Prove that $\int _{-\infty }^{+\infty }{\frac {\mathrm {d} z}{(\phi ^{n}z)^{2}+(F_{2n+1}-\phi F_{2n})(e^{\gamma }z^{2}+\zeta (3)z-\pi )^2}}=1$

Question

Recently while dealing with few interesting integrals, I was quite fascinated by this one:

$$\int _{-\infty }^{+\infty }{\dfrac {1}{(\phi ^{n}z)^{2}+(F_{2n+1}-\phi F_{2n})(e^{\gamma }z^{2}+\zeta (3)z-\pi )^2}}\,\mathrm {d} z=1$$

where $\phi$ represents the golden ratio, $\gamma$ represents the Euler-Mascheroni constant and $F_n$ denotes the $n$-th Fibonacci number. I have no idea on how to prove this but a possible approach would be using residue theorem and complex analysis. I'm looking for a real analytic approach.

Any help would be appreciated. Thanks for reading $:)$

Answer 1

The proposed identity is a disguised version of $$(*)\quad\int_{-\infty}^\infty \frac{dx}{x^2+(ax^2+bx-c)^2} = \frac{\pi}{c} \quad \text{with } a,b,c>0,$$ and a sketch proof will be given below. It uses contour integration because that's the easiest thing I could think of. It might be possible to prove the above ID with Glasser's theorem, or real integration, though you have to careful of the branch of the arctan function.

The important thing to notice is that the right-hand side is independent of $(a,b),$ once the positivity of those constants is assumed.

With $c=\pi,$ introduce a scaling parameter $\kappa$ and anequivalent expression is $$\int_{-\infty}^\infty \frac{dx}{x^2+(\kappa (ax^2+bx-\pi))^2} = \frac{1}{\kappa} $$ Let $\kappa=\phi^{-2n}$, $a=e^\gamma,$ $b=\zeta(3).$ Then $$\int_{-\infty}^\infty \frac{dx}{x^2+ \phi^{-4n} (e^\gamma x^2+\zeta(3)x-\pi)^2} = \phi^{2n} $$ Division by $\phi^{2n}$ and the result is $$\int_{-\infty}^\infty \frac{dx}{(\phi^n x)^2+\phi^{-2n} (e^\gamma x^2+\zeta(3)x-\pi)^2} =1 $$

It is easy to show $F_{2n+1}-\phi F_{2n}=\phi^{-2n},$ so once (*) is proved, the OP's integral follows.

It is well-known that an integral over $(-\infty,\infty)$ in which the integrand has isolated poles can be computed with the residue theorem. Factor the quartic polynomial; fortunately it is rater simple: $$ (ax^2+bx-c)^2 + x^2 =a^2 (x-x_1)(x-x_2)(x-x_3)(x-x_4) \text{where} $$ $$ x_{1,2} = \frac{1}{2a}(i-b \pm \sqrt{(i-b)^2 + 4ac} )$$ $$ x_{3,4} = \frac{1}{2a}(-i-b \pm \sqrt{(-i-b)^2 + 4ac} )$$ For $a,b,c>0, \ \ x_{1,2}$ contribute poles in the upper half plane. Break factored integrand into partial fractions. You'll only keep the terms with $(x-x_1)$ and $(x-x_2).$ Therefore the integral is $$I(a,b,c):=\quad\int_{-\infty}^\infty \frac{dx}{x^2+(ax^2+bx-c)^2} = $$ $$= \frac{2\pi i }{a^2}\Big(\frac{1}{(x_1-x_2)(x_2-x_3)(x_2-x_4)}- \frac{1}{(x_1-x_2)(x_1-x_3)(x_1-x_4)} \Big)$$ Substitute the $(a,b,c)$ dependent expression of the roots $x_{1,2}$ and simplify (I used Mathematica) to get (*).