$\int_0^{\pi/2}\int_0^{\pi/2}\frac{(\tan\alpha)(\tan\beta)}{\tan\alpha+\tan\beta} d\alpha d\beta=(0.9999999913...)(\pi/2)$? Seriously?

Question

In the diagram, $\alpha$ and $\beta$ are independent uniformly random real numbers in $\left(0,\frac{\pi}{2}\right)$.

What is $\mathbb{E}(h)$?

Superimposing a cartesian coordinate system, the equations of the lines are $y=(\tan\alpha)x$ and $y=(-\tan\beta)(x-1)$, so the $y$-coordinate of their intersection is $\frac{(\tan\alpha)(\tan\beta)}{\tan\alpha+\tan\beta}$. So we have

$$\mathbb{E}(h)=\frac{4}{\pi^2}\int_0^{\pi/2}\int_0^{\pi/2}\frac{(\tan\alpha)(\tan\beta)}{\tan\alpha+\tan\beta} d\alpha d\beta$$

Desmos says $I=\int_0^{\pi/2}\int_0^{\pi/2}\frac{(\tan\alpha)(\tan\beta)}{\tan\alpha+\tan\beta} d\alpha d\beta$ is $(0.9999999913...)(\frac{\pi}{2})$. Is that a computer error, and the result is exactly $\frac{\pi}{2}$? I don't know how to evaluate the integral. Wolfram evaluates the inside integral but doesn't evaluate both integrals.

If Desmos is not very reliable, then maybe my earlier weird conjecture is actually true.

(This question was inspired by a question about random points in a square.)

Edit: In the comments, @G.Gare notes that Mathematica says the integral $I$ is exactly $\pi/2$. Can we prove it? Maybe an intuitive geometrical argument?

Edit2: I seek to generalize this result here.

Answer 1

$$I=\int_0^{\pi/2}\int_0^{\pi/2}\frac{(\tan\alpha)(\tan\beta)}{\tan\alpha+\tan\beta} d\alpha d\beta\overset{\binom{\alpha=\arctan x}{\beta=\arctan y}}{=}\int_0^\infty\int_0^\infty\frac{xy}{(1+x^2)(1+y^2)(x+y)}dxdy$$ $$=\int_0^\infty\int_0^\infty\left(\frac1{1+x^2}-\frac1{1+y^2}\right)\frac{xy}{x+y}\frac{dxdy}{y^2-x^2}$$ $$=2\int_0^\infty\int_0^\infty\frac1{1+x^2}\frac{xy}{x+y}\frac{dxdy}{y^2-x^2}\overset{x=ty}{=}2\int_0^\infty dy\int_0^\infty\frac{t\,dt}{(1+t^2y^2)(1-t)(1+t)^2}$$ $$=2\int_0^\infty\frac{t\,dt}{(1-t)(1+t)^2}\int_0^\infty\frac{dy}{1+t^2y^2}$$ $$=\pi\int_0^\infty\frac{dt}{(1-t)(1+t)^2}\overset{x=\frac1t}{=}-\pi\int_0^\infty\frac{x\,dx}{(1-x)(1+x)^2}$$ $$\Rightarrow\,\,2I=\pi\int_0^\infty\frac{dx}{(1+x)^2}\,\,\Rightarrow\,\,I=\frac\pi2$$

Answer 2

The first antiderivative is quite simple.

After solving the inner integral, we are left with:

$$\int \sin (b) \left(\frac{1}{2} \pi \cos (b)+\sin (b) \,\log (\tan (b))\right)\,db$$ Now, there is a quite nasty antiderivative.

Using the bounds, the first term is $\frac \pi 4$.

So, for the total, $$I=\frac \pi 4+ \frac{1}{16} (4+i \pi ) \pi-i\frac{ \pi ^2}{16}=\frac \pi 2$$

Edit

$$I=\int \frac{\tan (a) \tan (b)}{\tan (a)+\tan (b)}\, da$$ $$I=\sin (b) (a \cos (b)-\sin (b) \log (\sin (a+b)))$$ Using the bounds, the integral written at the top.

$$J=\int \sin ^2(b) \log (\tan (b))\,db$$ $$J=\frac{1}{2} i \text{Li}_2\left(e^{2 i b}\right)-\frac{1}{8} i \text{Li}_2\left(e^{4 i b}\right)+\frac{b}{2}+b \tanh ^{-1}\left(e^{2 i b}\right)+$$ $$\frac{1}{2} b \log (\tan (b))-\frac{1}{4} \sin (2 b) \log (\tan (b))$$

Now, use the bounds.

