Integral of the day XXIII

Nice integral solved with contour integration

Today we show the proof of this fun integral posted by @integralsbot \[ \int_{0}^{\frac{\pi}{4}} \frac{1}{1+\sqrt{1-\tan^2 (x)}}dx = \frac{-1+2\sqrt{2}}{4}\pi -1 \] The proof will rely on contour integration round the unit complex circle.

Proof

\begin{align*} I = \int_{0}^{\frac{\pi}{4}} \frac{1}{1+\sqrt{1-\tan^2 (x)}}dx =& \int_{0}^{1} \frac{1}{(1+\sqrt{1-w^2})(1+w^2)}dw \quad \left(w \mapsto \tan x \right)\\ =& \int_{0}^{1} \left( \frac{1-\sqrt{1-w^2}}{1-\sqrt{1-w^2}}\right)\frac{1}{(1+\sqrt{1-w^2})(1+w^2)}dw \\ =& \int_{0}^{1} \frac{1-\sqrt{1-w^2}}{w^2(1+w^2)} dw\\ =& \underbrace{\int_{0}^{1} \frac{1-\sqrt{1-w^2}}{w^2} dw}_{J} - \underbrace{\int_{0}^{1} \frac{1}{1+w^2} dw}_{K} + \underbrace{\int_{0}^{1} \frac{\sqrt{1-w^2}}{1+w^2}dw}_{L} \end{align*} \begin{align*} J = \int_{0}^{1} \frac{1-\sqrt{1-w^2}}{w^2} dw \stackrel{IBP}{=}& \frac{\sqrt{1-w^2}-1}{w}\Big|_{0}^{1} + 2\int_{0}^{1} \frac{1}{\sqrt{1-w^2}} dx \\ =& -1 + 2\arcsin(1)\\ =& -1 + \frac{\pi}{2} \end{align*} \begin{align*} K = \int_{0}^{1} \frac{1}{1+w^2} dw = \arctan(1) = \frac{\pi}{4} \end{align*} \begin{align*} L = \int_{0}^{1} \frac{\sqrt{1-w^2}}{1+w^2}dw =& \frac{1}{2} \int_{-1}^{1} \frac{\sqrt{1-w^2}}{1+w^2}dw \quad \textrm{(the integrand is even)}\\ =& \int_{0}^{\pi} \frac{\sin^2 s}{1+ \cos^2 s} ds \\ =& \int_{0}^{\pi} \frac{1-\cos^2 s}{1+ \cos^2 s} ds \\ =& \int_{0}^{\pi} \frac{2-(1+\cos^2 s)}{1+ \cos^2 s} ds \\ =& 2\int_{0}^{\pi} \frac{1}{1+ \cos^2 s} ds- \int_{0}^{\pi} ds\\ =& 4\int_{0}^{\pi} \frac{1}{3+ \cos(2s)} ds- \pi \quad \left( \cos^2 \vartheta = \frac{1+\cos(2\vartheta)}{2} \right)\\ =& 2\underbrace{ \int_{0}^{2\pi} \frac{1}{3+ \cos\theta}d\theta}_{M} - \pi \\ \end{align*} To solve $M$ we can make the following change of variable \[ \cos \theta = \frac{z+z^{-1}}{2} \] \[ d\theta = \frac{dz}{zi} \] with this change the integral is transformed in a contour integral round the unit complex circle \begin{align*} M = \int_{0}^{2\pi} \frac{1}{3+ \cos\theta}d\theta = \frac{2}{i} \oint_{|z|=1} \frac{1}{z^2+6z+1} dz \end{align*} The only pole inside the unit circle is $ z= 2\sqrt{2}-3$. Therefore by the residue theorem: \begin{align*} \oint_{|z|=1} \frac{1}{z^2+6z+1} dz=& 2\pi i \operatorname{Res}\left(\frac{1}{z^2+6z+1},2\sqrt{2}-3\right) \\ =& \lim_{z \to 2\sqrt{2}-3} \frac{2\pi i}{z+3+2\sqrt{2}} \\ =& \frac{\pi i }{\sqrt{2}} \end{align*} Hence \[ M = \frac{2}{i} \oint_{|z|=1} \frac{1}{z^2+6z+1} dz = \frac{2 \pi}{\sqrt{2}} \] Therefore \[ L = \frac{2}{\sqrt{2}}\pi - \pi \] \[ I = J-k+L = -1 + \frac{\pi}{2}- \frac{\pi}{4} + \frac{2}{\sqrt{2}}\pi - \pi = \frac{-1+2\sqrt{2}}{4}\pi -1 \] Finally, we can conclude \[ \boxed{ \int_{0}^{\frac{\pi}{4}} \frac{1}{1+\sqrt{1-\tan^2 (x)}}dx = \frac{-1+2\sqrt{2}}{4}\pi -1 } \]