$\arctan{x}+\arctan{y}$ from integration

Question

I was trying to derive the property $$\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}$$ for $x,y>0$ and $xy<1$ from the integral representation $$ \arctan{x}=\int_0^x\frac{dt}{1+t^2}\,. $$ I am aware of "more trigonometric" proofs, for instance using that $\tan{(\alpha+\beta)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$, but I was willing to see if there is a proof that uses more directly the properties of the integral representation. For instance, if $x>0$, one immediately gets $$\begin{aligned} \arctan{x}+\arctan\frac{1}{x} &=\int_0^x\frac{dt}{1+t^2} + \int_0^{\frac{1}{x}}\frac{dt}{1+t^2}\\ &= \int_0^x\frac{dt}{1+t^2}+\int_x^\infty\frac{dt}{1+t^2}\\ &=\int_0^\infty \frac{dt}{1+t^2} = \frac{\pi}{2} \end{aligned}$$ sending $t\to\frac{1}{t}$ in the second integral. Similarly I tried considering $$ \int_0^x\frac{dt}{1+t^2} + \int_0^y\frac{dt}{1+t^2}=(x+y)\int_0^1\frac{1+xyt^2}{1+(x^2+y^2)t^2+x^2y^2t^4}\ dt $$ after rescaling $t\to xt$ and $t\to yt$. On the other hand, via a similar rescaling $t\to \frac{x+y}{1-xy}t$, we have $$ \int_0^\frac{x+y}{1-xy}\frac{dt}{1+t^2} = (x+y)\int_0^1\frac{1-xy}{(1-xy)^2+(x+y)^2t^2}\ dt\,. $$ By a clever choice of variable it should (must?) be possible to see that these integrals are actually the same, but I can't figure it out...

Answer 1

We want show that \begin{eqnarray*} \int_x^{ \frac{x+y}{1-xy}} \frac{dt}{1+t^2} = \int_{0}^{y} \frac{du}{1+u^2} \end{eqnarray*} that's to say the LHS is actually independent of $x$.

The substitution \begin{eqnarray*} t=x+ \frac{u(1+x^2)}{1-ux} \end{eqnarray*} will do the trick.

The limits are easily checked and we have \begin{eqnarray*} dt= \frac{1+x^2}{(1-ux)^2} du. \end{eqnarray*} The rest is a little bit of algebra.

Note the similarity with $ \ln(a)+\ln(b) = \ln(ab)$ \begin{eqnarray*} \int_{1}^{a} \frac{dt}{t} +\int_{1}^{b} \frac{dt}{t} = \int_{1}^{ab} \frac{dt}{t}. \end{eqnarray*} And $ u=at $ \begin{eqnarray*} \int_{1}^{b} \frac{dt}{t} = \int_{a}^{ab} \frac{du}{u}. \end{eqnarray*}

Answer 2

Another change of variables that works, very similar to that in the answer by @DonaldSplutterwit, is the following: $$ t=f(u)=\frac{1-u\sigma}{u+\sigma}\,,\qquad\text{with}\ \ \sigma(x,y)=\frac{1-xy}{x+y}\,. $$ It is more symmetric, since it works both for $$ \int_x^{1/\sigma}\frac{dt}{1+t^2}=\int_0^y\frac{du}{1+u^2} $$ and for $$ \int_y^{1/\sigma}\frac{dt}{1+t^2}=\int_0^x\frac{du}{1+u^2}\,. $$ Indeed, $$ f(0)=\frac{1}{\sigma}\,,\qquad f(x)=y\,,\qquad f(y)=x $$ and $$ dt=-\frac{1+\sigma^2}{(u+\sigma)^2}du\,,\qquad \frac{1}{1+t^2}=\frac{(u+\sigma)^2}{(1+\sigma^2)(1+u^2)}\,. $$ It also has the property of reducing to the inversion as $xy\to1^-$, namely $\sigma\to0^+$, since $$ f(u)\big|_{\sigma=0}=\frac{1}{u}\,, $$ and we get back $$ \int_{x}^\infty \frac{dt}{1+t^2} = \int_{0}^{\frac{1}{x}}\frac{du}{1+u^2}\,. $$ In fact, $f$ is also an involution $f(f(u))=u$ and also allows to run the proof "forward" in the following way $$ \int_0^x\frac{dt}{1+t^2}+\int_0^y\frac{dt}{1+t^2}=\int_0^x\frac{dt}{1+t^2}+\int_x^{\frac{1}{\sigma}}\frac{du}{1+u^2}=\int_0^{\frac{1}{\sigma}}\frac{dt}{1+t^2}\,, $$ where we let $t=f(u)$ in the second integral.

Answer 3

hint

If we put

$$t=\frac{x+y}{1-xy}u$$

the left integral becomes

$$\int_0^1\frac{1}{1+(\frac{x+y}{1-xy})^2u^2}\frac{x+y}{1-xy}du$$

Answer 4

Fixing $y$, define $f(x):=\arctan x+\arctan y-\arctan\frac{x+y}{1-xy}$ so $f(0)=0$ and$$\begin{align}f^\prime(x)&=\frac{1}{1+x^2}-\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\partial_x\frac{x+y}{1-xy}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1-xy-(x+y)(-y)}{(1-xy)^2}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1+y^2}{(1-xy)^2}\\&=0,\end{align}$$i.e. $f(x)=0$ for all $x$.