Hellinger integral for the Student/Cauchy family

Question

Let $p$ and $q$ be probability densities on $\mathbb R$, with respect to the Lebesgue measure $dx$. The corresponding Hellinger integral is $H(p,q):=\int_{\mathbb R}\sqrt{pq}\,dx$.

Let now $p$ be the density of Student's distribution with $d$ degrees of freedom, so that $$p(x)=C_d\,(1+x^2/d)^{-(d+1)/2}$$ for real $x$, where $d\in(0,\infty)$ and $C_d$ does not depend on $x$.

At least for $d=1$, is there a closed form expression for $H(p,p_t)$ for real $t\ne0$, where $p_t(x):=p(x-t)$ for real $x,t$?

An obviously equivalent form of this question: Is there a closed form expression for the integral $$\int_{\mathbb R}\frac{dx}{(1+x^2)^a\,(1+(x-t)^2)^a} $$ for all real $t\ne0$ and all real $a>1/4$ (or at least for $a=1/2$)? Of course, this is not a problem for natural $a$.

Answer 1

Denote $t=2\tau$. $$ \int_{\mathbb R}\frac{dx}{(1+x^2)^a\,(1+(x-2\tau)^2)^a}=\int_{\mathbb R}\frac{dx}{(1+(x+\tau)^2)^a\,(1+(x-\tau)^2)^a}. $$ We have $$ \left(1+(x+\tau)^2\right)\,\left(1+(x-\tau)^2\right)=\left(x^2+\tau^2+1\right)^2-4\tau^2x^2=x^4+2x^2(1-\tau^2)+(\tau^2+1)^2=:x^4+Ax^2+B, $$ so our problem for reduces to evaluation of integral $$ \int_{\mathbb{R}} \frac {dx}{(x^4+Ax^2+B)^a}=2\int_0^{\infty} \frac {dx}{(x^4+Ax^2+B)^a}. $$ Denote $x^2=y$, we get $$ \int_0^{\infty} \frac {y^{-1/2}dy}{(y^2+Ay+B)^a}. $$ Well, this is number 2.2.9.7 in Prudnikov-Brychkov-Marichev. In my home edition this specific integral is corrected by my father (what a coincidence!). I asked him, hopefully it was just a misprint in the exponent of $a$, and the result is $$ \int_0^{\infty} \frac {x^{\alpha-1}dx}{(ax^2+2bx+c)^\rho}= a^{-\alpha/2} c^{\alpha/2-\rho} B(\alpha,2\rho-\alpha)\,_2F_1(\frac{\alpha}2,\rho-\frac{\alpha}2;\rho+\frac12;1-\frac{b^2}{ac}) $$ for natural assumptions $a>0$, $b^2<ac$, $0<\Re \alpha<2\Re \rho$.