Questions tagged [hyperbolic-functions]

For questions related to hyperbolic functions: $\sinh$, $\cosh$, $\tanh$, and so on.

The hyperbolic functions are analogs of the usual trigonometric functions; we may define such functions as the hyperbolic sine

$$\sinh{x} = \frac{e^x - e^{-x}}{2}$$

and the hyperbolic cosine

$$\cosh{x} = \frac{e^x + e^{-x}}{2}$$

as well as the hyperbolic tangent

$$\tanh{x} = \frac{\sinh{x}}{\cosh{x}}=\frac{e^x - e^{-x}}{e^x + e^{-x}}$$

These functions are differentiable. More precisely, $\cosh'=\sinh$, $\sinh'=\cosh$, and $\tanh'=1-\tanh^2$.

Just as the point $(\cos{t}, \sin{t})$ describes a point on the unit circle $x^2 + y^2 = 1$, the point $(\cosh{t}, \sinh{t})$ defines a point on the unit hyperbola $x^2 - y^2 = 1$.

Reference: Hyperbolic function.

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Is $1 / \sinh (4x)$ equal to $\operatorname{csch} (4x)$?

I do know that $\operatorname{csch}(x) = \dfrac 1 { \sinh(x)}$, but I'm not sure if it applies to $x$ only. I don't know if it's applicable for $4x$ as well, or any other monomial.
Lowell0803
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$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx$

Is there any closed-form representation for the following integral? $$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx,$$ where $\mathrm{sech}\,x$ is the hyperbolic secant, $\mathrm{sech}\,x=\dfrac{2}{e^x+e^{-x}}$.
Marty Colos
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Why is $\frac{7 \cosh(\sqrt 6)}{13}$ near $\pi$?

$\frac{7 \cosh(\sqrt 6)}{13} = 3.1415926822 ...$ $\pi = 3.1415926535 ... $ Why are these numbers so close to each other? Is this just a coincidence? P.S. About ten days ago, I saw this question on twitter, but no one has answered it.
TOM
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Prove $\int_0^1 \frac{\tanh^{-1} (\beta t) dt}{t\sqrt{(1-t)(1- \alpha t)}}=\log (a) \log (b)$

If we set: $$\alpha= \frac{(ab-1)^2+(a-b)^2}{(ab+1)^2+(a+b)^2}$$ $$\beta= \frac{(ab+1)^2-(a+b)^2}{(ab+1)^2+(a+b)^2}$$ Then it follows that: $$\int_0^1 \frac{\tanh^{-1} (\beta t) dt}{t\sqrt{(1-t)(1- \alpha t)}}=\log (a) \log (b)$$ I have derived…
Yuriy S
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Series $\sum\limits_{n=1}^\infty \frac{1}{\cosh(\pi n)}= \frac{1}{2} \left(\frac{\sqrt{\pi}}{\Gamma^2 \left( \frac{3}{4}\right)}-1\right)$

I was playing around with Mathematica and found that $$\sum_{n=1}^\infty\frac1{\cosh(\pi n)} = \frac12\left(\frac{\sqrt{\pi}}{\Gamma \left(\tfrac34\right)^2}-1\right)$$ Does anybody know how to prove this manually?
Shobhit Bhatnagar
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Need help with $\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$

Please help me to evaluate this integral $$\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$$ I tried a change of variable $x=\tanh z$, that transforms it into the form $$\int_0^\infty\frac{4z}{\left(1+\log^2\tanh…
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Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$

While I was working on this question by @Vladimir Reshetnikov, I've conjectured the following closed-forms. $$ I_0(n)=\int_0^\infty \frac{1}{\left(\cosh x\right)^{1/n}} \, dx \stackrel{?}{=} \frac{\sqrt{\pi}}{2}…
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Geometric construction of hyperbolic trigonometric functions

If we have a circle we can geometrically construct the trigonometric functions as shown. The functions all derive from sin and cos. If we say that the circle is a conic section and imagine it on the cone we can draw a hyperbola perpendicular to it.…
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How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$?

I was recently searching for interesting looking integrals. In my search, I came upon the following result: $$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$$ and I wanted to…
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Integral (Tanh and Normal)

I am trying to evaluate the following: The expectation of the hyperbolic tangent of an arbitrary normal random variable. $\mathbb{E}[\mathrm{tanh}(\phi)]; \phi \sim N(\mu, \sigma^2)$ Equivalently: $\int_{-\infty}^{\infty}…
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Closed form for ${\large\int}_0^\infty\frac{x\,\sqrt{e^x-1}}{1-2\cosh x}\,dx$

I was able to calculate $$\int_0^\infty\frac{\sqrt{e^x-1}}{1-2\cosh x}\,dx=-\frac\pi{\sqrt3}.$$ It turns out the integrand even has an elementary antiderivative (see here). Now I'm interested in a similar…
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Geometric interpretation of hyperbolic functions

When proving identities like $$\cosh(2x)=\cosh^2(x)+\sinh^2(x)$$ $$\cosh^2(x)=\sinh^2(x)+1$$ algebraically, I am beset by the feeling that there should be a geometrical interpretation that makes them immediately obvious. Most of the analogous…
Meow
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Prove that $\sinh(\cosh(x)) \geq \cosh(\sinh(x))$

Prove that $$\sinh(\cosh(x)) \geq \cosh(\sinh(x))$$ I tried to tackle this problem by integrating both lhs and rhs, in order to get two functions who show clearly that inequality holds. I've struggled for this problem a little bit, i don't know if…
james42
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Geometric definitions of hyperbolic functions

I've learned in school that all the trigonometric functions can be constructed geometrically in terms of a unit circle: Can the hyperbolic functions be constructed geometrically as well? I know that $\sinh$ and $\cosh$ can be constructed based on…
st12
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Showing $\int_0^{\int_0^u{\rm sech}vdv}\sec vdv\equiv u$ and $\int_0^{\int_0^u\sec vdv}{\rm sech} vdv\equiv u$

The two following very weird-looking theorems $$\int_0^{\int_0^u\operatorname{sech}\upsilon d\upsilon}\sec\upsilon d\upsilon \equiv u$$ $$\int_0^{\int_0^u\sec\upsilon d\upsilon}\operatorname{sech}\upsilon d\upsilon \equiv u$$ are simple consequences…
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