Compute $\int_{-\infty}^{+\infty}\frac{x^2}{(x^2-\sqrt{2}x+1)(x^2+1)^{\alpha}}dx$

Question

As in the title, i have to compute $\int_{-\infty}^{+\infty}\frac{x^2}{(x^2-\sqrt{2}x+1)(x^2+1)^{\alpha}}dx$ for $\frac{1}{2}<\alpha<1$. I tried passing to the complex plane using a contour similar to the one in the following picture with the small circle around the branch point $i$. The problem is that by doing so, i get other integrals which i can't compute and I'm stuck. Should i choose an other path?

Answer 1

Somewhat more generally, write $$ \frac{x^2}{(x^2 - b x + 1)(x^2+1)^a} = \sum_{k=0}^\infty \frac{b^k x^{k+2}}{(x^2+1)^{a+k+1}}$$ for $|b| < 2$.
Note that $$\int_{-\infty}^\infty \frac{b^k x^{k+2}}{(x^2+1)^{a+k+1}} \; dx = \cases{0 & if $k$ is odd\cr \dfrac{b^k \Gamma(a+k/2 - 1/2) \Gamma((k+3)/2)}{\Gamma(a+k+1)} & otherwise}$$ Then we should have $$ \int_{-\infty}^\infty \frac{x^2}{(x^2-bx+1)(x^2+1)^a} \; dx = \sum_{j=0}^\infty \frac{b^{2j} \Gamma(a + j - 1/2) \Gamma(j+3/2)}{\Gamma(a+2j+1)}$$

This can be written as a hypergeometric

$$ \frac{\sqrt{\pi}\;\Gamma(a-1/2)}{2 \;\Gamma(a+1)} {}_3F_2(1,3/2,a-1/2; 1+a/2, (1+a)/2; b^2/4) $$

but I don't know if there is a closed-form expression in terms of more standard functions. Maple doesn't seem to find one.

Answer 2

Another solution from a CAS. $$I_\alpha=\int_{-\infty}^{+\infty}\frac{x^2}{(x^2-x\sqrt{2}+1)(x^2+1)^{\alpha}}dx$$ $$I_\alpha= 2^{\frac{1-\alpha}{2}}\pi \sin \left(\frac{\pi \alpha }{4}\right)+2 \sqrt{\pi }\,\frac{ \Gamma \left(\alpha -\frac{1}{2}\right) }{\Gamma (\alpha -1)}\,\,_3F_2\left(1,\frac{2\alpha-1 }{4},\frac{2\alpha+1 }{4};\frac{3}{4},\frac{5}{4};-1\right)$$ which, for sure, is discontinuous for $\alpha=\frac12$ since, in this case, the expansion of the integrand is $$\frac{1}{x}+\frac{\sqrt{2}}{x^2}+O\left(\frac{1}{x^3}\right)$$