Showing $\int_0^{\int_0^u{\rm sech}vdv}\sec vdv\equiv u$ and $\int_0^{\int_0^u\sec vdv}{\rm sech} vdv\equiv u$

Question

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.

Answer 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.

Answer 2

I've managed to arrive at some degree of generalisation of this - infact through that observation that $\operatorname{asech}$ & $\operatorname{asec}$ are both integrals of 'complementary' functions - complementary in the sense of having occurences of $1-x$ replaced with $x-1$ as the point $x=1$ is traversed. An observation that has arisen through this little exercise is that say we have a function $\operatorname{f}$, and we integrate it to get the primitive of $\operatorname{f}$, say $\operatorname{g}$, and then we take the inverse of that, say $\operatorname{ag}$, and integrate that, it is actually equivalent to integrating $\operatorname{f}$ multiplied by the ordinate. This is easy to see on a graph: the integral just described is the area enclosed under the curve cut-off by the horizontal line joining (0, $\operatorname{f}(x)$) to ($x$, $\operatorname{f}(x)$), which is

$$\int x.d(g(x)) = \int xf(x).dx .$$

. In this particular case and considering $0\leq x\leq 1$, we have

$$\int_1^x\frac{d\upsilon}{\upsilon\sqrt{1-\upsilon^2}}=\operatorname{asech}(x), $$

and

$$\int_1^x\frac{d\upsilon}{\sqrt{1-\upsilon^2}}=\operatorname{acos}(x), $$

and (using $y$, to emphasise that we are switching to viewlng the integration along the $y$ axis)

$$\int_0^y\operatorname{sech}\upsilon.d\upsilon=\operatorname{acos}x=\operatorname{acos}(\operatorname{sech}y) ,$$

which is indeed one of the formulæ arrived at in the main body of the question.

Likewise for $x\geq 1$ we have

$$\int_1^x\frac{d\upsilon}{\upsilon\sqrt{\upsilon^2-1}}=\operatorname{asec}(x), $$

and

$$\int_1^x\frac{d\upsilon}{\sqrt{\upsilon^2-1}}=\operatorname{acosh}(x), $$

and (using $y$, to emphasise ... again)

$$\int_0^y\operatorname{sec}\upsilon.d\upsilon=\operatorname{acosh}x=\operatorname{acosh}(\operatorname{sec}y) ,$$

Which is the corresponding formula arrived at in the main body.

It can be seen with a little consideration that the being inverses of each other of the two integrals in $y$ hinges on its being so that the functions

$$\operatorname{acosh} \& \operatorname{asech}$$

are functions $\operatorname{\phi} \& \operatorname{\psi}$ such that

$$\operatorname{\phi}(x)\equiv\operatorname{\psi}(1/x) ;$$

and likewise for

$$\operatorname{acos} \& \operatorname{asec} .$$

Translating this into terms of the origin of these functions as integrals of

$$\operatorname{f_l}(x)\equiv\frac{1}{x\sqrt{1-x^2}} \&$$

$$\operatorname{f_r}(x)\equiv\frac{1}{x\sqrt{x^2-1}}, $$

(l=left & b=right), that which is what it hinges on is that

$$\operatorname{f_a}(x).dx\equiv (1/x)\operatorname{f_b}(1/x) \&$$

$$\operatorname{f_a}(1/x).d(1/x)\equiv x\operatorname{f_b}(x), $$

where a=l or r & b=r or l. This is very fiddly to put into symbols: but basically what it hinges on is that if $1/x$ be substituted for $x$ in $\operatorname{f_l}.dx$ then $x.\operatorname{f_r}.dx$ should be obtained; if $1/x$ be substituted for $x$ in $x.\operatorname{f_l}.dx$ then $\operatorname{f_r}.dx$ should be obtained; if $1/x$ be substituted for $x$ in $\operatorname{f_r}.dx$ then $x.\operatorname{f_l}.dx$ should be obtained; if $1/x$ be substituted for $x$ in $x.\operatorname{f_r}.dx$ then $\operatorname{f_l}.dx$ should be obtained: basically just thorough reciprocity under those transformations.

