parametric integral relating to hyperbolic function Suppose that $a$ is real number such that $0<a<1$, how can we calculate
$$ I(a)=\int_0^\infty \big(1-\frac{\tanh ax}{\tanh x}\big)dx .$$
As for some speical cases, I can work out $I(1/2)=1$.  Any suggestion to the integral above? Tks.

Expanding @Lucian's comment, if $a=\frac{p}{q}\in\mathbb{Q}$, we have:
$$\begin{eqnarray*}I(a)&=&q\int_{0}^{+\infty}\left(1-\frac{\tanh px}{\tanh qx}\right)\,dx=q\int_{1}^{+\infty}\left(1-\frac{x^{2p}-1}{x^{2p}+1}\cdot\frac{x^{2q}+1}{x^{2q}-1}\right)\frac{dx}{x}\\&=&2q\int_{1}^{+\infty}\frac{x^{2q}-x^{2p}}{(x^{2p}+1)(x^{2q}-1)}\cdot\frac{dx}{x}=q\int_{0}^{+\infty}\frac{x^{2p}-x^{2q}}{(1+x^{2q})(1-x^{2p})}\cdot\frac{dx}{x}\\&=&\pi i q\sum_{z\in Z\cap \{\Im z>0\}}\operatorname{Res}\left(\frac{x^{2p}-x^{2q}}{x(1+x^{2q})(1-x^{2p})},x=z\right)\end{eqnarray*}$$
where $$Z=Z_p\cup Z_q=\left\{e^{\frac{\pi i}{p}},e^{\frac{2\pi i}{p}},\ldots,e^{\frac{(p-1)\pi i}{p}}\right\}\cup\left\{e^{\frac{\pi i}{4q}},e^{\frac{3\pi i}{4q}},\ldots,e^{\frac{(2q-1)\pi i}{4q}}\right\}.$$
However, I strongly doubt that a nice closed formula exists, since, for instance:
$$I(3/5)=\frac{\pi}{15} \sqrt{\frac{154}{3}+\frac{18}{\sqrt{5}}-4 \sqrt{30 \left(5+\sqrt{5}\right)}}.$$
