<s>[INST] Finding the anti-derivative of $ \frac{e^{-c y^2 }}{y\sqrt{y^2-1}}$ I am trying to evaluate the integral
\begin{align}
\frac{1}{2\sqrt{2}\pi}\int_{0^{-}}^{t} ds \ \frac{e^{-x^2/2S^2(t,s) }}{\Sigma(s) S(t,s)}
\end{align}
where $S(t, s) = 2D(t-s)+\frac{\Sigma(s)}{2}$ and $\Sigma(s)= \sigma^2+2Ds$. I found a rather neat change of variable by taking $\xi=S^{-1}(t,s)$ so that
\begin{align}
\Sigma(s)&=\sqrt{2} \ \xi^{-1}\sqrt{\xi^2\Sigma^2(t)-1}\\
2D(t-s)&=2 \ \xi^{-2}-\Sigma^2(t)\\
ds&=\frac{2}{D\xi^3}\ d\xi
\end{align}
Applying this to the integral gives the result
\begin{align}
\frac{1}{2 \pi D}\int_{\sqrt{2}/\Sigma (2 t)}^{\sqrt{2}/\Sigma (t)} d{\xi} \ \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}=\frac{1}{2 \pi D}{\int_{\xi_L}^{\xi_H} d{\xi} \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}}.
\end{align}
and at last applying the change of variable $y=\Sigma(t)\xi$ gives
\begin{align}
\frac{1}{2 \pi D}{\int_{\xi_L}^{\xi_H} d{\xi} \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}} = \frac{1}{2 \pi D}{\int_{\sqrt{2}\Sigma(t)/\Sigma(2t)}^{\sqrt{2}} d{y} \frac{e^{-(x^2/2\Sigma^2(t)) y^2 }}{y\sqrt{y^2-1}}} =\frac{1}{2 \pi D}\color{blue}{\int_{a}^{b} d{y} \frac{e^{-c y^2 }}{y\sqrt{y^2-1}}}
\end{align}
Above calculation indeed check out numerically.

It seems the integral in blue could have an anti-derivative. However, no luck just yet!  Someone who knows a way forward? Many thanks in advance!
 [/INST] As others have pointed out, no closed-form solution in terms of elementary functions can be found. However, if you can live with the error function and the Owen T function then a closed-form solution can be found in terms of these special functions.
I will assume the error function $\operatorname{erf} (x)$ is well known to you. An integral representation for the Owen T function we intent to use can be found here. It is
$$\operatorname{T}(h,a) = \frac{1}{2\sqrt{2\pi}} \int_h^\infty e^{-x^2/2} \operatorname{erf} \left (\frac{ax}{\sqrt{2}} \right )\, dx.$$
Enforcing a substitution of $x \mapsto x \sqrt{2}$ gives the following form that will be found most convenient for the integral considered here:
$$\int_z^\infty e^{-x^2} \operatorname{erf}(ax) \, dx = 2 \sqrt{\pi} \operatorname{T}(z \sqrt{2}, a). \tag1$$
Now let us look at the integral in question. Let
$$I(t) = \int_a^b \frac{e^{-tx^2}}{x\sqrt{x^2 - 1}} \, dx, \quad t > 0.$$
I will assume $1 < a < b = \sqrt{2}$. Note that $I(\infty) = 0$ and we are required to find $I(c)$ where $c > 0$. Using Feynman's trick of differentiating under the integral sign with respect to $t$ gives
$$I'(t) = - \int_a^b \frac{x e^{-tx^2}}{\sqrt{x^2 - 1}} \, dx.$$
Enforcing a substitution of $u^2 = x^2 - 1$ yields
\begin{align*}
I'(t) &= e^{-t} \int_{\alpha_1}^{\alpha_2} e^{-tu^2} \, du = -\frac{\sqrt{\pi} e^{-t}}{2\sqrt{t}} \operatorname{erf}(u\sqrt{t}) \Big{|}_{\alpha_1}^{\alpha_2}\\
&= -\frac{\sqrt{\pi} e^{-t}}{2\sqrt{t}} \left [\operatorname{erf}(\alpha_2 \sqrt{t}) - \operatorname{erf}(\alpha_1 \sqrt{t}) \right ].
\end{align*}
Here $\alpha_1 = \sqrt{a^2 - 1}$ and $\alpha_2 = \sqrt{b^2 - 1}$.
Now, as
$$\int_c^\infty I'(t) \, dt = I(\infty) - I(c) = -I(c),$$
since $I(\infty) = 0$, we have
$$I(c) = -\frac{\sqrt{\pi}}{2} \int_c^\infty \frac{e^{-t}}{\sqrt{t}} \left [\operatorname{erf} \left (\alpha_2 \sqrt{t} \right ) - \operatorname{erf} \left (\alpha_1 \sqrt{t} \right ) \right ] \, dt.$$
On enforcing a substitution of $t \mapsto t^2$ one obtains
$$I(c) = \sqrt{\pi} \int_{\sqrt{c}}^\infty e^{-t^2} \left [\operatorname{erf}(\alpha_2 t) - \operatorname{erf}(\alpha_1 t) \right ] \, dt.$$
Evaluating this integral in terms of the Owen T function, from (1) one immediately sees that
$$I(c) = 2\pi \left (\operatorname{T} \left (\sqrt{2c}, \alpha_2 \right ) - \operatorname{T} \left (\sqrt{2c}, \alpha_1 \right ) \right ),$$
or
$$\int_a^b \frac{e^{-cy^2}}{y\sqrt{y^2 - 1}} \, dy = 2\pi \left (\operatorname{T} \left (\sqrt{2c}, \sqrt{b^2 - 1} \right ) - \operatorname{T} \left (\sqrt{2c}, \sqrt{a^2 - 1} \right ) \right ),$$
the required closed-form expression for our integral.
Note, if $b = \sqrt{2}$, as it appears to be from the question, further simplification of one of the Owen T functions is possible. As can be seen here when the right argument of the Owen T function is unity
$$\operatorname{T} \left (\sqrt{2c}, \sqrt{b^2 - 1} \right ) = \operatorname{T} \left (\sqrt{2c}, 1 \right ) = \frac{1}{8} \left [1 - \operatorname{erf}^2 \left (\sqrt{c} \right ) \right ],$$
allowing one to write
$$\int_a^{\sqrt{2}} \frac{e^{-cy^2}}{y\sqrt{y^2 - 1}} \, dy = -\frac{\pi}{4} \left [8 \operatorname{T} \left (\sqrt{2c}, \sqrt{a^2 - 1} \right ) + \operatorname{erf}^2 \left (\sqrt{c} \right ) - 1 \right ].$$
</s>