How to prove this definite integral does not depend on the parameter?

Question

I am working on some development formulas for surfaces and as a byproduct of abstract theory i get that: $$\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{1+\sin^2\theta}{(\cos^4\theta+(\gamma\cos^2\theta-\sin\theta)^2)^\frac{3}{4}}d\theta$$ is independent on the parameter $\gamma\in\mathbb{R}$. I thought that there was something wrong with my calculations but actually turns out that using Mathematica that the value is somewhat near $5,24412$ independently on the $\gamma$ I plug in the calculation of the integral. Is there any way to verify that actually this is a constant by direct computations, complex analysis, or at least is this kind of integrals studied?

Edit:obviously differentiating in the integral does not help much

Answer 1

Put \begin{equation*} I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1+\sin^2\theta}{(\cos^4\theta +(\gamma\cos^2\theta-\sin \theta)^2)^{\frac{3}{4}}}\, d\theta \end{equation*} If $x = \dfrac{\sin\theta}{\cos^2\theta}$, $\, y = \gamma-x$ and $y = \sqrt{z}$ then \begin{equation*} dx = \dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta}\, d\theta \end{equation*} and \begin{gather*} I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta\left(1 +\left(\gamma-\frac{\sin \theta}{\cos^2\theta}\right)^2\right)^{\frac{3}{4}}}\, d\theta = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +\left(\gamma- x\right)^2\right)^{\frac{3}{4}}}\, dx = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +y^2\right)^{\frac{3}{4}}}\, dy = \\[2ex] \int_{0}^{\infty}\dfrac{z^{\frac{1}{2}-1}}{(1+z)^{\frac{1}{2}+\frac{1}{4}}}\, dz = {\rm B}\left(\frac{1}{4},\frac{1}{2}\right) \approx 5.244115109 \end{gather*} where ${\rm B}$ is the Beta function.

Answer 2

This is not an answer, but too long for a comment.

I still think it may be helpful to consider the derivative (I use $x$ as integration variable).

$$J=\frac{d}{d \gamma} I(\gamma)=\int_{-\pi/2}^{\pi/2} \frac{(1+\sin^2 x)(\gamma ~\cos^2 x-\sin x) \cos^2 x~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}$$

We need to prove $J \equiv 0$.

However, there's a stronger statement $^*$, which seems to be true numerically:

$$J_1=\int_{-\pi/2}^{\pi/2} \frac{\gamma (1+\sin^2 x) \cos^4 x~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}$$

$$J_2=\int_{-\pi/2}^{\pi/2} \frac{(1+\sin^2 x) \cos^2 x \sin x ~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}$$

$$J_1 \equiv J_2$$

Mathematica confirms it numerically for $\gamma \in (-1,1)$. For $|\gamma|>1$ there's some trouble with computing the integrals numerically.

Moreover, as we see, both integrals follow linear dependence on $\gamma$ with very good accuracy.

Using linear regression in Mathematica, I obtained, with amazing accuracy the following fit:

$$J_1(\gamma)= \frac{2}{3} L \gamma $$

Where $L$ is a lemniscate constant:

$$\frac{2}{3} L = \frac{1}{3 \sqrt{2 \pi}} \left( \Gamma \left( \frac{1}{4} \right) \right)^2 =\frac{4}{3} \int_0^1 \frac{dx}{\sqrt{1-x^4}}$$

Here are the plots with the proposed closed form (green) for comparison:

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This gives us another task, to prove the proposed closed form for both integrals.

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Note that for $J_1$ we can divide by $\gamma$ and obtain the following proposition:

$$\int_{-\pi/2}^{\pi/2} \frac{ (1+\sin^2 x) \cos^4 x~dx}{\left(\cos^4 x+(\gamma~ cos^2 x-\sin x)^2 \right)^{7/4}}=\frac{4}{3} \int_0^1 \frac{dx}{\sqrt{1-x^4}}$$

Which makes $J_1/ \gamma$ yet another integral which doesn't depend on $\gamma$.

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Edit, remark:

$^*$ this statement is not really stronger, because as we can see, the rest of the integrated function is non-negative for all $x$, which means that $J_1 \equiv J_2$ follows from $J \equiv 0$.