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

Solution 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.

Solution 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:

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