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Here we only verified numerically that

\begin{equation} \lim_{n \rightarrow \infty} {\mathfrak f}_{2n-1}^{(y,h,\sigma)} = \frac{e^{-\frac{y^2}{2 \left(h^2+\sigma ^2\right)}}}{\sqrt{h^2+\sigma ^2}} \end{equation}

$n$ limit of the integral in question." />

Update 1:

Here we provide a large-$n$ approximation of the result $(1)$. We define auxiliary functions as:

\begin{eqnarray} P^{(h,\sigma)}_1(y)&=&\frac{ h \sigma ^2 \sqrt{\frac{\left(h^2+\sigma ^2\right)^3}{h^2 \sigma ^4}} \left(h^4+h^2 y^2-\sigma ^4\right)}{2 \left(h^2+\sigma ^2\right)^{7/2}} \\ P^{(h,\sigma)}_2(y)&=&\frac{\left(5 h^8+h^6 \left(13 \sigma ^2+2 y^2\right)+h^4 \left(15 \sigma ^4+y^4+\sigma ^2 y^2\right)+h^2 \left(11 \sigma ^6-\sigma ^4 y^2\right)+4 \sigma ^8\right)}{8 \left(h^2+\sigma ^2\right)^4} \\ \vdots \end{eqnarray}

And now we are ready to give the result. Here you go:

\begin{eqnarray} &&{\mathfrak f}_{2n-1}^{(y,h,\sigma)} = \frac{ \exp\left(+\frac{h^2 (2 n-1)-y^2}{2 \sigma^2}\right) 2^{1} \left(\frac{h^2 (2 n-1)}{2\sigma ^2}\right)^n} { h (n-1)!} \cdot \\ && % \int\limits_0^1 \frac{ \exp \left( \frac{h^2}{2 \eta ^2 \sigma ^2}+\frac{\eta ^2 y^2}{2 \sigma ^2}\right) \cdot % % % } {\left(1-\eta ^2\right)} \cdot \exp\left[ n \left( \log\left(\frac{1-\eta ^2}{\eta ^2}\right) -\frac{h^2}{\eta ^2 \sigma ^2}\right) \right] d\eta \\ &&\simeq \frac{ \text{sgn}(h) \cdot e^{-\frac{y^2}{2 \left(h^2+\sigma ^2\right)}} }{ \sqrt{h^2+\sigma ^2} } \cdot % \underline{ \frac{\sqrt{2 \pi } \left(n-\frac{1}{2}\right)^n e^{-n+\frac{1}{2}}}{\sqrt{n} (n-1)! } } % \cdot \\ && \underline{\underline{ \int\limits_{-\frac{2 h \sqrt{\frac{n \left(h^2+\sigma ^2\right)^3}{h^2 \sigma ^4}}}{\sqrt{h^2+\sigma ^2}}}^{2 \left(1-\frac{h}{\sqrt{h^2+\sigma ^2}}\right) \sqrt{\frac{n \left(h^2+\sigma ^2\right)^3}{h^2 \sigma ^4}}} % \frac{e^{-\frac{\eta ^2}{2}}}{ \sqrt{2 \pi }} \cdot % \left(1+P^{(h,\sigma)}_1(y)\cdot \frac{\eta}{\sqrt{n}}+ % P^{(h,\sigma)}_2(y) \cdot \frac{\eta ^2}{n} +O(\frac{\eta^3}{n^{3/2}}) \right) % d\eta }} \end{eqnarray}

In the second line we expanded (to the second order) both the first term in the integrand and the term in the parentheses in the exponential about the stationary point $\eta_* := h/\sqrt{h^2+\sigma^2}$ and then we simplified the result. Note that in the limit $n \rightarrow \infty$ both the underlined and the doubly underlined terms go to unity and as such we the result is $\lim_{n\rightarrow \infty} {\mathfrak f}_{2n-1}^{(y,h,\sigma)} = \frac{ \text{sgn}(h) \cdot e^{-\frac{y^2}{2 \left(h^2+\sigma ^2\right)}} }{ \sqrt{h^2+\sigma ^2} }$ as expected.

