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I am studying the integral $$I=\int_{-\pi/2}^{\pi/2} \frac{28\cos^2(\theta)+10\cos(\theta)\sin(\theta)-28\sin^2(\theta)}{2\cos^4(\theta)+3\cos^2(\theta)\sin^2(\theta)+m\sin^4(\theta)}d\theta,$$ where $m>0$. In the problem I am working, it is very important to know what value of $m$ makes this integral positive, negative or equal to zero. I found that if $m=2$, then the integral is zero (introducing it in Wolfram Alpha). However, I don't know how to prove it formally.

On the other hand, I tried some values of $m$ in Wolfram and it seems that if $m<2$, then the integral is negative, and if $m>2$, the integral is positive. But again, I have no proof of this.

Any ideas of how to approach this problem?

Just in case, I found this alternative representation of the integral $$I=\int_{-\pi/2}^{\pi/2} \frac{8[5\sin(2\theta)+28\cos(2\theta)]}{\cos(4\theta)+15+8(m-2)\sin^4(\theta)}d\theta.$$

Any help would be appreciated.

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  • $\begingroup$ Surely it is $8(m-2)\sin^4\theta$ not $(m-2)\sin^4\theta$ in the denominator of the last equation? $\endgroup$ – user10354138 Jun 20 at 19:57
  • $\begingroup$ Yes. I've corrected it $\endgroup$ – user326159 Jun 20 at 20:04
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    $\begingroup$ Writing $\alpha^2 = m/2$, we get $$ I(\alpha) = 14 \pi \left( 1-\frac{1}{\alpha}\right)\sqrt{\frac{2}{4\alpha+3}}. $$ From this, we easily deduce the sign of $I$ as function of $\alpha$, i.e., $\operatorname{sign}(I(\alpha)) = \operatorname{sign}(\alpha - 1)$. Now if you ask me how I arrived this expression, it is basically from an ugly residue computation applied to $$I(\alpha) = \int_{-\infty}^{\infty} \frac{28(1-t^2)}{2+3t^2+2\alpha^2 t^4}\, \mathrm{d}t, $$ but I have not enough energy to write up all the intermediate steps... $\endgroup$ – Sangchul Lee Jun 20 at 22:27
  • $\begingroup$ @SangchulLee Thank you for your answer. It's amazing that you find a closed form for this integral. I've tried to compute it by the residue theorem but I cannot get this formula. Could you be more explicit on how did you do it? To know how to calculate this integral is very important for me, and I've been working many hours on it without success... $\endgroup$ – user326159 Jun 24 at 15:08
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Define the function $\mathcal{I}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the trigonometric integral

$$\begin{align} \mathcal{I}{\left(\mu\right)} &:=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{28\cos^{2}{\left(\theta\right)}+10\cos{\left(\theta\right)}\sin{\left(\theta\right)}-28\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}.\\ \end{align}$$


Let $\mu\in\mathbb{R}_{>0}$. Since the integral $\mathcal{I}$ has a symmetric interval of integration, the integral of the odd component of the integrand vanishes identically:

$$\begin{align} \mathcal{I}{\left(\mu\right)} &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{28\cos^{2}{\left(\theta\right)}+10\cos{\left(\theta\right)}\sin{\left(\theta\right)}-28\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}\\ &=\int_{-\frac{\pi}{2}}^{0}\mathrm{d}\theta\,\frac{28\cos^{2}{\left(\theta\right)}+10\cos{\left(\theta\right)}\sin{\left(\theta\right)}-28\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}\\ &~~~~~+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{28\cos^{2}{\left(\theta\right)}+10\cos{\left(\theta\right)}\sin{\left(\theta\right)}-28\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{28\cos^{2}{\left(\theta\right)}-10\cos{\left(\theta\right)}\sin{\left(\theta\right)}-28\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}};~~~\small{\left[\theta\mapsto-\theta\right]}\\ &~~~~~+\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{28\cos^{2}{\left(\theta\right)}+10\cos{\left(\theta\right)}\sin{\left(\theta\right)}-28\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{56\cos^{2}{\left(\theta\right)}-56\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}\\ &=56\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{\cos^{2}{\left(\theta\right)}-\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}.\\ \end{align}$$


