Integral by the residue theorem

Check that the value of $$I(\alpha)=28\int_{-\infty}^\infty \frac{1-x^2}{2+3x^2+2\alpha^2x^4}\text{d}x$$ is $$I(\alpha)=14\pi\left(1-\frac{1}{\alpha}\right)\sqrt{\frac{2}{4\alpha+3}}$$ for some $$\alpha>0$$.

I've tried to apply the Residue Theorem to this integral with the contour $$C_R\cup C$$, where $$C_R$$ is the semicircle of radius $$R$$ centered at the origin (with initial point $$(R,0)$$ and final point $$(-R,0)$$) and $$C$$ is the segment $$(-R,R)$$ (with initial point $$(-R,0)$$ and final point $$(R,0)$$).

$$\textbf{Lemma}$$. Suppose $$f(z)$$ is defined in the upper half-plane. If there is an $$a > 1$$ and $$M > 0$$ such that $$|f(z)| <\frac{M}{|z|^a}$$ for large $$|z|$$. Then, $$\lim_{R\to \infty} \int_{C_R} f(z)\text{d}z=0$$

The roots of the polynomial $$p(x)=2+3x^2+2\alpha^2x^4$$ are $$x_1=-\frac{1}{2}\sqrt{\frac{-\sqrt{9-16\alpha^2}-3}{\alpha^2}}, \qquad x_2=\frac{1}{2}\sqrt{\frac{-\sqrt{9-16\alpha^2}-3}{\alpha^2}}$$ $$x_3=-\frac{1}{2}\sqrt{\frac{\sqrt{9-16\alpha^2}-3}{\alpha^2}}, \qquad x_4=\frac{1}{2}\sqrt{\frac{\sqrt{9-16\alpha^2}-3}{\alpha^2}}$$ By the previous lemma and the residue theorem, we have that $$I(\alpha)=\int_{-\infty}^\infty \frac{28(1-x^2)}{2+3x^2+2\alpha^2x^4}\text{d}x=\lim_{R\to \infty} \int_C \frac{28(1-x^2)}{2+3x^2+2\alpha^2x^4}\text{d}x=2\pi i \sum_{i=1}^n\text{Res}(f,x_i)$$ However, I don't know how to discuss what roots are in the upper half-plane due to the parameter $$\alpha$$.

Can anyone continue from here or give some hints to follow?

• The choice depends on the branch of $\sqrt{}$ you take, but exactly one of $x_1,x_2$ and exactly one of $x_3,x_4$ is in the upper half plane, assuming $\alpha\neq\frac34$. – user10354138 Jun 24 '19 at 13:33

You are on the right track, and only needs a bit of extra input to get to the answer. Assume for a moment that $$\alpha > 3/4$$. Then the zeros of $$2+3t+3\alpha^2 t^2 = 0$$ are

$$t = \frac{-3 \pm i\sqrt{16\alpha^2-9}}{4a^2} = \left( \frac{\sqrt{4\alpha-3} \pm i\sqrt{4\alpha+3}}{\sqrt{8}\alpha} \right)^2.$$

Now, among the square roots of these values, the only ones with positive imaginary parts are

$$x_{(1)} = \frac{\sqrt{4\alpha-3} + i\sqrt{4\alpha+3}}{\sqrt{8}\alpha} \qquad \text{and} \qquad x_{(2)} = \frac{-\sqrt{4\alpha-3} + i\sqrt{4\alpha+3}}{\sqrt{8}\alpha}.$$

Then, after some painful algebra,

$$I(\alpha) = 2\pi i \sum_{k=1}^{2} \operatorname{Res}(f, x_{(k)}) = 14\pi \left(1 - \frac{1}{\alpha}\right) \sqrt{\frac{2}{4\alpha+3}}.$$

At this point, this equality is established only for $$\alpha > 3/4$$. But since both sides are analytic near $$(0, \infty)$$, it extends to all of $$(0, \alpha)$$ by the principle of analytic continuation.

• Amazing! But what "both sides" are you refering the last part? – user326159 Jun 24 '19 at 16:29
• @user326159, Oh, by 'both sides' I was referring $I(\alpha)$ and $14\pi(1-\alpha^{-1})\sqrt{2/(4\alpha+3)}$ as functions of $\alpha$. – Sangchul Lee Jun 24 '19 at 17:02
• Great! I understand now. Thank you very much – user326159 Jun 24 '19 at 17:06