Solving the integral $\int_{0}^{1} d{v} \frac{e^{-y^2(1+v^2)}}{1+v^2}e^{\frac{2 t v^2}{v^2+1}}$

Question

For my research, I have to solve many integrals of the Owen's T function. As such, I am having struggles in calculating the integral $$ \int_{0^{-}}^{t}d{s} \ e^{(t+s)} \ \operatorname{T}\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right). $$ I asked a similar question in [1] for the integral $$ \int_{0^{-}}^{t}d{s} \ \operatorname{T}\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right), $$ which was beautifully answered. Nevertheless, if one follows the same procedure the exponential complicates things. What I have done until know is following [1] which yields \begin{align} f(x,t)=\int_{0^{-}}^{t}d{s} \ e^{(t+s)} \ T\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right)=t\int_{0}^{1} d{u} \ e^{t(1+u)} \ T\left(\sqrt{2}y, \sqrt{\frac{u}{2-u}}\right) \end{align} where we performed the change of variable $s=tu$ and took $y=\frac{x}{2\sqrt{t}}$. Substituting the definition for Owen's T function and using Tonelli theorem gives \begin{align} f(x,t)&= \frac{1}{2\pi}t \int_{0}^{1}d{u}\int_{0}^{\sqrt{\frac{u}{2-u}}}d{v} \ e^{t(1+u)} \ \frac{e^{-y^2(1+v^2)}}{1+v^2}\\ &=\frac{1}{2\pi}t \int_{0}^{1}d{v} \ \int_{\frac{2v^2}{1+v^2}}^{1}d{u} \ e^{t(1+u)} \ \frac{e^{-y^2(1+v^2)}}{1+v^2}\\ &= \frac{1}{2\pi}t \int_{0}^{1} d{v} \ \frac{e^{-y^2(1+v^2)}}{1+v^2}\left(\frac{e^{2 t}}{t}-\frac{e^{\frac{2 t v^2}{v^2+1}+t}}{t}\right)\\ &= \frac{1}{2\pi} e^{2 t}\int_{0}^{1} d{v} \ \frac{e^{-y^2(1+v^2)}}{1+v^2}-\frac{1}{2\pi}e^{t} \int_{0}^{1} d{v} \ \frac{e^{-y^2(1+v^2)}}{1+v^2}e^{\frac{2 t v^2}{v^2+1}}\\ &= T\left(\tfrac{x}{\sqrt{2t}}, 1\right)e^{2 t}+\frac{1}{2\pi} e^{t} \color{blue}{\int_{0}^{1} d{v} \frac{e^{-y^2(1+v^2)}}{1+v^2}e^{\frac{2 t v^2}{v^2+1}}}\\ \end{align} The Owen T term, i.e. the first term, can be written in terms of Error function. However, I am struggling in solving the integral in the second term. I have tried integrate by part, but this just yield a more complicated expression. Someone who knows a way forward?

Also I checked above calculation numerically and it indeed holds up!

In[270]:= ClearAll["Global`*"];
int1[x_, t_]:=OwenT[x/Sqrt[2t],1]*Exp[2t]+1/(2\[Pi])*Exp[t]*NIntegrate[ Exp[-(x/(2Sqrt[t]))^2*(1+v^2)]/(1+v^2)*Exp[(2tv^2)/(1+v^2)], {v,0,1},WorkingPrecision->7] ;
int2[x_, t_]:=NIntegrate[Exp[(t+s)]*OwenT[x/Sqrt[2t],Sqrt[s/(2t-s)]], {s,0,t},WorkingPrecision->7] ;
{x,t}=RandomReal[{0,1},2,WorkingPrecision->50]; 
int1[x, t]
int2[x, t]
Out[274]= 0.01602865
Out[275]= 0.01603027

Answer 1

The key difference between this integral and the one you linked is the presence of the reciprocal term $(v^2+1)^{-1}$ inside the exponential. Without it, we can easily rely on error-type functions such as Owen's $T$, but as we will see later its presence leads to incomplete gamma-/Bessel-type functions at the very least.

