A beautiful integral which yields a solution to the Basel problem

Recently, I came across a powerful elementary integral, 011(1+yx)1x2dx\int_{0}^1 \frac{1}{(1+yx)\sqrt{1-x^2}} \, dx where y(1,1).y\in (-1, 1).

We will start out by finding the value of the integral above, letting x=sin(t)x=\sin (t),

011(1+yx)1x2dx=0π211+ysintdt=0π21sin2(t2)+cos2(t2)+2ysin(t2)cos(t2)dt=20π2(tan(t2))tan2(t2)+2ytan(t2)+1dt=2011u2+2yu+1du                                                                       [u=tan(t2)]=2011(u+y)2+1y2du=2y1+y1v2+1y2dv                                                                        [v=u+y]=21y2tan1(y1y2)tan1(1+y1y2)dθ                                                               [v=1y2tanθ]=21y2[tan1(1+y1y2)tan1(y1y2)]=21y2tan1(1y1+y)                                                             [tan1(x)tan1(y)=tan1(xy1+xy)]=arccosy1y2                                                                                            [arccos(α)=2arctan(1α1+α), α(1,1]].\begin{aligned} \int_{0}^1 \frac{1}{(1+yx)\sqrt{1-x^2}} \, dx =& \int_{0}^{\frac{\pi}{2}} \frac{1}{1+y\sin t} \, dt \\ =& \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin^2 (\frac{t}{2})+\cos^2 (\frac{t}{2})+2y\sin (\frac{t}{2})\cos (\frac{t}{2})} \, dt \\ =& 2\int_{0}^{\frac{\pi}{2}} \frac{(\tan (\frac{t}{2}))'}{\tan^2 (\frac{t}{2})+2y\tan (\frac{t}{2})+1} \, dt \\ =& 2\int_{0}^{1} \frac{1}{u^2+2yu+1} \, du \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[u=\tan (\frac{t}{2})\right] \\ =& 2\int_{0}^{1} \frac{1}{(u+y)^2+1-y^2} \, du \\ =& 2\int_{y}^{1+y} \frac{1}{v^2+1-y^2} \, dv \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[v=u+y\right] \\ =& \frac{2}{\sqrt{1-y^2}}\int_{\tan^{-1} \left(\frac{y}{\sqrt{1-y^2}}\right)}^{\tan^{-1} \left(\frac{1+y}{\sqrt{1-y^2}}\right)} \, d\theta \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[v=\sqrt{1-y^2} \tan \theta \right] \\ =& \frac{2}{\sqrt{1-y^2}} \left[\tan^{-1} \left(\frac{1+y}{\sqrt{1-y^2}}\right)-\tan^{-1} \left(\frac{y}{\sqrt{1-y^2}}\right)\right] \\ =& \frac{2}{\sqrt{1-y^2}} \tan^{-1} \left(\sqrt{\frac{1-y}{1+y}}\right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[\tan^{-1} (x)-\tan^{-1}(y)=\tan^{-1} \left(\frac{x-y}{1+xy} \right) \right]\\ =& \frac{\arccos y}{\sqrt{1-y^2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[\arccos (\alpha) = 2 \arctan \left(\sqrt{\frac{1-\alpha}{1+\alpha}}\right), \ \alpha \in (-1,1] \right]\end{aligned} .

Now, we can solve the Basel problem using this integral, by integrating both sides with respect to yy from 1-1 to 11, the RHS is just 11arccosy1y2dy=arccos2y211=π22.\int_{-1}^1 \frac{\arccos y}{\sqrt{1-y^2}} \, dy = \left. -\frac{\arccos^2 y}{2}\right|_{-1}^1 =\frac{\pi ^2}{2}. For the LHS, we switch the order of integration to get

11011(1+yx)1x2dxdy=01111(1+yx)1x2dydx=01ln(1+xy)x1x211dx=01ln(1x1+x)x1x2dx=401lny1y2dy                                               [Let  y2=1x1+x]=401n=0y2nlnydy=4n=001y2nlnydy=4n=0(y2n+1lny2n+101  =001y2n2n+1dy)=4n=0y2n+1(2n+1)201=4n=01(2n+1)2\begin{aligned} \int_{-1}^{1} \int_{0}^1 \frac{1}{(1+yx)\sqrt{1-x^2}} \, dx \, dy =& \int_{0}^{1} \int_{-1}^1 \frac{1}{(1+yx)\sqrt{1-x^2}} \, dy \, dx \\ =& \int_{0}^1 \left. \dfrac{\ln (1+xy)}{x\sqrt{1-x^2}}\right|_{-1}^{1} \, dx \\ =& -\int_{0}^{1} \frac{\ln \left(\frac{1-x}{1+x}\right)}{x\sqrt{1-x^2}} \, dx \\ =& -4\int_{0}^{1} \frac{\ln y}{1-y^2} \, dy \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \textrm{Let} \ \ y^2 = \frac{1-x}{1+x}\right]\\ =& -4\int_{0}^{1} \sum_{n=0}^\infty y^{2n} \ln y dy \\ =& -4\sum_{n=0}^\infty \int_{0}^1 y^{2n} \ln y \, dy \\ =& -4\sum_{n=0}^\infty \left(\cancelto{=0}{\left. \dfrac{y^{2n+1} \ln y}{2n+1} \right|_{0}^{1}}- \int_{0}^1 \frac{y^{2n}}{2n+1} \, dy \right) \\ =& 4 \sum_{n=0}^\infty \left. \dfrac{y^{2n+1}}{(2n+1)^2} \right|_{0}^1 \\ =& 4\sum_{n=0}^\infty \frac{1}{(2n+1)^2} \end{aligned}

Therefore, by these two values, we get n=11(2n1)2=n=01(2n+1)2=π28.\sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi ^2}{8}. If we let S=n=11n2S=\displaystyle{\sum_{n=1}^\infty \frac{1}{n^2}}, we have S=n=11(2n1)2+n=11(2n)2=n=11(2n1)2+14n=11n2=n=11(2n1)2+S4    3S4=n=11(2n1)2=π28    S=π26S=\sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \sum_{n=1}^\infty \frac{1}{(2n)^2} = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \frac{S}{4} \implies \frac{3S}{4}=\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi ^2}{8} \implies \boxed{S=\frac{\pi ^2}{6}}

Note by ChengYiin Ong
1 week, 3 days ago

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I vaguely remember this integral coming up in one of my quantum mechanics courses - the professor dedicated the entire class to talking about the Basel problem and how it can be solved this way. It's nice to see it on here!

Levi Walker - 1 week, 2 days ago

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Wow, in a quantum mechanics course! I was also surprised when I first came across this integral.

ChengYiin Ong - 1 week, 1 day ago

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