About Integration $ \int \frac{\tanh(\sqrt{1+z^2})}{\sqrt{1+z^2}}dz $

Is there any ways to calculate those integral in analytic?

No. Letting $x=\sinh t$, we have $~I=\displaystyle\int\tanh(\cosh x)~dx$, which cannot be expressed in terms of elementary functions. In fact, it cannot even be expressed in terms of special Bessel functions.

Is $[0,\infty]$, case the integral is possible?

No. The integral diverges, since the numerator $\simeq1$, and the denominator $\simeq x$. However,

$\displaystyle\int_0^\infty\bigg[1-\tanh(\cosh x)\bigg]~dx$ does converge to a value of about $0.20769508925321053$.

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Integration

Improper Integrals

Indefinite Integrals

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