Using the double-angle formulas for sine and cosine,

$$\begin{align} \sin{\left(2\theta\right)} &=2\sin{\left(\theta\right)}\cos{\left(\theta\right)},\\ \cos{\left(2\theta\right)} &=\cos^{2}{\left(\theta\right)}-\sin^{2}{\left(\theta\right)}\\ &=2\cos^{2}{\left(\theta\right)}-1\\ &=1-2\sin^{2}{\left(\theta\right)},\\ \end{align}$$

we can rewrite the integral as

$$\begin{align} \mathcal{I}{\left(\mu\right)} &=56\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{\cos^{2}{\left(\theta\right)}-\sin^{2}{\left(\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}\\ &=56\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{\cos{\left(2\theta\right)}}{2\cos^{4}{\left(\theta\right)}+3\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+\mu\sin^{4}{\left(\theta\right)}}\\ &=56\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{4\cos{\left(2\theta\right)}}{8\cos^{4}{\left(\theta\right)}+12\cos^{2}{\left(\theta\right)}\sin^{2}{\left(\theta\right)}+4\mu\sin^{4}{\left(\theta\right)}}\\ &=56\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{4\cos{\left(2\theta\right)}}{2\left[1+\cos{\left(2\theta\right)}\right]^{2}+3\sin^{2}{\left(2\theta\right)}+\mu\left[1-\cos{\left(2\theta\right)}\right]^{2}}\\ &=56\int_{0}^{\pi}\mathrm{d}\theta\,\frac{2\cos{\left(\theta\right)}}{2\left[1+\cos{\left(\theta\right)}\right]^{2}+3\sin^{2}{\left(\theta\right)}+\mu\left[1-\cos{\left(\theta\right)}\right]^{2}};~~~\small{\left[\theta\mapsto\frac12\theta\right]}\\ &=112\int_{0}^{\pi}\mathrm{d}\theta\,\frac{\cos{\left(\theta\right)}}{2\left[1+\cos{\left(\theta\right)}\right]^{2}+3\sin^{2}{\left(\theta\right)}+\mu\left[1-\cos{\left(\theta\right)}\right]^{2}}.\\ \end{align}$$

Using the tangent half-angle substitution, the trigonometric integral transforms as

$$\begin{align} \mathcal{I}{\left(\mu\right)} &=112\int_{0}^{\pi}\mathrm{d}\theta\,\frac{\cos{\left(\theta\right)}}{2\left[1+\cos{\left(\theta\right)}\right]^{2}+3\sin^{2}{\left(\theta\right)}+\mu\left[1-\cos{\left(\theta\right)}\right]^{2}}\\ &=112\int_{0}^{\infty}\mathrm{d}t\,\frac{2}{1+t^{2}}\cdot\frac{\left(\frac{1-t^{2}}{1+t^{2}}\right)}{2\left(1+\frac{1-t^{2}}{1+t^{2}}\right)^{2}+3\left(\frac{2t}{1+t^{2}}\right)^{2}+\mu\left(1-\frac{1-t^{2}}{1+t^{2}}\right)^{2}};~~~\small{\left[\theta=2\arctan{\left(t\right)}\right]}\\ &=\int_{0}^{\infty}\mathrm{d}t\,\frac{56\left(1-t^{2}\right)}{2+3t^{2}+\mu\,t^{4}}.\\ \end{align}$$

Setting $\sqrt{\frac{2}{\mu}}=:a\in\mathbb{R}_{>0}$ and $\frac34a=:b\in\mathbb{R}_{>0}$, we then have

$$\begin{align} \mathcal{I}{\left(\mu\right)} &=\mathcal{I}{\left(\frac{2}{a^{2}}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}t\,\frac{56\left(1-t^{2}\right)}{2+3t^{2}+2a^{-2}t^{4}}\\ &=\sqrt{a}\int_{0}^{\infty}\mathrm{d}u\,\frac{56\left(1-au^{2}\right)}{2+3au^{2}+2u^{4}};~~~\small{\left[t=u\sqrt{a}\right]}\\ &=\frac13\sqrt{a}\int_{0}^{\infty}\mathrm{d}u\,\frac{28\left(3-4bu^{2}\right)}{1+2bu^{2}+u^{4}}\\ &=\frac{28}{3}\sqrt[4]{\frac{2}{\mu}}\int_{0}^{\infty}\mathrm{d}u\,\frac{\left(3-4bu^{2}\right)}{1+2bu^{2}+u^{4}}\\ &=\frac{28}{3}\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{\left(3-4bu^{2}\right)}{1+2bu^{2}+u^{4}}\\ &~~~~~+\frac{28}{3}\sqrt[4]{\frac{2}{\mu}}\int_{1}^{\infty}\mathrm{d}u\,\frac{\left(3-4bu^{2}\right)}{1+2bu^{2}+u^{4}}\\ &=\frac{28}{3}\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{\left(3-4bu^{2}\right)}{1+2bu^{2}+u^{4}}\\ &~~~~~+\frac{28}{3}\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{\left(3u^{2}-4b\right)}{1+2bu^{2}+u^{4}};~~~\small{\left[u\mapsto\frac{1}{u}\right]}\\ &=\frac{28}{3}\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{\left(3-4b\right)\left(1+u^{2}\right)}{1+2bu^{2}+u^{4}}\\ &=28\left(1-\frac43b\right)\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{1+u^{2}}{1+2bu^{2}+u^{4}}\\ &=28\left(1-a\right)\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{1+u^{2}}{1+2bu^{2}+u^{4}}\\ &=28\left(1-\sqrt{\frac{2}{\mu}}\right)\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{1+u^{2}}{1+2bu^{2}+u^{4}}\\ &=\frac{28\left(\mu-2\right)}{\left(\sqrt{\mu}+\sqrt{2}\right)\sqrt{\mu}}\sqrt[4]{\frac{2}{\mu}}\int_{0}^{1}\mathrm{d}u\,\frac{1+u^{2}}{1+2bu^{2}+u^{4}}.\\ \end{align}$$

