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The KoenigsFunction of $f(x) = \tanh(\ln(1+x^2))$

The generating function for the coefficients $a_n$ of the Koenigs function of $f(x) = \tanh(\ln(1+x^2))$ is:

$$A(z) = \sum_{n=1}^{\infty} a_n z^n = \frac{2z}{(e^{2z}-1)\sqrt{3}}$$

To calculate its Mellin transform, we first define the Mellin transform of a function $f(x)$ as:

$$\mathcal{M}f(x) = \int_0^{\infty} x^{s-1} f(x) dx$$

where $s$ is a complex variable.

Substituting $f(x) = A(x)$, we have:

$$\mathcal{M}A(x) = \int_0^{\infty} x^{s-1} A(x) dx = \int_0^{\infty} x^{s} \frac{2x}{(e^{2x}-1)\sqrt{3}} dx$$

Using the change of variables $u = 2x$, we obtain:

$$\mathcal{M}A(x) = \frac{1}{\sqrt{3}} \int_0^{\infty} \frac{u^s}{e^u-1} du$$

To evaluate this integral, we use the identity:

$$\frac{1}{e^u-1} = \sum_{n=1}^{\infty} e^{-nu}$$

which holds for $\operatorname{Re}(u)>0$. Substituting this identity and changing the order of integration and summation, we obtain:

$$\mathcal{M}A(x) = \frac{1}{\sqrt{3}} \sum_{n=1}^{\infty} \int_0^{\infty} u^s e^{-nu} du = \frac{1}{\sqrt{3}} \Gamma(s+1) \sum_{n=1}^{\infty} \frac{1}{n^{s+1}}$$

where $\Gamma(s)$ is the gamma function. This is a well-known result for the Mellin transform of the Riemann zeta function $\zeta(s)$.

Therefore, the Mellin transform of the generating function $A(z)$ is:

$$\mathcal{M}A(x) = \frac{1}{\sqrt{3}} \Gamma(s+1) \zeta(s+1)$$