Is there an analytical solution to $\int_{0^{-}}^{t} d{s} \ T\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right)$?

Question

I am trying to solve the integral \begin{align} \int_{0^{-}}^{t} d{s} \ \ T\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right) \end{align} where $T$ is Owen's T function. I have been trying to substitute some integral definition of the Owen T function, e.g. \begin{align} T(h, a) & =\frac{1}{2 \pi} \int_{0}^{a} d{x} \ \frac{e^{-\frac{1}{2} h^{2}\left(1+x^{2}\right)}}{1+x^{2}} \label{eq: core def OwenT}\\ & =\frac{1}{2\sqrt{2 \pi}} \int_{h}^{\infty} d{\xi} \ e^{-\xi^{2}/2} \ \operatorname{Erf}\left(\frac{a \xi}{\sqrt{2}}\right) \end{align} which I found in [1] and [2]. However, this yields integrals over Error functions. Some sources one can check for integrals over Error function are [3], [4] and [5]. Nevertheless, so far, no luck. Is there someone who knows a way to tackle integrals over Owen's T functions?

Answer 1

Let $$ f \colon \mathbb{R} \times (0,\infty) \to (0,\infty) \, , \, f(x,t) = \int \limits_0^t \operatorname{T} \left(\frac{x}{\sqrt{2 t}},\sqrt{\frac{s}{2t-s}}\right) \, \mathrm{d} s \, .$$ The substitution $s = t u$ yields $f(x,t) = t g\left(\frac{x}{2\sqrt{t}}\right)$ with $g \colon \mathbb{R} \to (0,\infty)$ given by \begin{align} g(y) &= \int \limits_0^1 \operatorname{T} \left(\sqrt{2} y, \sqrt{\frac{u}{2-u}}\right) \, \mathrm{d} u = \frac{1}{2 \pi} \int \limits_0^1 \int \limits_0^{\sqrt{\frac{u}{2-u}}} \frac{\mathrm{e}^{- y^2 (1 + v^2)}}{1+v^2} \, \mathrm{d} v \, \mathrm{d} u \\ &\!\!\!\!\overset{\text{Tonelli}}{=} \frac{1}{2\pi} \int \limits_0^1 \int \limits_{\frac{2v^2}{1+v^2}}^1 \frac{\mathrm{e}^{- y^2 (1 + v^2)}}{1+v^2} \, \mathrm{d} u \, \mathrm{d} v = \frac{1}{2\pi} \int \limits_0^1 \frac{1-v^2}{(1+v^2)^2} \, \mathrm{e}^{- y^2 (1 + v^2)} \, \mathrm{d} v \\ &\!\overset{\text{IBP}}{=} \frac{1}{2\pi} \left[\frac{1}{2} \mathrm{e}^{-2y^2} + 2 y^2 \int \limits_0^1 \frac{v^2}{1+v^2} \, \mathrm{e}^{- y^2 (1 + v^2)} \, \mathrm{d} v\right] \\ &= \frac{1}{2\pi} \left[\frac{1}{2} \mathrm{e}^{-2y^2} + 2 y^2 \left(\frac{\sqrt{\pi}}{2y} \mathrm{e}^{-y^2} \operatorname{erf}(y)-2 \pi \operatorname{T} (\sqrt{2} y,1)\right)\right] \\ &= \frac{1}{2\pi} \left[\frac{1}{2} \mathrm{e}^{-2y^2} + 2 y^2 \left(\frac{\sqrt{\pi}}{2y} \mathrm{e}^{-y^2} \operatorname{erf}(y)-\frac{\pi}{2} \operatorname{erfc}(y) \left(1 - \frac{1}{2} \operatorname{erfc}(y)\right)\right)\right] \\ &= \frac{\mathrm{e}^{-2 y^2}}{4 \pi} + \frac{y \, \mathrm{e}^{-y^2}}{2 \sqrt{\pi}} \operatorname{erf}(y) - \frac{y^2}{4} \left(1 - \operatorname{erf}^2(y)\right) = \frac{\left(y \operatorname{erf}(y) + \frac{\mathrm{e}^{-y^2}}{\sqrt{\pi}}\right)^2 - y^2}{4} \, . \end{align}