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How to calculate the following integral

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

Is there any ways to calculate those integral in analytic? (Is $[0,\infty]$, case the integral is possible?)

How about using numerical method? Is there are good numerical scheme to complete this integral?

From the answer by @Lucian, The integral of $\displaystyle\int_0^\infty\bigg[1-\tanh(\cosh x)\bigg]~dx$ is converges.

How one can evaluate this integral?

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    $\begingroup$ Maybe you could try the obvious substitution and look if you can find something there. Take $x = \sqrt{1+z^2}$. $\endgroup$ – Karl Oct 8 '14 at 17:10
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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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  • $\begingroup$ Okay, Then how we obtain $\int_0^{\infty}[1−\tanh(coshx)] dx$ has that value? can you explain me more detail? $\endgroup$ – phy_math Oct 9 '14 at 4:56
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    $\begingroup$ @phy_math: $\tanh(\cosh x)$ approaches $1$ at an exponential rate: we can see this from the definition of the two functions, so the integral definitely converges, and it does so rather rapidly. $($Parents are moving furniture around, gotta go$)$. $\endgroup$ – Lucian Oct 9 '14 at 5:29
  • $\begingroup$ Thanks for your comment @Lucian, i understood $\endgroup$ – phy_math Oct 9 '14 at 11:07
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i think it is not possible to find an antiderivative in the known elementary functions

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For $\int_0^\infty(1-\tanh\cosh x)~dx$ ,

Similar to How to evaluate $\int_{0}^\infty \frac{{x^2}}{e^{\beta {\big(\sqrt{x^2 + m^2}}- \nu\big)} + 1} dx$,

$\int_0^\infty(1-\tanh\cosh x)~dx$

$=\int_0^\infty\left(1-\dfrac{1-e^{-2\cosh x}}{1+e^{-2\cosh x}}\right)~dx$

$=\int_0^\infty\dfrac{2e^{-2\cosh x}}{1+e^{-2\cosh x}}~dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty2(-1)^ne^{-2(n+1)\cosh x}~dx$

$=\sum\limits_{n=0}^\infty2(-1)^nK_0(2(n+1))$

For $\int\dfrac{\tanh\sqrt{1+z^2}}{\sqrt{1+z^2}}~dz$ ,

$\int\dfrac{\tanh\sqrt{1+z^2}}{\sqrt{1+z^2}}~dz$

$=\int\dfrac{1-e^{-2\sqrt{1+z^2}}}{(1+e^{-2\sqrt{1+z^2}})\sqrt{1+z^2}}~dz$

$=\int\dfrac{1}{\sqrt{1+z^2}}~dz-\int\dfrac{2e^{-2\sqrt{1+z^2}}}{(1+e^{-2\sqrt{1+z^2}})\sqrt{1+z^2}}~dz$

$=\sinh^{-1}z+\int\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n+1}e^{-2(n+1)\sqrt{1+z^2}}}{\sqrt{1+z^2}}~dz$

$=\sinh^{-1}z+\int\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n+1}e^{-2(n+1)\sqrt{1+\sinh^2u}}}{\sqrt{1+\sinh^2u}}~d(\sinh u)$ $(\text{Let}~z=\sinh u)$

$=\sinh^{-1}z+\int\sum\limits_{n=0}^\infty2(-1)^{n+1}e^{-2(n+1)\cosh u}~du$

$=\sinh^{-1}z+\sum\limits_{n=0}^\infty2(-1)^{n+1}J(2(n+1),0,u)+C$ (according to https://www.cambridge.org/core/services/aop-cambridge-core/content/view/9C572E5CE44E9E0DE8630755DF99ABAC/S0013091505000490a.pdf/incomplete-bessel-functions-i.pdf)

$=\sinh^{-1}z+\sum\limits_{n=0}^\infty2(-1)^{n+1}J(2(n+1),0,\sinh^{-1}z)+C$

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