Since the integral factor in the last line above is a positive quantity, we don't need to actually solve the integral to determine the sign of $\mathcal{I}$. We have

$$\operatorname{sgn}{\left(\mathcal{I}{\left(\mu\right)}\right)}=\operatorname{sgn}{\left(\mu-2\right)},$$

as you originally conjectured.


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This problem is "nice" in the sense that the integrand is really trig function of $2\theta$ $$ I=\int_{-\pi/2}^{\pi/2}\frac{28\cos 2\theta+5\sin2\theta}{\frac12(1+\cos2\theta)^2+\frac34\sin^2 2\theta+\frac{m}4(\cos2\theta-1)^2}\,\mathrm{d}\theta $$ So $$ I=\frac12\int_{-\pi}^\pi\frac{28\cos\phi+5\sin\phi}{\frac12(1+\cos\phi)^2+\frac34\sin^2 \phi+\frac{m}4(\cos\phi-1)^2}\,\mathrm{d}\phi $$ which we can rewrite as a contour integral of a rational function over the unit circle $z=e^{i\phi}$, $-\pi\leq\phi\leq\pi$ in $\mathbb{C}$, hence it is just a matter of computing residues.

So $$ I=\frac12\int_{\mathbb{T}}\frac{28\cdot\frac12(z+z^{-1})+5\cdot\frac1{2i}(z-z^{-1})}{\frac12(1+\frac12(z+z^{-1}))^2+\frac34(-\frac14(z-z^{-1})^2)+\frac{m}4(\frac12(z+z^{-1})-1)^2}\,\frac{\mathrm{d}z}{iz} $$ which simplifies to $$ I=\frac1{2i}\int_{\mathbb{T}} \frac{8 ((28 + 5 i) + (28 - 5 i) z^2)\,\mathrm{d}z}{(m-1)(z^4+1)-4(m-2)(z^3+z)+6(m-3)z^2} $$ The poles are at, if $m\neq 1$, $$ z+z^{-1}=\frac{2(m-2\pm\sqrt{m})}{m-1}. $$ so, since $m>0$ $$\label{eq:poles} z= \begin{cases} 0,\frac12(3\pm\sqrt5)& m=1\\ \frac{2\pm\sqrt{m}\pm\sqrt{3\pm 2\sqrt{m}}}{1\pm\sqrt{m}}&m\neq 1. \end{cases}\tag{$\star$} $$ So all poles are at worst simple unless $m=\frac49$ (in which case we get a double pole at $z=4$ which we can ignore), and $$ I=\pi\sum_{\substack{\lvert z_i\rvert<1\\z_i\in\eqref{eq:poles}}}\operatorname{res}_{z_i}(\dots)+(\text{correction if }\lvert z_i\rvert=1). $$

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  • $\begingroup$ Can you give more details of this method? $\endgroup$ – user326159 Jun 20 at 20:30
  • $\begingroup$ @user326159 It's quite a standard method which you can find in any textbook dealing with complex integration. The key point here is the so-called Residue Theorem which I suggest you to look up. $\endgroup$ – mrtaurho Jun 21 at 7:59
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note that since the function part of the function is odd i.e: $$f(x)=\frac{28\cos^2x+10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$ $$f(-x)=\frac{28\cos^2x-10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$ you could notice that the integral can be simplified to: $$\int_{-\pi/2}^{\pi/2}\frac{28\cos^2x+10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$ $$=\int_{-\pi/2}^{\pi/2}\frac{28\cos^2x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$ $$=2\int_0^{\pi/2}\frac{28\cos^2x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$ $$=56\int_0^{\pi/2}\frac{\cos^2x-\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}dx$$


One route you could try to take is Tangent half-angle substitution, which yields: $$112\int_0^1(1+t^2)\frac{(1-t^2)^2-(2t)^2}{2(1-t^2)^4+3(1-t^2)^2(2t)^2+m(2t)^4}dt$$ the bottom of this fraction can be expanded to: $$2t^8+4t^6+16t^4m-12t^4+4t^2+2$$ this may be factorisable for certain values of $m$

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