Again, the code snippet below verifies all the steps numerically:

{h, y, \[Sigma]} = RandomReal[{1, 2}, 3, WorkingPrecision -> 50];
n = RandomInteger[{10, 15}];
b = 2 n - 1;
CC = h y/(Sqrt[2] \[Sigma]^2);


( 2^1 E^(((2 n - 1) h^2 - y^2)/(
  2 \[Sigma]^2)) (((2 n - 1) h^2)/(2 \[Sigma]^2))^n)/(h (-1 + n)!)
   NIntegrate[((-\[Eta]^2 + 1)/\[Eta]^2)^n E^((
   y^2 \[Eta]^4 - (2 n - 1) h^2)/(2 \[Eta]^2 \[Sigma]^2))/(
   1 - \[Eta]^2) , {\[Eta], 0, 1}]
( 2^1 E^(((2 n - 1) h^2 - y^2)/(
  2 \[Sigma]^2)) (((2 n - 1) h^2)/(2 \[Sigma]^2))^n)/(h (-1 + n)!)
   NIntegrate[E^(
  h^2/(2 \[Eta]^2 \[Sigma]^2) + (y^2 \[Eta]^2)/(2 \[Sigma]^2) + 
   n (-(h^2/(\[Eta]^2 \[Sigma]^2)) + Log[(1 - \[Eta]^2)/\[Eta]^2]))/(
  1 - \[Eta]^2) , {\[Eta], 0, 1}]
(*Stationary phase approximation.*)
( 2^1 E^(((2 n - 1) h^2 - y^2)/(
  2 \[Sigma]^2)) (((2 n - 1) h^2)/(2 \[Sigma]^2))^n)/(h (-1 + n)!)  (
 E^((h^4 + \[Sigma]^4 + h^2 (y^2 + 2 \[Sigma]^2))/(
  2 \[Sigma]^2 (h^2 + \[Sigma]^2))) (h^2 + \[Sigma]^2))/\[Sigma]^2 \
NIntegrate[(1 + (7/2 + y^2  (h^2/\[Sigma]^4 - 1/(2 \[Sigma]^2)) + (
        5 h^4)/(2 \[Sigma]^4) + (4 h^2)/\[Sigma]^2 + (2 \[Sigma]^2)/
        h^2 + (h^2 y^4)/(2 \[Sigma]^4 (h^2 + \[Sigma]^2)))  (\[Eta] - 
        h/Sqrt[h^2 + \[Sigma]^2])^2 + (\[Eta] - h/Sqrt[
        h^2 + \[Sigma]^2])  ((
        h y^2)/(\[Sigma]^2 Sqrt[
         h^2 + \[Sigma]^2]) + ((h^2 - \[Sigma]^2) Sqrt[
         h^2 + \[Sigma]^2])/(h \[Sigma]^2))) E^(
   n ((-h^2 - \[Sigma]^2)/\[Sigma]^2 - (
      2 (h^2 + \[Sigma]^2)^3 (\[Eta] - h/Sqrt[h^2 + \[Sigma]^2])^2)/(
      h^2 \[Sigma]^4) + Log[\[Sigma]^2/h^2])) , {\[Eta], 0, 1}]
(Sign[h] E^(-y^2/(2 (h^2 + \[Sigma]^2))))/(h^2 + \[Sigma]^2)^(1/2) (
 Sqrt[2 \[Pi]]  (E^(-n + 1/2))  (( -(1/2) + n)^n) )/( (-1 + n)! Sqrt[
  n]) NIntegrate[(1 + (\[Eta])^2 (
      5 h^8 + 4 \[Sigma]^8 + h^6 (2 y^2 + 13 \[Sigma]^2) + 
       h^4 (y^4 + y^2 \[Sigma]^2 + 15 \[Sigma]^4) + 
       h^2 (-y^2 \[Sigma]^4 + 11 \[Sigma]^6))/(
      8  n (h^2 + \[Sigma]^2)^4) + (\[Eta]) (
      h \[Sigma]^2 Sqrt[(h^2 + \[Sigma]^2)^3/(
       h^2 \[Sigma]^4)] (h^4 + h^2 y^2 - \[Sigma]^4))/(
      Sqrt[4 n] (h^2 + \[Sigma]^2)^(7/2)) ) E^(-1/2 (\[Eta])^2)/Sqrt[
   2 \[Pi]] , {\[Eta], -(h/Sqrt[h^2 + \[Sigma]^2]) Sqrt[(
    4  n ((h^2 + \[Sigma]^2)^3) )/(
    h^2 \[Sigma]^4)], (-(h/Sqrt[h^2 + \[Sigma]^2]) + 1) Sqrt[(
    4  n ((h^2 + \[Sigma]^2)^3) )/(h^2 \[Sigma]^4)]}]