Answer 3

A variant of Svyatoslav's solution, using polar coordinates $(x, y) = (r \cos \phi, r \sin\phi)$:

$$ \begin{align} I &=\int_0^{\pi/2}\int_0^{\pi/2}\frac{\tan\alpha \tan\beta}{\tan\alpha+\tan\beta} d\alpha d\beta \\ &=\int_0^\infty\int_0^\infty\frac{xy}{(x+y)(1+x^2)(1+y^2)}dxdy \\ &= \int_0^{\pi/2} \frac{\cos\phi \sin\phi}{\cos \phi + \sin\phi}\left( \int_0^\infty \frac{r^2}{(1+r^2 \cos^2\phi)(1+r^2\sin^2\phi)} \, dr \right) d\phi \, . \end{align} $$ For $ 0 < \phi < \pi/2$ the inner integral is $$ \begin{align} &\frac{1}{ \sin^2 \phi-\cos^2\phi } \int_0^\infty \left( \frac{1}{1+r^2 \cos^2 \phi)} - \frac{1}{1+r^2 \sin^2\phi}\right) dr \\ &\qquad = \frac{1}{ \sin^2 \phi-\cos^2\phi } \left[ \frac{\arctan(r \cos\phi)}{\cos \phi} - \frac{\arctan(r \sin\phi)}{\sin\phi}\right]_{r=0}^{r=\infty} \\ &\qquad = \frac{1}{\cos \phi \sin\phi (\cos\phi + \sin\phi)} \frac{\pi}{2} \, . \end{align} $$ It follows that $$ I = \frac{\pi}{2} \int_0^{\pi/2} \frac{1}{(\cos\phi + \sin\phi)^2} d\phi = \frac{\pi}{2} \left[ \frac{\sin \phi}{\cos\phi+\sin\phi}\right]_{\phi=0}^{\phi=\pi/2}= \frac{\pi}{2} \, . $$

Answer 4

Long comment

Related result: The expected length of one of the sides of the triangle (not the base) is

$$\frac{4}{\pi^2}\int_0^{\pi/2}\int_0^{\pi/2}\frac{\sin \beta}{\sin (\alpha+\beta)}d\alpha d{\beta}$$

$$=\frac{4}{\pi^2}\int_0^{\pi/2}\left[(\sin \beta)\log\left(\tan\left({\frac{\alpha+\beta}{2}}\right)\right)\right]_0^{\pi/2}d\beta$$

$$=\frac{4}{\pi^2}\int_0^{\pi/2}(\sin \beta)\log\left(\frac{\tan\left({\frac{\frac{\pi}{2}+\beta}{2}}\right)}{\tan\left({\frac{\beta}{2}}\right)}\right)d\beta$$

$$=\frac{4}{\pi^2}\left[\beta-\log\left(\frac{\sin b}{2}\right)-(\cos\beta)\log \left(\frac{\cos\left(\frac{\beta}{2}\right)\left(\cot\left(\frac{\beta}{2}\right)+1\right)}{\cos\left(\frac{\beta}{2}\right)-\sin\left(\frac{\beta}{2}\right)}\right)\right]_0^{\pi/2}$$

$$=\frac{2\pi+\log 16}{\pi^2}$$

(with help from Wolfram)

Answer 5

$$\frac{\tan(\alpha)\tan(\beta)}{\tan(\alpha)+ \tan(\beta)}= -\cot(\alpha+ \beta) +\frac{1}{\tan(\alpha)+ \tan(\beta)}$$

$$\displaystyle I:=\int _0^{\frac{\pi}2}\int _0^{\frac{\pi}2}\frac{\tan(\alpha)\tan(\beta)}{\tan(\alpha)+ \tan(\beta)} d \alpha d \beta=\int _0^{\frac{\pi}2}\int _0^{\frac{\pi}2}-\cot(\alpha+ \beta) +\frac{1}{\tan(\alpha)+ \tan(\beta)} d \alpha d \beta$$

