Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$?

Question

I suspect that

$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$$

for $C>0$.

I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ with Wolfram Alpha. But how to prove it?

Answer 1

Let $f : \Bbb{R} \to \Bbb{C}$ be an integrable even function. Then we claim that

$$ \int_{0}^{\infty} f\left( y - \frac{1}{y} \right) \, dy = \int_{0}^{\infty} f(x) \, dx. \tag{1} $$

Indeed, let $I$ denote the LHS of $(1)$. Then by the substitution $y \mapsto y^{-1}$ we have

$$ I = \int_{0}^{\infty} f\left( \frac{1}{y} - y \right) \, \frac{dy}{y^2} = \int_{0}^{\infty} \frac{1}{y^2} f \left( y - \frac{1}{y} \right) \, dy. $$

Thus with the substitution $x = y - y^{-1}$, we have

$$ 2I = \int_{0}^{\infty} \left( 1 + \frac{1}{y^2} \right) f\left( y - \frac{1}{y} \right) \, dy = \int_{-\infty}^{\infty} f(x) \, dx = 2 \int_{0}^{\infty} f(x) \, dx $$

and the conclusion follows.

Now you can plug

$$ f(x) = \frac{1}{C + x^2} $$

to $(1)$ to obtain the conclusion:

$$ \int_{0}^{\infty} \frac{y^2}{Cy^2 + (y^2 - 1)^2} \, dy = \int_{0}^{\infty} \frac{dy}{C + (y - y^{-1})^2} \, dy = \int_{0}^{\infty} \frac{dx}{C + x^2} = \frac{\pi}{2\sqrt{C}}. $$

Answer 2

Note that $$\frac{y^2}{y^4-(2-C)y^2+1}=\frac{1}{\alpha_+-\alpha_-}\left(\frac{\alpha_+}{y^2+\alpha_+}-\frac{\alpha_-}{y^2+\alpha_-}\right),$$ where $-\alpha_{\pm}$ denote two roots of the equation $z^2-(2-C)z+1=0$. For the initial integral to converge, $\alpha_{\pm}$ should be either both positive, either conjugate to each other. Consider for simplicity the former case. Then \begin{align} \int_{0}^{\infty}\frac{y^2dy}{y^4-(2-C)y^2+1}&=\frac{\alpha_+}{\alpha_+-\alpha_-}\int_0^{\infty}\frac{dy}{y^2+\alpha_+}-\frac{\alpha_-}{\alpha_+-\alpha_-}\int_0^{\infty}\frac{dy}{y^2+\alpha_-}=\\ &=\left(\frac{\sqrt{\alpha_+}}{\alpha_+-\alpha_-}-\frac{\sqrt{\alpha_-}}{\alpha_+-\alpha_-}\right)\int_0^{\infty}\frac{dt}{t^2+1}=\\ &=\frac{1}{\sqrt{\alpha_+}+\sqrt{\alpha_-}}\frac{\pi}{2}. \end{align} But we have $\left(\sqrt{\alpha_+}+\sqrt{\alpha_-}\right)^2=(\alpha_++\alpha_-)+2\sqrt{\alpha_+\alpha_-}=(C-2)+2\sqrt{1}=C$, so the statement follows.

Answer 3

So, you can start out by writing the denominator like this: $ y^4+y^2(C-2)+1=(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})$. This is simply the solution of a quadratic equation (ugly, but it will do). So, now, we want to simplify this expression somehow, like by writing:

$ \frac{y^2}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{A}{y^2-\frac{2-C-\sqrt{C(C-4)}}{2}}+\frac{B}{y^2-\frac{2-C+\sqrt{C(C-4)}}{2}}=\frac{A(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})+B(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{y^2(A+B)-1(A(\frac{2-C-\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}$.

Therefore,we want $A+B=1 \wedge A(\frac{2-C+\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4}}{2})=0$. This is a linear system, that you can solve to find A and B, and then calculate the integral.