$n$ approximation." />

As being said above this integral does a closed form solution yet it can be simplified considerably to a form which, for big values of $b$ can be approximated by a closed form. The idea is simple; we express the t-Student as a continuous mixture of zero-mean normal distributions with the inverse variance conforming to a Gamma distribution, change the order of integration, carry out the integral over $x$ as a Gaussian integral and then simplify the remaining integral as much as possible. Here you go:

As being said above this integral does not a closed form solution yet it can be simplified considerably to a form which, for big values of $b$ can be approximated by a closed form. The idea is simple; we express the t-Student as a continuous mixture of zero-mean normal distributions with the inverse variance conforming to a Gamma distribution, change the order of integration, carry out the integral over $x$ as a Gaussian integral and then simplify the remaining integral as much as possible. Here you go:

\begin{eqnarray} \int\limits_{-\infty}^\infty \frac{1}{\sqrt{2 \pi } \sqrt{\sigma ^2}} e^{-\frac{(y-h x)^2}{2 \sigma ^2}} \left(\frac{x^2}{b}+1\right)^{\frac{1}{2} (-b-1)} dx= \\ \frac{\exp(-\frac{y^2}{2 \sigma ^2})}{\sqrt{\sigma ^2} (n-1)! } \int\limits_0^\infty \frac{\xi ^{n-1} \exp \left(\frac{h^2 y^2}{2 \sigma ^4 \left(\frac{2 \xi }{b}+\frac{h^2}{\sigma ^2}\right)}-\xi \right)}{\sqrt{\frac{2 \xi }{b}+\frac{h^2}{\sigma ^2}}} d\xi =\\ \frac{2}{h} \exp\left( +\frac{b h^2-y^2}{2 \sigma^2}\right) \int\limits_0^1 \frac{\left(\frac{1-\eta ^2}{2\eta ^2}\right)^{n-1} \left(\frac{b h^2}{\sigma ^2}\right)^n }{(n-1)!} \exp \left( \frac{\eta ^4 y^2-b h^2}{2 \eta ^2 \sigma ^2} \right) \frac{d\eta}{\eta^2} \end{eqnarray}

\begin{eqnarray} {\mathfrak f}_b^{(y,h,\sigma)} := \int\limits_{-\infty}^\infty \frac{1}{\sqrt{2 \pi } \sqrt{\sigma ^2}} e^{-\frac{(y-h x)^2}{2 \sigma ^2}} \left(\frac{x^2}{b}+1\right)^{\frac{1}{2} (-b-1)} dx= \\ \frac{\exp(-\frac{y^2}{2 \sigma ^2})}{\sqrt{\sigma ^2} (n-1)! } \int\limits_0^\infty \frac{\xi ^{n-1} \exp \left(\frac{h^2 y^2}{2 \sigma ^4 \left(\frac{2 \xi }{b}+\frac{h^2}{\sigma ^2}\right)}-\xi \right)}{\sqrt{\frac{2 \xi }{b}+\frac{h^2}{\sigma ^2}}} d\xi =\\ \frac{2}{h} \exp\left( +\frac{b h^2-y^2}{2 \sigma^2}\right) \int\limits_0^1 \frac{\left(\frac{1-\eta ^2}{2\eta ^2}\right)^{n-1} \left(\frac{b h^2}{\sigma ^2}\right)^n }{(n-1)!} \exp \left( \frac{\eta ^4 y^2-b h^2}{2 \eta ^2 \sigma ^2} \right) \frac{d\eta}{\eta^2} \tag{1} \end{eqnarray}

Update:

It is interesting to analyze the large $n$ limit of our result above. Below we plot the integrand of the last equation in $(1)$ as a function of the integration variable $\eta$.

$n=2,\cdots,20$ from violet to red respectively." />

As we can see this integrand turns into a narrow bell shaped curve located at some $\eta_* \in (0,1)$ and as such we can use the stationary point approximation in this regime.

Here we only verified numerically that

\begin{equation} \lim_{n \rightarrow \infty} {\mathfrak f}_{2n-1}^{(y,h,\sigma)} = \frac{e^{-\frac{y^2}{2 \left(h^2+\sigma ^2\right)}}}{\sqrt{h^2+\sigma ^2}} \end{equation}

$n$ limit of the integral in question." />