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# Hyperbolic Integral

We will prove:

$$\forall$$ $$a$$,$$b$$ $$\in \mathbb{R}$$:

$$\displaystyle \int_{a}^b \sqrt{1+(\frac{d}{dx} \cosh(x))^2}\,dx = \displaystyle \int_{a}^b \cosh(x) \,dx$$

In words: "The arc length of $$\cosh(x)$$ in a finite interval is always equal to the area under the curve in the same interval."

Proof:

To be as complete as possible, we will prove two things first, and the reader will see that the proof for the above will follow directly:

(1) $$\cosh(x) = \cos(ix)$$

(2) $$\sinh(x) = -i\sin(ix)$$

Proof for (1):

$$\cosh(x)$$ is defined:

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

Let us present an equivalent formulation:

$$\cosh(x) = \frac{ e^{i^4x} + e^{i^2x}}{2}$$

Then by Euler's Formula, we have:

$$\cosh(x) = \frac{1}{2} \left[\cos(i^3x) +i\sin(i^3x) +\cos(ix) + i\sin(ix) \right]$$

$$\Rightarrow$$ $$\cosh(x) = \frac{1}{2} \left[ \cos(-ix) + i\sin(-ix) +\cos(ix) +i\sin(ix)\right]$$

$$\Rightarrow$$ $$\cosh(x) = \frac{1}{2} \left[ 2\cos(ix) \right] = \cos(ix)$$

This proves (1)

Proof for (2):

Begin with Euler's Formula with a slight modification:

$$e^{i(\phi i)} = \cos(\phi i) + i\sin(\phi i )$$

$$\Rightarrow$$ $$e^{-\phi} -\cos(\phi i) = i\sin(\phi i)$$

$$\Rightarrow$$ $$\cos(\phi i) - e^{- \phi} =-i\sin(\phi i)$$

By (1):

$$\Rightarrow$$ $$\cosh(\phi) - e^{- \phi} =-i\sin(\phi i)$$

$$\Rightarrow$$ $$\frac{e^{\phi} + e^{-\phi}}{2} - e^{-\phi}=-i\sin(\phi i)$$

$$\Rightarrow$$ $$\frac{e^{\phi} - e^{-\phi}}{2} = -i\sin(\phi i)$$

$$\Rightarrow$$ $$\sinh(\phi) = -i\sin(\phi i)$$

This proves (2).

Now, begin with the arc length integral:

$$\displaystyle \int_{a}^b \sqrt{1+(\frac{d}{dx} \cosh(x))^2}\,dx = \displaystyle \int_{a}^b \sqrt{1+\sinh^2(x)}\,dx$$

By (2):

$$= \displaystyle \int_{a}^b \sqrt{1+(-i\sin(ix))^2}\,dx =\displaystyle \int_{a}^b \sqrt{1-\sin^2(ix)}\,dx$$

$$=\displaystyle \int_{a}^b \sqrt{\cos^2(ix)}\,dx =\displaystyle \int_{a}^b \cos(ix)\,dx$$

And finally, by (1):

$$=\displaystyle \int_{a}^b \cosh(x)\,dx$$

QED

Note by Ethan Robinett
3 years, 2 months ago

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This is one of the interesting properties of the catenary curve, as the only plane curve besides a horizontal line to have this property, i.e., its arc length is proportional to the area under it.

- 3 years, 1 month ago