It can also be seen that this happens if the function $\operatorname{f}$ is of the form reciprocal of ($x$×function that changes $1-x$ into $x-1$ or $x-1$ into $1-x$ when $1/x$ is substituted for $x$ and it is multiplied by $x$ from $d(1/x)=-dx/x^2$). As I said, this is terribly fiddly to put into words & symbols; but the underlying idea is really quite an elementary one.

So we could construct pairs of functions of the following form:

$$\frac{1}{x((1-x)^k.(1+x)^{n-k})^{1/n}} $$

&

$$\frac{1}{x((x-1)^k.(1+x)^{n-k})^{1/n}} ,$$ where $n$ is a natural number, and $k$ is one of the Euler totient set of $n$; and the functions obtained by taking the primitives of these ought to have this reciprocal property we have been discussing of the integrals of their inverses being mutual inverses.

It actually works beautifully for the case n=1, even though that results in integrals that have infinities; and it's clear that it works for n=2, as that is the case that prompted this post in the first place. As for higher values of n ... anything > 2 results in thoroughly diabolickal hypergeometric functions that have who-knows-what inverses, and I shall have to feed them into some kind of mathematics package to test them for this property. It's actually not so bad though for mutual inverses (a bit like it's not so bad testing whether two numbers are coprime): you don't have to actually compute the inverses - you can just feed one into the other as an argument & see whether the identity function drops out.

For n=1, the pair is

$$-\ln(e^{-y}+1)$$

&

$$-\ln(e^{-y}-1) .$$

The former is from

$$\frac{-1}{x(1-x)}$$

... integral ...

$$\ln\frac{1-x}{x}$$

... inverse ...

$$\frac{1}{1+e^y}$$

... integral ...

$$-\ln(e^{-y}+1) ;$$

& the latter from

$$\frac{1}{x(x-1)}$$

... integral ...

$$\ln\frac{x-1}{x}$$

... inverse ...

$$\frac{1}{1-e^y}$$

... integral ...

$$-\ln(e^{-y}-1) .$$

This isn't as much of a generalisation as I was hoping for; but it does at least exhibit the original pair of functions being queried as being part of some kind of pattern, and not just an isolated special case that just happens to have that property because it does!

The functions gotten in $n=1$ régime are worthy of some comment, I think, on various grounds. For one thing, it will be observed that in this case $\operatorname{f_l} \& \operatorname{f_r}$ are not actually different atall! However ... the integrals are different functions. And also we get infinities in this case, but fortunately it doesn't foil the reasoning. If we look at the whole matter graphically, we see that in our familiar $n=2$ case, the two integrals $\operatorname{g_l} \& \operatorname{g_r}$ spring from the point (1,0) square-root-wise; whereas the functions under $n=1$ proceed asymptotically from (1, -∞); and for $n>1$ they will proceed from the point (1,0) with behaviour tending to linearity - ^1-1/n, to be more precise - so the vertical & then curving down section will get progressively tighter with increasing $n$. This brings to the next point: the functions for $n=1$ positively abound in negativity; normally I would want to translate the whole affair into a region better-populated by positive numbers ... but I think in this case they are better left as they are, as they better-exhibit that continuity & progression of behaviour just described. It applies also to the functions that constitute the final result: like any bona-fide self-respecting mutual inverses they ought to be mutual refections in the line $x=y$; in the $n=2$ case they spring from the origin & 'splay-out' to become asypmtotic one to the line $x=\pi/2$ the other to the line $y=\pi/2$, whereas in $n=1$ case they approach asymptotically from (-∞, -∞) along the negative branch of the identity-function graph, and splay-out to become asymptotic, one to the $x$-axis, the other to the $y$, on the negative faces of them: this might incur that most of the affair is in terms of negative numbers ... but it the better shows that continuity of evolution of behaviour with increasing $n$.

Images for Case $n=2$ (the original subject of the post)

The upper diagram shows the functions $y=\operatorname{asech}x$ & $y=\operatorname{asec}x$, and the lower the integrals of these along the $y$-axis $x=\operatorname{acos}\operatorname{sech}y$ & $x=\operatorname{acosh}\operatorname{sec}y$.

Images for Case $n=1$

The upper diagram shows the functions $y=\operatorname{ln}((1-x)/x)$ & $y=\operatorname{ln}((x-1)/x)$, and the lower the integrals of these along the $y$-axis $x=-\operatorname{ln}(e^{-y}+1)$ & $x=-\operatorname{ln}(e^{-y}-1)$.