Return to Revisions

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 should ask him tonight, but if we trust them four (well, you may check this numerically using some computer stuff), 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 $a>0$, $b^2<ac$, $0<\Re \alpha<2\Re \rho$.