I am trying to evaluate the integral
$$
\int_0^1 \frac{e^{-y^2(1+v^2)}}{(1+v^2)^n}dv = e^{-y^2}\int_0^1 \frac{e^{-y^2v^2}}{(1+v^2)^n}dv
$$
for $n\in \mathbb{N}$.For n=1 one finds Owen's T function, i.e.
\begin{align}
\int_0^1 \frac{e^{-y^2(1+v^2)}}{(1+v^2)}dv=2\pi \operatorname{T}\left(\sqrt{2} y,1\right) = \frac{\pi}{2} \operatorname{erfc}(y) \left(1 - \frac{1}{2} \operatorname{erfc}(y)\right)
\end{align}
A nice source on the Owen's T function is [2]. In [3] they state that
\begin{align}
\int \frac{e^{-v^2}}{v^2 + 1} dv ,
\end{align}
has no anti-derivative. Hence, I do not suspect one can find one for our integral.
This integral occurs in a series I am integrating over for a approximation I am performing. Hence, it would already be nice if I could find the second (n=2) and third (n=3) term. Has someone an idea how to evaluate the integral. Many thank in advance!
I transformed your integral. Maybe this is helpful. Note that: $$\int_{0}^{\infty }x^{n-1}e^{-(1+v^{2})x}dx=\frac{\Gamma(n) }{(1+v^{2})^{n}}$$$$\int_{0}^{1}e^{-v^{2}a}dv=\frac{\sqrt{\pi}\operatorname{erf}\sqrt{a}}{2\sqrt{a}}.$$ Therefore the solution is, $$\int_{0}^{1 }\frac{e^{-y^{2}(1+v^{2})}}{(1+v^{2})^{n}}dv=\frac{1}{\Gamma(n) }\int_{0}^{1}\int_{0}^{\infty }e^{-(1+v^{2})y^{2}}x^{n-1}e^{-(1+v^{2})x}dvdx$$$$\frac{e^{-y^{2}}}{\Gamma(n) }\int_{0}^{\infty }x^{n-1}e^{-x}(\int_{0}^{1}e^{-v^{2}(x+y^{2})}dv)dx=\frac{\sqrt{\pi}e^{-y^{2}}}{2\Gamma(n) }\int_{0}^{\infty }\frac{x^{n-1}e^{-x}\operatorname{erf}\sqrt{x+y^{2}}}{\sqrt{x+y^{2}}}dx.$$