Let $\theta = \tan^{-1}(t)$. Nowadays it is taught:

1º that $$ \frac{d\theta}{dt} = \frac 1{dt\,/\,d\theta} = \frac 1{1+t^2}, \tag1 $$

2º that, via the fundamental theorem of calculus, this is equivalent to $$ \theta = \int_0^t\frac{du}{1+u^2}, \tag2 $$

3º that, expanding the integrand in a geometric series and integrating term by term, this becomes the ~~Nilakantha ~~Madhava-Gregory-Leibniz formula
$$
\theta = t - \frac{t^3}3+\frac{t^5}5-\frac{t^7}7+\dots.
\tag3
$$

Question:Who first proved $(1)$ in print as we do, by deriving an inverse function?

I can’t find it in Nilakantha:

According to Ranjan Roy (1990, p.300), Nilakantha first published $(3)$ without proof in his *Tantrasangraha* (1501); a later commentary known as *Yuktibhasa* contains a proof by rectification of an arc of circle, which is beautiful but certainly not quite the same as $(1)$.

I can’t find it in Gregory:

According to Dehn & Hellinger (1943, p.149), Gregory communicated $(3)$ in a 1671 letter to Collins, and never published a proof; speculation exists that he found it by deriving $\tan^{-1}$ enough times to figure out its Taylor series at $0$, but in any event, he left no trace of how he may have computed these derivatives.

I can’t find it in Leibniz:

According to González-Velasco (2011, p.347), Leibniz communicated $(3)$ in 1674 letters to Oldenburg and Huygens, and later published the case $t=1$ in *Acta Eruditorum* (1682, pp.41-46); his unpublished proof is available (many times over) in the 700+ pages of his Collected Works, Vol. VII,6. There, or in the nice exposition given in Hairer & Wanner (1996, 2nd printing, pp.49-50), one sees that he was squaring the circle in an elaborate way which has nothing to do with $(1)$.

Of course Leibniz must have become aware of $(1)$ and $(2)$ *at some point*, as (later!) inventor of the notation that makes them almost automatic. Unfortunately, I can’t find any written evidence of that. Maybe someone else will have better luck!

(The closest I can find is a 1707 letter of Wolff to Leibniz, where the new notation is used to write in effect that $d\theta = dt:(1+t^2)$, and then deduce $(2)$ and $(3)$. The two correspondents may well have had in mind the modern proof $(1)$ of this differential relation, but neither says so.)

I can’t find it in Jacob Bernoulli:

With Leibniz notation spreading, one might think that a disciple would write $(1)$ at the first opportunity. But that’s not what Jacob B. does to rectify a unit circle in *Positionum de Seriebus Infinitis Pars Tertia* (Basel, 1696, Prop. XLV): instead, he parametrizes one with $(x,y)=$ $\bigl(x,\sqrt{2x-x^2}\bigr)$ and then expresses the resulting arc length differential — also seen in Leibniz (1686) — as
$$
d\theta
=\sqrt{\smash{dx^2+dy^2}\vphantom{a^2}}
=\frac{dx}{\sqrt{2x-x^2}}
=\frac{2d\mathsf t}{1+\mathsf t^2}
\tag{$*$}
$$
$(=\mathrm{LH}$ on his Fig. 3$)$ by introducing a “diophantine” (a.k.a. Weierstraß) substitution $\smash{\mathsf t=\frac xy}$ $=\smash{\tan\frac\theta2}$ $(=\mathrm{BI}$ on the figure, as he notes; so his $\mathsf t=\tan\mathrm{BAI}$ is not our $t=\tan\mathrm{BAH})$. While this still proves $(2)$ and $(3)$ for the halved angle and its tangent, the argument definitely isn’t $(1)$.

$\hspace{8.5em}$

I can’t find it in Johann Bernoulli:

When faced with the task of integrating $\smash{\frac{dt}{1+t^2}}$ in his paper on rational integrals (1702), Johann B. proposes two substitutions:

The first (in Probl. I, Corol.) comes from the partial fraction decomposition $\frac1{1+t^2}=\frac{1/2}{1+it} + \frac{1/2}{1-it}$, and consists in putting $u = \frac{1+it}{1-it}$ so that $ \frac{dt}{1+t^2}=\frac{du}{2iu}=d\left[\frac1{2i}\log\frac{1+it}{1-it}\right] $.

The second (in Probl. II) consists in putting $u=\frac1{1+t^2}$ so that $\frac{dt}{1+t^2} = \frac{-du}{\sqrt{4(u-u^2)}}$, which he knows (perhaps by recognizing half $(*)$ with $x=2(1-u)$?) is a “circular sector or arc differential”.

Neither of these is the substitution $\theta = \tan^{-1}(t)$, which *via* $(1)$ would have led directly to $\frac{dt}{1+t^2} = d\theta$. And in later papers (1712, 1719) Bernoulli is content to describe this relation as “well-known”.

I can’t find it in de Moivre:

Schneider (1968, footnotes 248 & 250) seems to claim that a 1708 letter of de Moivre to Bernoulli has the proof $(1)$ and also the “Euler” formula $\theta=\smash{\frac1i}\log(\cos\theta+i\sin\theta)$. But this is not borne out by the letter’s text in (1931, pp.241-257): there, as also in his paper (1703, p.1124) and book (1730, p.44), de Moivre simply states $\smash{d\theta=\frac{dt}{1+t^2}}$ without proving it anew.

I can’t find it in Euler:

Euler was of course well aware of $(2)$, which appears for instance in his fifth paper (1729, pp.93, 95) and in his later E60, 65, 66, 125, 129, 130, 162, 217, 391, 475, 482, etc. But when it comes to *proving* $(2)$ or $(3)$, then again he seems to eschew $(1)$:

In his precalculus book (1748, §§139-140) he chooses to first establish Bernoulli’s above formula $$ \theta = \frac1{2i}\log\frac{1+it}{1-it} \tag4 $$ (this he does by multiplying numerator and denominator by $\cos\theta$ so they become $e^{\pm i\theta}$), and then to deduce $(3)$ by plugging $it$ in the series for $\smash{\log\frac{1+x}{1-x}}$. None of this requires $(1)$, $(2)$, or any calculus.

In his differential calculus book (1755, §§194-197) he first differentiates a similar logarithmic formula for $\theta = \sin^{-1}(s)$, namely $\theta = -i\log(\sqrt{1-s^2} + is)$, to obtain $$ d\theta = \frac{ds}{\sqrt{1-s^2}}. \tag5 $$ Plugging $s = t/\sqrt{1+t^2}$ into $(5)$ then gives him $(2)$. He might as well have differentiated $(4)$ directly! Either way, $(1)$ is not used, although to be fair, Euler at least gives (§195) an alternative proof of $(5)$ using

**the argument**$(1)$, but applied to $\smash{\sin^{-1}}$ instead of $\smash{\tan^{-1}}$.

So where can you find it?

It’s in Lacroix (1797, pp.113-114) and its progeny. Still I have trouble believing it took over 100 years for $(1)$ to become the standard proof — hence my question.

De Analysi(written c. 1669, published 1711, Opera Omnia I, p. 264) obtains the value (3) of the integral (2) by method 3º. I see no mention of (1), which is what the present question is about. $\endgroup$ – Francois Ziegler Jun 6 '17 at 7:53