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

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...

• This identity holds for $xy<1$. In the other cases, you need to add or subtract a multiple of $\pi$ to the RHS. – bjorn93 Mar 19 at 23:28
• True, probably I should add that in the statement of the problem, thanks – Brightsun Mar 20 at 7:33

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 $$u=at$$ $$\begin{eqnarray*} \int_{1}^{b} \frac{dt}{t} = \int_{a}^{ab} \frac{du}{u}. \end{eqnarray*}$$

• Nice, thank you. I think the substitution is slightly more insightful written as $t=\frac{x+u}{1-ux}$ due to the similarity with $\frac{x+y}{1-xy}$. Just a curiosity, did you arrive at the result by trial and error or did you have some guiding principle? – Brightsun Mar 19 at 22:14
• Don't tell anyone ... but the first guess written in my notebook is \begin{eqnarray*} t=x + \frac{y(1+u^2)}{1-uy}. \end{eqnarray*} "Trial and error" or "guiding principle"? A bit of both. Thanks, this is a cool problem. – Donald Splutterwit Mar 19 at 22:21

hint

If we put

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

the left integral becomes

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

• I think that's what I did in the last step of my question..? – Brightsun Mar 19 at 20:59
• Sorry, i was thinking about corona, i did not read your question as i had to do. the problem is that i cannot delete from mobile. i hope no one will downvote. – hamam_Abdallah Mar 19 at 21:04

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 $$x$$.

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.