I was trying to solve a probability question involving two dimensional joint pdf. I need to find $f_X(x)$
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{I \equiv
\int_{0}^{1}{1 \over \root{\vphantom{\large a}y}}
\,\expo{-\pars{x^{2} + y^{2}}/2y}\,\,\dd y =
\int_{0}^{1}\exp\pars{-\,\half\,x^{2}\,{1 \over y} - \half\,y}{\dd y \over \root{\vphantom{\large a}y}}:\ ?}$.

\begin{align}
\color{#f00}{I} & =
\int_{0}^{1}\exp\pars{-\,\half\,\verts{x}\bracks{{\verts{x} \over y} +
{y \over \verts{x}}}}{\dd y \over \root{\vphantom{\large a}y}}
\\[5mm] & \stackrel{y\,\, \equiv\,\, \verts{x}\,t^{2}}{=}\,\,\,\,\,\,\,\,
2\root{\verts{x}}\int_{0}^{1/\root{\verts{x}}}
\exp\pars{-\,\half\,\verts{x}\bracks{{1 \over t^{2}} + t^{2}}}\,\dd t
\\[5mm] & \stackrel{t\,\, \equiv\,\, \expo{-\theta}}{=}\,\,\,\,\,
2\root{\verts{x}}\int^{\infty}_{\ln\pars{\verts{x}}/2}
\exp\pars{-\verts{x}\cosh\pars{2\theta}}\expo{-\theta}\,\dd\theta
\\[1cm] & =
2\root{\verts{x}}\expo{-\verts{x}}\int^{\infty}_{\ln\pars{\verts{x}}/2}
\exp\pars{-2\verts{x}\sinh^{2}\pars{\theta}}\cosh\pars{\theta}\,\dd\theta
\\[5mm] & -
2\root{\verts{x}}\expo{\verts{x}}\int^{\infty}_{\ln\pars{\verts{x}}/2}
\exp\pars{-2\verts{x}\cosh^{2}\pars{\theta}}\sinh\pars{\theta}\,\dd\theta
\\[1cm] & =
\color{#f00}{\root{\pi \over 2}\bracks{%
\expo{-\verts{x}}\,\mrm{erfc}\pars{\verts{x} - 1 \over \root{2}}
-
\expo{\verts{x}}\,\mrm{erfc}\pars{\verts{x} + 1 \over \root{2}}}}
\end{align}
$\ds{\mrm{erfc}}$ is the Complementary Error Function. 

OP: Could you please ckeck it ?.

Note that
$$
\left\lbrace\begin{array}{rclcl}
\ds{\expo{-\theta}} & \ds{=} &
\ds{{\expo{-\theta} + \expo{\theta} \over 2} +
{\expo{-\theta} - \expo{\theta} \over 2}} & \ds{=} &
\ds{\cosh\pars{\theta} - \sinh\pars{\theta}}
\\[2mm]
\ds{\cosh\pars{2\theta}} & \ds{=} & \ds{2\sinh^{2}\pars{\theta} + 1}&&
\\[2mm]
\ds{\cosh\pars{2\theta}} & \ds{=} & \ds{2\cosh^{2}\pars{\theta} - 1}&& 
\end{array}\right.
$$
A: Let $I(x)$ be the integral given by
$$I(x)=\int_0^1 \frac{e^{-(x^2+y^2)/2y}}{\sqrt y}\,dy \tag 1$$
Note that $I(0)=\int_0^1 \frac{e^{-\frac12 y}}{\sqrt{y}}\,dy=\sqrt{2\pi}\text{erf}\left(\frac{\sqrt 2}{2}\right)$.
Since $I(x)$ is an even function of $x$, we may assume without loss of generality that $x> 0$.
We first enforce the substitution $y\to y^2$ in $(1)$ to obtain
$$\begin{align}
I(x)&=2\int_0^1 e^{-(x^2+y^4)/2y^2}\,dy \tag 2
\end{align}$$
Next, enforce the substitution $y=\sqrt{x}y$ to find 
$$\begin{align}
I(x)&=2\sqrt{x} \int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y^2+\frac{1}{y^2}\right)}\,dy \\\\
&=2\sqrt{x} e^{x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y+\frac{1}{y}\right)^2}\,dy \\\\
&=2\sqrt{x} e^{x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y+\frac{1}{y}\right)^2}\,\left(\frac12 +\frac1{2y^2}+\frac12-\frac{1}{2y^2}\right)\,dy \\\\
&=\sqrt{x} e^{x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y+\frac{1}{y}\right)^2}\,\left(1-\frac{1}{y^2}\right)\,dy\\\\
&+\sqrt{x} e^{-x}\int_0^{1/\sqrt{x}} e^{-\frac12 x\left(y-\frac{1}{y}\right)^2}\,\left(1+\frac{1}{y^2}\right)\,dy\\\\
&=-\sqrt{x} e^{x}\int_{\sqrt{x}+1/\sqrt{x}}^\infty e^{-\frac12 xu^2}\,du\\\\
&+\sqrt{x} e^{-x}\int_{-\infty}^{-\left(\sqrt{x}-1/\sqrt{x}\right)} e^{-\frac12 xu^2}\,du\\\\
&=-\sqrt{2}e^x\int_{(x+1)/\sqrt{2}}^\infty e^{-t^2}\,dt\\\\
&+\sqrt{2}e^{-x}\int_{-\infty}^{-(x-1)/\sqrt{2}} e^{-t^2}\,dt\\\\
&=\sqrt{\frac{\pi}{2}}e^{-x}\text{erfc}\left(\frac{x-1}{\sqrt{2}}\right)-\sqrt{\frac{\pi}{2}}e^x\text{erfc}\left(\frac{x+1}{\sqrt{2}}\right)
\end{align}$$
which after exploiting the evenness of $I(x)$ (i.e., replace $x$ with $|x|$) agrees with the result posted earlier by @Felix Marin!