The intuitive substitution $x=v^2+1$ leads to $$\int_0^1\frac{e^{-y^2(v^2+1)+2tv^2/(v^2+1)}}{v^2+1}\,dv=\frac{e^{2t}}2\int_1^2\frac{e^{-y^2x-2t/x}}{x\sqrt{x-1}}\,dx.$$ The issue with this integrand is the $\sqrt{x-1}$ term; without it, we would have $$\int_1^2\frac{e^{-y^2x-2t/x}}x\,dx=\frac{e^{2t}}2\left(\Gamma(0,y^2;2ty^2)-\Gamma(0,2y^2;2ty^2)\right)$$ where $\Gamma(\alpha,x;b):=\int_x^\infty t^{\alpha-1}e^{-t-b/t}\,dt$ is quite a common function in statistical physics (Harris, 2008). In our case, we could try to exploit its inverse Laplace transform $${\cal L}^{-1}\left[\frac1{\sqrt{s-1}}\right]=\frac{e^t}{\sqrt{\pi t}}$$ to write the integral as \begin{align}\int_1^2\frac{e^{-y^2x-2t/x}}{x\sqrt{x-1}}\,dx&=\int_1^2\int_0^\infty\frac{e^{-y^2x-2t/x}}x\cdot\frac{e^{-xu}e^u}{\sqrt{\pi u}}\,du\,dx\\&=\int_0^\infty\frac{e^u}{\sqrt{\pi u}}\int_1^2\frac{e^{-(y^2+u)x-2t/x}}x\,dx\,du\\&=\int_0^\infty\frac{e^u(\Lambda_2(u)-\Lambda_1(u))}{\sqrt{\pi u}}\,du\end{align} where $\Lambda_k(u):=\Gamma(0,k(y^2+u);2t(y^2+u))$. This puts the integral in a $(0,\infty)$-form where there are many more integral tricks, but the nature of the integrand makes the existence of a closed form rather hard to believe.

Answer 2

Let $I$ be the following:

$$I=\int_{0}^{1} \frac{e^{-y^2(1+v^2)}}{1+v^2}e^{\frac{2 t v^2}{1+v^2}} \textrm{d}v$$

Now use the substitution $v=\tan{u}$:

$$I=\int_{0}^{\frac{\pi}{4}} \frac{e^{-y^2(1+\tan^2u)}}{1+\tan^2u}e^{\frac{2 t \tan^2u}{1+\tan^2u}} \sec^2(u) \textrm{d}u$$

If you are familiar with simple trig identities, you may also notice the following:

$$I=\int_{0}^{\frac{\pi}{4}} \frac{e^{-y^2\sec^2u}}{\sec^2u}e^{\frac{2 t \tan^2u}{\sec^2u}} \sec^2(u) \textrm{d}u$$

$$I=\int_{0}^{\frac{\pi}{4}} e^{-y^2\sec^2u}e^{\frac{2 t \tan^2u}{\sec^2u}} \textrm{d}u$$

$$I=\int_{0}^{\frac{\pi}{4}} e^{-y^2\sec^2u}e^{2 t \tan^2(u)\cos^2u} \textrm{d}u$$

$$I=\int_{0}^{\frac{\pi}{4}} e^{-y^2\sec^2u}e^{2 t \sin^2(u)} \textrm{d}u$$

$$I=\int_{0}^{\frac{\pi}{4}} e^{2 t \sin^2(u)-y^2\sec^2u} \textrm{d}u$$

$$I=\int_{0}^{\frac{\pi}{4}} e^{a\sin^2(v)-b\sec^2v} \textrm{d}v$$

Unfortunately, I don't know if I could do any better than that. I'll leave an update if I do find something, though.

Edit 1: One thing I have noticed about the function $f(x)=e^{a\sin^2(x)-b\sec^2x}$ is that it is a periodic function and that $f(x)=f(\pi-x)$.

Edit 2: I think I'm on to something:

$$I=\int_{0}^{1} \frac{e^{-y^2(1+v^2)}}{1+v^2}e^{\frac{2 t v^2}{1+v^2}} \textrm{d}v = 2\pi\int_{0}^{1} \dfrac{\textrm{d}}{\textrm{d} v}\left(T(y\sqrt{2},v)\right) e^{\frac{2 t v^2}{1+v^2}} \textrm{d}v$$

If we shorten $\dfrac{\textrm{d}}{\textrm{d} v}T(y\sqrt{2},v)$ to just $\dfrac{\textrm{d}T}{\textrm{d}v}$, we can convert the integral to the following:

$$I = 2\pi\int_{0}^{1} \dfrac{\textrm{d}T}{\textrm{d}v}e^{\frac{2 t v^2}{1+v^2}} \textrm{d}v$$

$$I = 2\pi\int_{0}^{1} e^{\frac{2 t v^2}{1+v^2}} \textrm{d}T$$

Then use IBP:

$$ I = 2\pi T(y\sqrt{2},1)e^{t} - 2\pi\int_{0}^{1} T(y\sqrt{2},v)\dfrac{4tv}{\left(1+v^2\right)^2}e^{\frac{2 t v^2}{1+v^2}} \textrm{d}v $$