A mysterious limit related to the integral $\int_{0}^{+\infty}\left(1-\frac{\tanh(ax)}{\tanh x}\right)\,dx$ I have to show that the following limit:
$$ \lim_{a\rightarrow0}\Big[\sin(\pi a)\int_0^\infty \left(1-\frac{\tanh ax}{\tanh x}\right)dx\Big]=\pi\ln2$$
holds.
This problem relates to my previous question parametric integral relating to hyperbolic function.
 A: We have:
$$I_n=\int_{0}^{+\infty}\left(1-\frac{\tanh(x/n)}{\tanh(x)}\right)\,dx=\int_{0}^{+\infty}\left(1-\frac{e^{2x/n}-1}{e^{2x/n}+1}\cdot\frac{e^{2x}+1}{e^{2x}-1}\right)\,dx$$
hence:
$$ I_n = \int_{0}^{+\infty}\frac{e^{x}-e^{x/n}}{(e^{x/n}+1)(e^{x}-1)}\,dx=n\int_{1}^{+\infty}\frac{z^n-z}{z(z+1)(z^n-1)}\,dz=n\int_{0}^{1}\frac{z^{n-1}-1}{z^n-1}\frac{dz}{z+1}$$
If we consider the sequence of functions given by $f_n=\frac{z^{n-1}-1}{z^n-1}$, we have that $f_n\to 1$ pointwise on $[0,1)$, hence:
$$\int_{0}^{1}\frac{z^{n-1}-1}{z^n-1}\frac{dz}{z+1}\to \int_{0}^{1}\frac{dx}{x+1}=\log 2$$
holds by the dominated convergence theorem and it is enough to prove our claim, since $\lim_{a\to 0}\frac{\sin(a)}{a}=1.$ We can also avoid the dominated convergence theorem by estimating how much $f_n$ deviates from $1$ on $[0,1)$:
$$ 1-f_n(z) = \frac{1-z}{1-z^n}z^{n-1}\leq z^{n-1}$$
then applying a trivial inequality:
$$ \log 2-\int_{0}^{1}\frac{f_n(z)\,dz}{z+1}\leq\int_{0}^{1}\frac{z^{n-1}}{z+1}\,dz\leq\int_{0}^{1}z^{n-1}\,dz = \frac{1}{n}.$$
