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

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?

• Maybe you could try the obvious substitution and look if you can find something there. Take $x = \sqrt{1+z^2}$. – Karl Oct 8 '14 at 17:10

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 $$0.20769508925321053$$.

• Okay, Then how we obtain $\int_0^{\infty}[1−\tanh(coshx)] dx$ has that value? can you explain me more detail? – phy_math Oct 9 '14 at 4:56
• @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$)$. – Lucian Oct 9 '14 at 5:29
• Thanks for your comment @Lucian, i understood – phy_math Oct 9 '14 at 11:07

i think it is not possible to find an antiderivative in the known elementary functions

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

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

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

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

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

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

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)

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