Integral of the day IX

Another singular integral of @integralsbot

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{1} \frac{x^2}{1+x^2} \sqrt{\frac{1-x}{1+x}}dx = \frac{\ln(1+\sqrt{2})}{\sqrt{2}} + \left(1-\frac{1}{\sqrt{2}}\right)\frac{\pi}{2}-1$$

Proof

Firt, if we make the substitution $$w = \sqrt{\frac{1-x}{1+x}} \Longrightarrow x = \frac{1-w^2}{1+w^2} \Longrightarrow dx = -\frac{4w}{(1+w^2)^2}dw$$ Hence $$\begin{align*} I =& \int_{0}^{1} \frac{x^2}{1+x^2} \sqrt{\frac{1-x}{1+x}}dx = -\int_{0}^{1} \frac{2w^2(w-1)^2(1+w)^2}{(1+w^2)^2(1+w^4)}dw\\ =& 4\underbrace{\int_{0}^{1} \frac{1}{1+w^2}dw}_{J} -4\underbrace{\int_{0}^{1} \frac{1}{(1+w^2)^2} dw}_{K} - 2 \underbrace{\int_{0}^{1} \frac{w^2}{1+w^4}dw}_{Q} \end{align*}$$ $$J = \int_{0}^{1} \frac{1}{1+w^2}dw = \arctan(w)\Big|_{0}^{1} = \frac{\pi}{4}$$ $$\begin{align*} K =& \int_{0}^{1} \frac{1}{(1+w^2)^2} dw \\ =& \int_{0}^{\frac{\pi}{4}} \frac{\sec^2(r)}{(1+\tan^2(r))^2} dr \quad (w \mapsto \tan(r))\\ =& \int_{0}^{\frac{\pi}{4}} \cos^2(r) dr \quad ( 1+\tan^2(r) = \sec^2(r))\\ =& \frac{1}{2}\int_{0}^{\frac{\pi}{4}}\cos(2r) dr + \frac{1}{2}\int_{0}^{\frac{\pi}{4}}dr \quad \left(\cos^2(r)=\frac{\cos(2r)+1}{2}\right)\\ =& \frac{1}{4}\int_{0}^{\frac{\pi}{2}}\cos(t) dt + \frac{\pi}{8} \quad (t \mapsto 2r)\\ =& \frac{1}{4} + \frac{\pi}{8} \end{align*}$$ Now for $Q$, note that $$\underbrace{\int_{0}^{\infty} \frac{w^2}{1+w^4} dw}_{L} - \underbrace{\int_{1}^{\infty} \frac{w^2}{1+w^4} dw}_{M} = \underbrace{\int_{0}^{1} \frac{w^2}{1+w^4} dw}_{Q}$$ $$L = \int_{0}^{\infty} \frac{w^2}{1+w^4} dw = \frac{1}{2}\int_{0}^{\infty} \frac{\sqrt{s}}{1+s^2} ds\quad (s^2 \mapsto w^4)$$ Recall the integral representation of the $\sec(x)$ function: $$\sec(x) = \frac{2}{\pi} \int_{0}^{\infty} \frac{t^{2x/\pi}}{t^2+1}dt \quad |x|< \frac{\pi}{2}$$ If we put $\displaystyle x = \frac{\pi}{4}$ $$\sec\left(\frac{\pi}{4}\right) = \frac{2}{\pi} \int_{0}^{\infty} \frac{\sqrt{t}}{t^2+1}dt \Longrightarrow \int_{0}^{\infty} \frac{\sqrt{t}}{t^2+1}dt = \frac{\pi}{2} \sec\left(\frac{\pi}{4}\right) = \frac{\pi}{\sqrt{2}}$$ Then $$L =\frac{1}{2}\int_{0}^{\infty} \frac{\sqrt{s}}{1+s^2} ds = \frac{\pi}{2\sqrt{2}}$$ For $M$: $$M=\int_{1}^{\infty} \frac{w^2}{1+w^4} dw = \int_{0}^{1} \frac{1}{1+r^4} dr \quad \left(r \mapsto \frac{1}{w}\right)$$ Hence $$M= \int_{0}^{1} \frac{1}{1+r^4} dr = \frac{1}{2}\int_{0}^{1} \frac{(1+r^2)+(1-r^2)}{1+r^4} dr = \frac{1}{2} \underbrace{\int_{0}^{1} \frac{1+r^2}{1+r^4}dr}_{F} + \frac{1}{2}\underbrace{\int_{0}^{1} \frac{1-r^2}{1+r^4}dr}_{G}$$ $$\begin{align*} F= \int_{0}^{1} \frac{1+x^2}{1+x^4}dx =& \int_{0}^{1} \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}dx\\ =& \int_{0}^{1} \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}dx \\ =& \int_{0}^{1} \frac{\frac{1}{x^2}+1}{\left(x-\frac{1}{x}\right)^2+2}dx\\ =& \int_{0}^{\infty} \frac{1}{w^2+2}dx \quad \left(w \mapsto x-\frac{1}{x}\right) \\ =& \frac{\sqrt{2}}{2}\int_{0}^{\infty} \frac{1}{r^2+1}dx \quad \left(r^2 \mapsto \frac{w^2}{2}\right)\\ =& \frac{\pi}{2\sqrt{2}} \end{align*}$$ $$\begin{align*} G = \int_{0}^{1} \frac{1-x^2}{1+x^4}dx =& \int_{0}^{1} \frac{1-\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx\\ =& \int_{0}^{1} \frac{1-\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx \\ =& \int_{0}^{1} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-2}dx\\ =& -\int_{2}^{\infty} \frac{1}{w^2-2} dx \quad \left(w \mapsto x+\frac{1}{x}\right) \\ =& \frac{1}{2\sqrt{2}}\int_{2}^{\infty} \left[ \frac{1}{w+\sqrt{2}}-\frac{1}{w-\sqrt{2}} \right] dx\\ =& \frac{1}{2\sqrt{2}} \left[\ln(w+\sqrt{2}) - \ln(w-\sqrt{2})\right]_{2}^{\infty}\\ =& \frac{1}{\sqrt{2}} \left[\ln\left(\sqrt{\frac{\frac{w}{\sqrt{2}}+1}{\frac{w}{\sqrt{2}}-1}}\right) \right]_{2}^{\infty}\\ =& \frac{1}{\sqrt{2}} \left[\operatorname{arccoth}\left(\frac{w}{\sqrt{2}}\right)\right]_{2}^{\infty}\\ =& \frac{\operatorname{arccoth}(\sqrt{2})}{\sqrt{2}} \end{align*}$$ Hence $$M = \frac{1}{2}F+\frac{1}{2}G = \frac{\pi}{4\sqrt{2}}+ \frac{\operatorname{arccoth}(\sqrt{2})}{2\sqrt{2}}$$ $$Q = L - M = \frac{\pi}{4\sqrt{2}} -\frac{\operatorname{arccoth}(\sqrt{2})}{2\sqrt{2}}$$ Therefore $$I = 4J - 4K - 2Q = \pi-1 - \frac{\pi}{2} - \frac{\pi}{2\sqrt{2}} + \frac{\operatorname{arccoth}(\sqrt{2})}{\sqrt{2}}$$ Finally, using the following identity: $$\operatorname{arccoth}(x) = \operatorname{sign}(x)\operatorname{arcsinh}\left(\frac{1}{\sqrt{x^2-1}}\right)$$ We have $$\operatorname{arccoth}(\sqrt{2})= \operatorname{arcsinh}(1) = \ln(1+\sqrt{2})$$ Therefore $$\boxed{\int_{0}^{1} \frac{x^2}{1+x^2} \sqrt{\frac{1-x}{1+x}}dx = \frac{\ln(1+\sqrt{2})}{\sqrt{2}} + \left(1-\frac{1}{\sqrt{2}}\right)\frac{\pi}{2}-1}$$