Integral of $ \mathrm{d}z/(a+\sin^2{z}) $

Question

I am trying to compute $$\int_0^{\pi/2} \frac{\mathrm{d}z}{a+\sin^2{z}},\,a>0 $$ for a physics problem. I've managed to expand the integrand to $$\frac{1}{a} - \frac{z^2}{a^2}+\frac{(3+a)z^4}{3a^3} + O(z^6)$$ but there are no terms of order $z^{-1}$ and so I don't see how to apply Cauchy's remainder theorem. Any help would be appreciated; the expected result is $$\frac{\pi\sqrt{1+1/a}}{2(1+a)}$$ obtained via Mathematica.

Answer 1

I assume $a>0$. We have $$I=\int_{0}^{\pi/2}\frac{dz}{a+\sin^{2}\left(z\right)}=\int_{0}^{\pi/2}\frac{\sec^{2}\left(z\right)}{a\sec^{2}\left(z\right)+\tan^{2}\left(z\right)}dz=\int_{0}^{\pi/2}\frac{\sec^{2}\left(z\right)}{\left(a+1\right)\tan^{2}\left(z\right)+a}dz $$ now put $\tan\left(z\right)=u .$ We have $$I=\int_{0}^{\infty}\frac{1}{\left(a+1\right)u^{2}+a}du=\frac{1}{a}\int_{0}^{\infty}\frac{1}{\frac{\left(a+1\right)u^{2}}{a}+1}du $$ now if put $\sqrt{\frac{a+1}{a}}u=v $ we get $$I=\frac{1}{\sqrt{a\left(a+1\right)}}\int_{0}^{\infty}\frac{1}{v^{2}+1}du=\color{red}{\frac{\pi}{2\sqrt{a\left(a+1\right)}}}.$$

Answer 2

Call the answer $J$. By symmetry, $$ J = \dfrac{1}{4} \int_0^{2\pi} \dfrac{dt}{a + \sin^2(t)} $$ Now express this as a contour integral around the unit circle $\Gamma$: $z = e^{it}$, $dz = i e^{it}\; dt$, $\sin(t) = (z - 1/z)/(2i)$:

$$ J = \dfrac{1}{4} \oint_\Gamma \dfrac{dz}{i z (a + ((z-1/z)/(2i))^2)} =-i \oint_\Gamma \dfrac{z\; dz}{4a z^2 - (z^2-1)^2} $$

This can be done using residues. The integrand has poles at $\pm \sqrt{2a+1 \pm 2\sqrt{a^2+a}}$. The question is whether these are inside or outside the unit circle. Note that $4 a w - (w-1)^2 = -w^2 + (4a+2) w - 1$. Its two roots have product $1$, so either both are on the unit circle or one is inside and the other outside. They are both on the unit circle when $-1 \le a \le 0$ (in which case the integral doesn't exist). For $a > 0$ the roots inside the unit circle are $\pm \sqrt{2a+1 - 2 \sqrt{a^2+a}}$.

Answer 3

Note that the integral of interest diverges for $-1<a<0$. We assume tacitly, therefore, that $a>0$. Then, we can write

$$\begin{align} a+\sin^2(z)&=(1+a)-\cos^2(z)\\\\ &=(1+a)-\left(\frac12+\frac12\cos(2z)\right)\\\\ &=\frac12\left((1+2a)-\cos(2z)\right) \end{align}$$

Then, we have

$$\begin{align} \int_0^{\pi/2} \frac{1}{a+\sin^2(z)}\,dz&=\frac12\int_{-\pi/2}^{\pi/2}\frac{1}{a+\sin^2(z)}\,dz\\\\ &=\int_{-\pi/2}^{\pi/2}\frac{1}{(1+2a)-\cos(2z)}\,dz\\\\ &=\frac12 \int_{-\pi}^\pi \frac{1}{(1+2a)-\cos(z)}\\\\ &=\frac12 \oint_{|z|=1}\frac{1}{iz\left((1+2a)-\frac12(z+z^{-1})\right)}\,dz\\\\ &=i\oint_{|z|=1}\frac{1}{z^2-2(1+2a)z+1}\,dz\\\\ &=(2\pi i) (i) \text{Res}\left(\frac{1}{z^2-2(1+2a)z+1}, z=(1+2a)+2\sqrt{a(a+1)}\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi}{2\sqrt{a(a+1)}}} \end{align}$$