$\\[4pt]$

for $\displaystyle \int \frac{1}{\tan x+ c}dx$ let $\tan x = t $ then $$\int \frac{1}{\tan x+ c}dx= \int \frac{1}{(t^2+1)(t+c)}dt$$ $$=\frac{c}{c^2+1}\int\frac{1}{1+t^2} dt -\frac{1}{2(c^2+1)}\int \frac{2t}{t^2+1}dt + \frac{1}{c^2+1}\int\frac{1}{c+t} dt $$ $$\frac{c}{c^2+1} \arctan(t) - \frac{1}{2(c^2+1)}\ln|t^2+1| +\frac{1}{c^2+1} \ln|t+c|$$ $$=\frac{c}{c^2+1} x +\frac{1}{(c^2+1)}\ln|\cos(x)| +\frac{1}{c^2+1} \ln|\tan(x)+c|$$ $\\[3pt]$ $$I=\int _0^{\frac{\pi}2} \biggr|-\ln|\sin(\alpha + \beta)| + \frac{\tan(\beta)}{\sec^2(\beta)} \alpha +\frac{1}{(\sec^2(\beta))}\ln|\cos(\alpha)| +\frac{1}{\sec^2(\beta)} \ln|\tan(\alpha)+\tan(\beta)|\biggr|_0 ^{\frac{\pi}{2}} d\beta$$ $$=\int _0^{\frac{\pi}2} \ln|\tan(x)| +\frac{\pi}{4}\sin(2\beta )+ \cos^2(\beta )(-\ln(\tan(\beta))) d\beta$$

first we can see that $\displaystyle \int _0 ^ \frac{\pi}{2}\ln(\tan(\alpha)) d\alpha= \int _{0}^ \frac{\pi}{2} \ln ( \cot \alpha)du =0 $ because $\displaystyle \int _0 ^ \frac{\pi}{2}\ln(\tan(\alpha)) d\alpha =-\int _0 ^ \frac{\pi}{2}\ln(\tan(\alpha)) d\alpha$ by king's rule

and $\displaystyle \int_0^{\frac{\pi}{2}} \sin(2\beta) d \beta =-\frac{\cos(2\beta)}{2} \biggr|_0^{\frac{\pi}{2}} = 1$ so that means $$I = \frac{\pi}{4} - \int _0^{\frac{\pi}{2}}\cos^2(\beta)\ln(\tan(\beta)) d\beta$$

$$J:=\int _0^{\frac{\pi}{2}}\cos^2(\beta)\ln(\tan(\beta)) d\beta=\int _0^{\frac{\pi}{2}}\cos^2(\beta)\ln(\sin(\beta)) d\beta-\int _0^{\frac{\pi}{2}}\cos^2(\beta)\ln(\cos(\beta)) d\beta $$

$$=\int _0^{\frac{\pi}{2}}\cos^2(\beta)\ln(\sin(\beta)) d\beta+\int _0^{\frac{\pi}{2}}\ln(\cos(\beta)) (1-\cos^2(\beta)) d\beta -\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta $$

since $\displaystyle \int _0^{\frac{\pi}{2}}\cos^2(\beta)\ln(\sin(\beta)) d\beta=\int _0^{\frac{\pi}{2}}\ln(\cos(\beta)) (\sin^2(\beta)) d\beta$ by king's rule so

$$J = 2\int _0^{\frac{\pi}{2}}\cos^2(\beta)\ln(\sin(\beta)) d\beta-\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta $$

let $\displaystyle y= \sin(\beta)$

$$=2\int _0^{1}\ln(y)\sqrt{1-y^2}dy -\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta $$ here I will use integration by parts and integrate $\sqrt{1-y^2} $ $$J=y \ln(y)\sqrt{1-y^2} +\arcsin(y)\ln(y)\biggr|_0^1 -\int_0^1 \sqrt{1-y^2} dy - \int_0^1 \frac{\arcsin{y}}{y} dy -\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta $$

since $\displaystyle \lim\limits_{y \to 0}y \ln (y) =0$(why ? use L'Hôpital's rule) and $\lim\limits_{y \to 0}\arcsin{y} \ln (y) =0$(why ? substitute $\sin(t) = x$ and use L'Hôpital's rule )

