Integral of the day XXIX

Integral representation of the $\arcsin(a)$ function

Today we show the proof of this integral posted by @integralsbot \[ \int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = \pi \arcsin(a) \quad |a|\leq 1 \] In the proof we use the residue theorem to evaluate it.

Proof:

Recall the logarithmic definition of $\operatorname{arcanh}(x)$ \[\operatorname{arcanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x} \right) \quad |x|\lt 1 \] Hence \[ I(a) = \int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = 2\int_{0}^{1} \frac{\operatorname{arcanh}(ax)}{x\sqrt{1-x^2}} dx \] Hence \begin{align*} I'(a) = 2\int_{0}^{1} \frac{1}{\sqrt{1-x^2}(1-a^2x^2)} dx = & 2\int_{0}^{\frac{\pi}{2}} \frac{1}{1-a^2\sin^2(w)} dw \quad (x \mapsto \sin(w)) \\ =& 4\int_{0}^{\frac{\pi}{2}} \frac{1}{a^2\cos(2w)-a^2+2} dw \\ =& 2\int_{0}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta \quad (\theta \mapsto 2w)\\ =& \int_{-\pi}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta \quad \textrm{(The integrand is even)} \end{align*} If we make the following change of variable \[ \cos\theta = \frac{z+z^{-1}}{2} \] \[ d\theta = \frac{dz}{zi} \] \[ I'(a) = \int_{-\pi}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta = \frac{2}{i}\oint_{|z|=1} \frac{dz}{a^2z^2+(4-2a^2)z+a^2} \] The integrand has two poles: \[ z_{1} = \frac{a^2-2+2\sqrt{1-a^2}}{a^2} \] \[ z_{2} = \frac{a^2-2-2\sqrt{1-a^2}}{a^2} \] Only $z_{1}$ is inside the unit circle. Suppose $|a|\leq 1$: \begin{align*} \left| \frac{a^2-2+2\sqrt{1-a^2}}{a^2}\right|\lt 1 \Longrightarrow& -a^2\lt a^2-2+2\sqrt{1-a^2}\lt a^2 \\ \Longrightarrow& -2a^2\lt-2+2\sqrt{1-a^2}\lt0\\ \Longrightarrow& 1-a^2\lt \sqrt{1-a^2}\lt 1\\ \Longrightarrow& a\neq 0 \; \wedge \; a\neq 1 \end{align*} Then, if we add the assumption $|a| \lt 1 \;\wedge\; a\neq 0 \Longrightarrow |z_{1}|\lt1$

By the residue theorem: \begin{align*} I'(a) = \int_{-\pi}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta =& \frac{2}{i}\oint_{|z|=1} \frac{dz}{a^2z^2+(4-2a^2)z+a^2}\\ =& \frac{4\pi}{a^2} \lim_{z \to \frac{a^2-2+2\sqrt{1-a^2}}{a^2}} \frac{1}{\left(z-\frac{a^2-2-2\sqrt{1-a^2}}{a^2}\right)}\\ =& \frac{4\pi}{\left(a^2-2+2\sqrt{1-a^2}-\left(a^2-2-2\sqrt{1-a^2}\right)\right)}\\ =& \frac{\pi}{\sqrt{1-a^2}} \end{align*} Therefore \[ I(a) = \int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = \pi \arcsin(a) + C \] If we let $ a \to 0$ then $C \to 0$.

The case when $a = 1$ can be obtained as a limitng case easily.

Therefore \[ \boxed{ \int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = \pi \arcsin(a) \quad |a|\leq 1} \]