How to integrate $\int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x}{y} + \frac{y}{x}\right) \,dx\,dy$

Question

how to integrate $$\int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x}{y} + \frac{y}{x}\right) \,dx\,dy$$

My attempt

$$\int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x}{y} + \frac{y}{x}\right) \,dx\,dy = \int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x^2 + y^2}{xy}\right) \,dx\,dy$$

$$\int_{0}^{1} \int_{0}^{1} \frac{1}{2} \ln\left[\frac{1 + \frac{x^2 + y^2}{xy}}{1 - \frac{x^2 + y^2}{xy}}\right] \,dx\,dy$$

$$\int_{0}^{1} \frac{1}{2} \int_{0}^{1} \ln\left[\frac{xy + x^2 + y^2}{xy - (x^2 + y^2)}\right] \,dx\,dy$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \ln\left[\frac{x^2 + yx + y^2}{-(x^2 - yx + y^2)}\right] \,dx\,dy$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \{\ln(x^2 + yx + y^2) \,dx\,dy - \ln(-1)\left[(x^2 - yx + y^2)\right]\}$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \ln(x^2 + yx + y^2) - \ln(-1) - \ln(x^2 - yx + y^2) \,dx \,dy$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \left[\ln(x^2 + yx + y^2) - \ln(x^2 - yx + y^2) - \ln(e^{i\pi})\right] \,dx\,dy$$

$$\frac{1}{2} \int_{0}^{1} \left[\int_{0}^{1} [\ln(x^2 + yx + y^2) - \ln(x^2 - yx + y^2)] \,dx - i\pi \int_{0}^{1} 1 \,dx\right] \,dy$$

$$\frac{1}{2} \int_{0}^{1} \left[\int_{0}^{1} \ln(x^2 + yx + y^2) \,dx - \int_{0}^{1} \ln(x^2 - yx + y^2) \,dx - i\pi \big[x\big]_{0}^{1}\right] \,dy$$

$$\frac{1}{2} \int_{0}^{1} \left[ \ln(1 + y + y^2) - \int_{0}^{1} \frac{2x^2 + yx}{x^2 + yx + y^2} \,dx - \ln(1 - y + y^2) + \int_{0}^{1} \frac{2x^2 - yx}{x^2 - yx + y^2} \,dx - ix \right]\,dy$$

Answer 1

Hint:

Per symmetry

$$\int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x}{y} + \frac{y}{x}\right) dxdy= 2\int_{0}^{1} \int_{0}^{y} \tanh^{-1}\left(\frac{x}{y} + \frac{y}{x}\right) \,dxdy $$ Then, substitute $x=yt$ to reduce the integral to $$\int_0^1 \tanh^{-1}\left(t+\frac1t\right)dt$$

Answer 2

There is no indication that complex numbers should be used in this exercise. Hence it is tacitly implied that this is a calculation on the real numbers. Now the function $\tanh$ yields a value in the range $(-1,1)$. Hence the inverse function only accepts values in this interval; other values are outside of its domain. In this case we can quickly establish that the argument lies outside, since $x/y + y/x \ge 2$ for all positive values of $x$ and $y$. Therefore the integrand does not exist and the integral is undefined.