and since $\displaystyle \int _0 ^1 \sqrt{1-y^2} dy = \frac{\pi}{4}$

$$J= -\frac{\pi}{4 } + \int_0^1 \frac{\arcsin{y}}{y} dy -\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta $$ $$-\frac{\pi}{4 } -\int _0^{\frac{\pi}{2}} \beta \cot(\beta) d\beta -\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta $$ $$=-\frac{\pi}{4 } - \beta \ln(\sin ( \beta))\biggr|_{0}^{\frac{\pi}{2}} +\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta -\int _0^{\frac{\pi}{2}} \ln(\cos(\beta)) d\beta $$

$$= -\frac{\pi}{4 }$$

so $$\color{red}{I = \frac{\pi}{4} -J =\frac{\pi }{2}}$$

Answer 6

Expanding on an earlier comment: let $(u,v)=\left(\dfrac{1-\tan \alpha}{1+\tan \alpha}, \dfrac{1-\tan \beta}{1+\tan \beta}\right)$, i.e. $(\alpha,\beta) = \left(\dfrac\pi4-\arctan u,\dfrac\pi4-\arctan v\right)$, to rewrite the integral as

$$\begin{align*} I &= \int_0^{\tfrac\pi2} \int_0^{\tfrac\pi2} \frac{\tan \alpha \tan \beta}{\tan \alpha + \tan\beta} \, d\alpha\,d\beta \\ &= \frac12 \int_{-1}^1 \int_{-1}^1 \frac{\frac{1-u}{1+u} \cdot \frac{1-v}{1+v}}{\frac{1-u}{1+u} + \frac{1-v}{1+v}} \, \frac{du}{1+u^2} \, \frac{dv}{1+v^2} \\ &= \frac12 \underbrace{\int_{-1}^1 \int_{-1}^1 \frac{1-u}{1+u^2} \, \frac{1-v}{1+v^2} \, \frac{du\,dv}{1-uv}}_{=\pi} \end{align*}$$

The inner integral is elementary:

$$\begin{align*} & \int_{-1}^1 \frac{1-u}{1+u^2} \, \frac{du}{1-uv} \\ &= \frac{1+v}{1+v^2} \int_{-1}^1 \frac{du}{1+u^2} + \frac{1-v}{1+v^2} \int_{-1}^1 \frac{u}{1+u^2} \, du - \frac{v(1-v)}{1+v^2} \int_{-1}^1 \frac{du}{1-uv} \\ &= \frac\pi2 \frac{1+v}{1+v^2} - 2 \frac{1-v}{1+v^2} \operatorname{artanh}v \end{align*}$$

and we're left with

$$\begin{align*} & \frac\pi2 \int_{-1}^1 \frac{1-v^2}{\left(1+v^2\right)^2} \, dv - 2 \int_{-1}^1 \frac{(1-v)^2}{\left(1+v^2\right)^2} \operatorname{artanh} v \, dv \\ &= \frac\pi2 + \underbrace{\int_{-1}^1 \frac{(1-v)^2}{\left(1+v^2\right)^2} \log \frac{1-v}{1+v} \, dv}_{=:J} \end{align*}$$

Showing $J=\dfrac\pi2$ can be done by enforcing $v\mapsto\dfrac{1-v}{1+v}$, folding up the subsequent integral at $v=1$ and expanding into partial fractions, exploiting power series, and integrating by parts:

$$\begin{align*} J &= 2 \int_0^\infty \frac{\left(1-\frac{1-v}{1+v}\right)^2}{\left(1+\frac{(1-v)^2}{(1+v)^2}\right)^2} \log v \, \frac{dv}{(1+v)^2} \\ &= 2 \int_0^\infty \frac{v^2}{\left(1+v^2\right)^2} \log v \, dv \\ &= 2 \int_0^1 \left(\frac1{1+v^2} - \frac2{\left(1+v^2\right)^2}\right) \log v \, dv \\ &= 2 \sum_{n\ge0} (-1)^{n+1} (2n+1) \int_0^1 v^{2n} \log v \, dv \\ &= 2 \sum_{n\ge0} \frac{(-1)^n}{2n+1} = 2\arctan 1 = \frac\pi2 \end{align*}$$