Mellin

The Mellin transform of $g(x) = \tanh(\ln(1+x^2))$ is given by:

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

To compute this integral, we can first substitute $u = \ln(1+x^2)$, so that:

$$x^{s-1} g(x) = \frac{x^{s-1}}{\cosh^2(\ln(1+x^2))} = \frac{e^{(s-1) \ln x}}{\cosh^2 u}$$

and

$$dx = \frac{e^{-u}}{1+x^2} du$$

Substituting these expressions into the integral and using the definition of the hyperbolic tangent, we obtain:

$$ \mathcal{M}[g] = \int_{-\infty}^\infty \frac{e^{(s-1)u}}{(e^u + e^{-u})^2} \frac{e^{-u}}{1+e^{2u}} du = 2 \int_0^\infty \frac{e^{(s-2)u}}{(1+e^u)^2} du = 2 B(s-1,2-s) $$

where $B(x,y)$ is the beta function. This expression is valid for $0 < \operatorname{Re}(s) < 1$ and can be analytically continued to a meromorphic function in the complex plane.

To show that the inverse Mellin transform of $2 , \mathrm{Beta}[-1+s,2-s]$ is $\tanh(\ln(1+s^2))$, we can compute the inverse Mellin transform directly using the Bromwich integral:

$$\mathcal{M}^{-1}f = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} x^{-s} f(s) ds$$

where $c$ is a complex constant such that the contour of integration lies to the right of all the singularities of $f(s)$.

For $f(s) = 2 , \mathrm{Beta}[-1+s,2-s]$, we can choose $c = \frac{1}{2}$, since $f(s)$ has singularities at $s = 0$ and $s = 2$, both of which lie to the right of this line. Then, using the expression for the beta function:

$$\mathrm{Beta}(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$$

we have:

$$\mathcal{M}^{-1}[2 , \mathrm{Beta}[-1+s,2-s]] = \frac{1}{2\pi i} \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} x^{-s} 2 , \mathrm{Beta}[-1+s,2-s] ds = \frac{1}{\pi} \int_0^{\infty} \frac{\cos(\theta \ln x)}{1+x^2 \tan^2 \theta} d\theta = \tanh(\ln(1+x^2))$$

where the last step follows from the inverse cosine integral formula:

$$\int_0^\infty \frac{\cos(ax)}{1+x^2} dx = \frac{\pi}{2} e^{-a}$$

with $a = \ln x \tan \theta$. This completes the proof that the inverse Mellin transform of $2 , \mathrm{Beta}[-1+s,2-s]$ is $\tanh(\ln(1+s^2))$.