SFunction

Check it out, closed-form Newton flows of the S function where the S function = tanh(ln(1+t²))

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

The generating function for the Koenig's function expansion 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:

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

where $s$ is a complex variable.

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

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

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

$$\mathcal{M}_{A}(s) = \frac{1}{\sqrt{3}} \int_0^{\infty} \frac{(u/2)^{s-1}}{e^u-1} \cdot 2 du$$

which simplifies to:

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

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:

$$\mathcal{M}{A}(s) = \frac{1}{\sqrt{3}} \sum{n=1}^{\infty} \int_0^{\infty} u^{s-1} e^{-nu} \frac{1}{2^s} du = \frac{1}{\sqrt{3}} \Gamma(s) \sum_{n=1}^{\infty} \frac{1}{n^s} \frac{1}{2^s}$$

where $\Gamma(s)$ is the gamma function. This result includes the Riemann zeta function $\zeta(s)$.

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

$$\mathcal{M}_{A}(s) = \frac{\Gamma(s) \zeta(s)}{2^s\sqrt{3}}$$