A trigonometric integral via a semicircular contour

Integrals containing nested inverse trigonometric functions often simplify dramatically after a change of variables that unwraps the inner function, followed by a passage to the unit circle in the complex plane. We will prove that for every \(n \in \mathbb{N}\),

\[ \int_0^{\frac{\pi}{2}}\frac{\cos\left((2n-1)\arcsin\left(\frac{\sin x}{\sqrt{2}}\right) \right)}{\sqrt{1-\frac{\sin^2 x}{2}}}\,dx = \frac{\sqrt{\pi}}{2}\sin\left(\frac{n\pi}{2}\right)\frac{\Gamma\left(\frac{n}{2} \right)}{\Gamma\left(\frac{n+1}{2}\right)}. \]

Unwrapping the inverse sine

Let \(I\) denote the integral. We begin with the substitution \(\sin t = \frac{\sin x}{\sqrt{2}}\), so that \(\cos t\,dt = \frac{\cos x}{\sqrt{2}}\,dx\). Since \(\cos x = \sqrt{1 - 2\sin^2 t} = \sqrt{\cos(2t)}\) and \(t\) runs from \(0\) to \(\frac{\pi}{4}\) as \(x\) runs from \(0\) to \(\frac{\pi}{2}\), the factor \(\sqrt{1 - \frac{\sin^2 x}{2}} = \cos t\) cancels against \(\cos t\,dt\):

\[ I = \sqrt{2}\int_0^{\frac{\pi}{4}}\frac{\cos\bigl((2n-1) t\bigr)}{\sqrt{1-2\sin^2 t}}\,dt = \sqrt{2}\int_0^{\frac{\pi}{4}}\frac{\cos\bigl((2n-1) t\bigr)}{\sqrt{\cos(2t)}}\,dt. \]

Next, substitute \(\theta = 2t\) and use the evenness of the integrand on \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\):

\[ \begin{aligned} I &= \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}}\frac{\cos\left(\frac{2n-1}{2}\theta\right)}{\sqrt{\cos \theta}}\,d\theta \\ &= \frac{1}{2\sqrt{2}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos\left(\frac{2n-1}{2}\theta\right)}{\sqrt{\cos \theta}}\,d\theta \\ &= \frac{1}{2\sqrt{2}} \operatorname{Re}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\exp\left(\frac{2n-1}{2}i\theta\right)}{\sqrt{\cos \theta}}\,d\theta. \end{aligned} \tag{1} \]

Integrating around the right half of the unit disk

Now integrate the principal branch of

\[ f(z) = \frac{z^{n-1}}{\sqrt{1+z^2}} \]

around the positively oriented contour in the right half-plane shown below:

Right half-unit-disk contour bounded by the unit semicircle in the right half-plane and the imaginary axis with small circular indents at plus i and minus i

The contour consists of the right half of the unit circle \(|z|=1\) (\(-\frac{\pi}{2} < \arg z < \frac{\pi}{2}\)), the segment of the imaginary axis between \(+i\) and \(-i\), and two small circular indents of radius \(\epsilon\) avoiding the branch points at \(z = \pm i\). Near \(z = \pm i\), we have \(|1+z^2| = |z-i||z+i| \sim 2\epsilon\), so \(|f(z)| = O(\epsilon^{-1/2})\) while the arc length of each indent is \(O(\epsilon)\). Thus the integrals along both indents are \(O(\epsilon^{1/2})\) and vanish as \(\epsilon \to 0^+\).

Since \(f(z)\) is analytic inside the indented region, Cauchy's theorem equates the integral over the right semicircle to the integral upwards along the imaginary axis from \(-i\) to \(+i\). Parameterizing the imaginary axis by \(z = iy\) with \(y \in (-1, 1)\) and \(dz = i\,dy\) gives

\[ \begin{aligned} \int_{\left\{z\in \mathbb{C} :\; |z|=1, \; -\frac{\pi}{2}<\arg z < \frac{\pi}{2} \right\}}\frac{z^{n-1}}{\sqrt{1+z^2}}\,dz &= \int_{-i}^i \frac{z^{n-1}}{\sqrt{1+z^2}}\,dz \\ &= i^{n}\int_{-1}^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy \\ &= \begin{cases} 2 i^n \int_0^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy, & n\text{ odd},\\ 0, & n\text{ even}. \end{cases} \end{aligned} \tag{2} \]

Notice that when \(n = 2k+1\) is odd, \(i^n = i(-1)^k = i\sin\left(\frac{n\pi}{2}\right)\), whereas when \(n\) is even, \(\sin\left(\frac{n\pi}{2}\right) = 0\). Hence, for every \(n \in \mathbb{N}\), equation (2) can be written in the unified form

\[ \int_{\left\{z\in \mathbb{C} :\; |z|=1, \; -\frac{\pi}{2}<\arg z < \frac{\pi}{2} \right\}}\frac{z^{n-1}}{\sqrt{1+z^2}}\,dz = 2i\sin\left(\frac{n\pi}{2}\right)\int_0^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy. \tag{2'} \]

Parameterizing the semicircular arc

On the right unit semicircle, set \(z = e^{i\theta}\) for \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), with \(dz = i e^{i\theta}\,d\theta\). Since \(1 + e^{2i\theta} = 2e^{i\theta}\cos\theta\) and \(\cos\theta > 0\) on \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), the principal square root is \(\sqrt{1+e^{2i\theta}} = \sqrt{2\cos\theta}\,e^{\frac{i\theta}{2}}\). Therefore,

\[ \begin{aligned} \int_{\left\{z\in \mathbb{C} :\; |z|=1, \; -\frac{\pi}{2}<\arg z < \frac{\pi}{2} \right\}}\frac{z^{n-1}}{\sqrt{1+z^2}}\,dz &= i\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{e^{in \theta}}{\sqrt{1+e^{2i\theta}}}\,d\theta \\ &= i\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{e^{\frac{2n-1}{2}i \theta}}{\sqrt{e^{i\theta}+e^{-i\theta}}}\,d\theta \\ &= \frac{i}{\sqrt{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{e^{\frac{2n-1}{2}i \theta}}{\sqrt{\cos \theta}}\,d\theta. \end{aligned} \tag{3} \]

Combining equations (1), (2'), and (3), and finally setting \(\xi = y^2\) (\(d\xi = 2y\,dy\)) to reduce the remaining integral to Euler's beta function \(B\left(\frac{n}{2}, \frac{1}{2}\right)\), we conclude that

\[ \begin{aligned} I &= \sin\left(\frac{n\pi}{2}\right)\int_0^1 \frac{y^{n-1}}{\sqrt{1-y^2}}\,dy \\ &= \frac{1}{2}\sin\left(\frac{n\pi}{2}\right)\int_0^1 \xi^{\frac{n}{2}-1}(1-\xi)^{-\frac{1}{2}}\,d\xi \\ &= \frac{\sqrt{\pi}}{2}\sin\left(\frac{n\pi}{2}\right)\frac{\Gamma\left(\frac{n}{2} \right)}{\Gamma\left(\frac{n+1}{2}\right)}. \end{aligned} \]