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 of the rather remarkable (in my opinion) fact that

$$\int_0^u\operatorname{sech}\upsilon d\upsilon$$

$$\int_0^u\operatorname{sec}\upsilon d\upsilon$$

are inverse functions of each other (specifically

$$2\operatorname{atn}\exp u-\frac{\pi}{2}$$&$$\ln\tan(\frac{u}{2}+\frac{\pi}{4})$$

respectively, or $\operatorname{gd}^{-1}u$ and $\operatorname{gd}u$ if the convention of using "$\operatorname{gd}$" to stand for gudermannian be received).

The relations between integrals of circular and hyperbolic functions, and what an extraordinarily tight 'system' they all seem to form is a constant source of fascination to me - and the particular relation cited here is probably the strangest of all, to my mind.

It's probably too much to expect that a group-under-composition of functions could be formed out of them, or anything quite so neat ... but anyway:

The "weird theorem" (really just one theorem, of course) that is the nominal subject of this post: can it be proven directly from the properties of hyperbolic and circular functions and their relations amongst each other, rather than by the crude expedient of just evaluating the integrals and simply exhibiting them as mutual inverses?

Update

I've just been brewing some thoughts, and I can't help thinking now that it might have something to do with the fact that

$$\int\frac{dy}{y\sqrt{1-y^2}}=\operatorname{asech}{y}$$

&

$$\int\frac{dy}{y\sqrt{y^2-1}}=\operatorname{asec}{y} .$$

(I tend to use "y" in these kinds of integral, as when I was shown them the very first time it was by differentiating $y$ = circular or hyperbolic function of $x$ & expressing the result in terms of $y$; & I've just never gotten out of the habit.)

@ Michael Hoppe

Let's see ... they are effectively compositions so

$$\operatorname{sech}u.\operatorname{sec}\int_0^u\operatorname{sech}\upsilon d\upsilon\equiv 1$$

$$\operatorname{sec}u.\operatorname{sech}\int_0^u\operatorname{sec}\upsilon d\upsilon\equiv 1 ... $$

I think that's what they would be in differentiated form. Is there any lead in that?

$$\operatorname{sec}\int_0^u\operatorname{sech}\upsilon d\upsilon\equiv \operatorname{cosh}u \cdot\cdot\cdot\cdot\cdot\cdot(\operatorname{i})$$

$$\operatorname{sech}\int_0^u\operatorname{sec}\upsilon d\upsilon\equiv \operatorname{cos}u\cdot\cdot\cdot\cdot\cdot\cdot(\operatorname{ii})$$

they certainly look less formidable in that form.

Manipulating

$$\operatorname{sech}{u} + i \operatorname{tanh}{u} = \exp{\left [i \arctan{\left ( \sinh{u} \right )} \right ]}$$

a little, we get

$$\operatorname{sech}u =\operatorname{cos}\operatorname{atn}\operatorname{sinh}u\cdot\cdot\cdot\cdot\cdot\cdot(\operatorname{iii})$$

$$\operatorname{sec}u =\operatorname{cosh}\operatorname{asinh}\operatorname{tan}u\cdot\cdot\cdot\cdot\cdot\cdot(\operatorname{iv})$$

Replacing $\operatorname{sec}()$ in (i) with its identity in (iv), & $\operatorname{sech}()$ in (ii) with its identity in (iii), we get

$$\int_0^u\operatorname{sech}\upsilon d\upsilon\equiv \operatorname{atn}\operatorname{sinh}u$$

$$\int_0^u\operatorname{sec}\upsilon d\upsilon\equiv \operatorname{asinh}\operatorname{tan}u$$

Looks a bit like going round in circles. No, it's not quite that - it's more like tracing the threads allover the place & observing how wonderfully they join up end-to-end, no matter how crazy an excursion they make. well it does show that the functions in question are indeed inverses - but - we have still done it by solving the integrals ... really. Is this an improvement? I'm not sure. But I have certainly tried to incorporate the various advice that the contibutors have most graciously dispensed. And we've gotten another rather curioferous theorem into the bargain.

$$\operatorname{sech}{u} + i \operatorname{tanh}{u} = \exp{\left [i\int_0^u\operatorname{sech}{\upsilon.d\upsilon} \right ]}$$

$$\operatorname{sech}{u}= \cos{\left [\int_0^u\operatorname{sech}{\upsilon.d\upsilon} \right ]}$$

$$\operatorname{sec}{u}= \cosh{\left [\int_0^u\operatorname{sec}{\upsilon.d\upsilon} \right ]}$$

And it's also quite possible that I've missed the point or lost the plot somewhere along the line!

Nor have I forgotten my line of thought either - the way that if you integrate

$$\frac{dy}{y\sqrt{1-y^2}}$$

and the range of integration straddles the point $y=1$, it's beautifully incorporated by $1/i = -i$ going outside; and the $\operatorname{asec}$ function 'splices' onto the $\operatorname{asech}$ - as both have a √ singularity at that point - & turns through a right-angle.

Solution 1:

I've always loved the Gudermannian function. The least mystifying way to start any discussion of it is to note that the functions $f(t):=\frac{2t}{1-t^2},\,g(t):=\frac{1-t^2}{1+t^2}$ satisfy not only $f(\tan\tfrac{x}{2})=\sin x,\,g(\tan\tfrac{x}{2})=\cos x$, but also $f(\tanh\tfrac{x}{2})=\tanh x,\,g(\tanh\tfrac{x}{2})=\operatorname{sech}x$. From this it follows that the definition $\operatorname{gd}x:=2\arctan\tanh\frac{x}{2}$ satisfies results such as $\tan\tfrac{\operatorname{gd}x}{2}=\tanh\frac{x}{2},\,\sin \operatorname{gd}x=\tanh x$ etc.

Now for the integrals. Note that $\tfrac{\operatorname{d}}{\operatorname{d}x}\operatorname{gd}x=\frac{\operatorname{sech}^2\frac{x}{2}}{1+\tanh^2\frac{x}{2}}=\operatorname{sech}x$. But to go from the Gudermannian to its inverse, all I have to do is swap the two kinds of "tangent". So look what happens to the derivative: $$\tfrac{\operatorname{d}}{\operatorname{d}x}\operatorname{gd}^{-1}x=\frac{\sec^2\frac{x}{2}}{1-\tan^2\frac{x}{2}}=\sec x.$$You know that sign difference when you compare $1=\cos^2\tfrac{x}{2}+\sin^2\tfrac{x}{2}$ to $1=\cosh^2\tfrac{x}{2}-\sinh^2\tfrac{x}{2}$, or $\cos x=\cos^2\tfrac{x}{2}-\sin^2\tfrac{x}{2}$ to $\cosh x=\cosh^2\tfrac{x}{2}+\sinh^2\tfrac{x}{2}$? It perfectly balances the sign difference when comparing $y=\tan x\implies y'=1+y^2$ to $y=\tanh x\implies y'=1-y^2$, i.e. comparing $\tfrac{\operatorname{d}}{\operatorname{d}x}\tan x=\tfrac{1}{1+x^2}$ to $\tfrac{\operatorname{d}}{\operatorname{d}x}\tanh x=\tfrac{1}{1-x^2}$. In fact, it causes that result about derivatives.