# Proving $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} =\frac{\pi^{2}}{6}$ using calculus

Most of us know of the fact that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$ . Euler has calculated the exact this sum in his time. Eulers solution to Basel problem. Today we are going to prove it using calculus, first by showing that a certain double integral of a function over a region is equivalent to $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ when expressed as its infinite series , then by directly finding the exact value of double integral using transformations . Since both forms are equivalent we can prove the sum.

Consider the double integral $\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dydx$ over the rectangular region bounded by points $\left(0,0\right),\left(0,1\right),\left(1,1\right),\left(1,0\right)$. Now we express this function as an infinite series $\displaystyle \sum_{n=0}^{\infty} \left(xy\right)^{n}$ , so the integral becomes $\displaystyle \int_{0}^{1} \int_{0}^{1} \displaystyle \sum_{n=0}^{\infty} \left(xy\right)^{n} dy dx$

$\displaystyle \int_{0}^{1} \displaystyle \sum_{n=0}^{\infty} \left(\frac{x^{n} y^{n+1}}{n+1}\right) dx \Bigr|_{0}^{1}$

$\displaystyle \int_{0}^{1} \displaystyle \sum_{n=0}^{\infty} \left(\frac{x^{n}}{n+1}\right) dx$

$\displaystyle \sum_{n=0}^{\infty} \left(\frac{x^{n}}{\left(n+1\right)^{2}}\right)\Bigr|_{0}^{1} =\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ .

Now we are going to find the definite integral of this function in another way: First we transform the following the region to another space which we will call $u ,v$ space using the following equations $\displaystyle x=\frac{u-v}{\sqrt{2}} , y=\frac{u+v}{\sqrt{2}}$. This transformation is linear therefore the transformed region will also be a rectangular region. The only difference is that it is rotated . The jacobian for the above transformation is $\det \begin{vmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{vmatrix} =1$

The points bounding the transformed regions : $\displaystyle\left( 0,0\right) in (x,y) =\left( 0,0\right) in (u,v)$

$\displaystyle\left( 0,1\right) in (x,y) =\left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) in (u,v)$

$\displaystyle\left( 1,0\right) in (x,y )=\left( \frac{\sqrt{2}}{2},\frac{-\sqrt{2}}{2}\right) in (u,v)$

$\displaystyle\left( 1,1\right) in (x,y) =\left( \sqrt{2}, 0\right) in (u,v)$

Now we rewrite the double integral:

$\displaystyle \int\int \frac{1}{1-\left(\left(\frac{u-v}{\sqrt{2}}\right)\left(\frac{u+v}{\sqrt{2}}\right)\right)} du dv$

$\displaystyle \int \int \frac{2}{2-u^{2} +v^{2}}dv du$

from the transformation we can split this integral into 2 as following

$$\displaystyle \int_{0}^{\frac{\sqrt{2}}{2}} \int_{-u}^{u} \frac{2}{2-u^{2} +v^{2}}dv du +\displaystyle \int_{\frac{\sqrt{2}} {2}}^{\sqrt{2}} \int_{u-\sqrt{2}}^{\sqrt{2}-u} \frac{2}{2-u^{2} +v^{2}}dv du$$

This is because the region up to point $u=\frac{\sqrt{2}}{2}$ is enclosed within lines $v=u$ and $v =-u$ and the region from this point to $u=\sqrt{2}$ is enclosed within lines $v=\sqrt{2} -u$ and $v =u-\sqrt{2}$

Now we can iterate through this double integral first w.r.t v and then u

$$\displaystyle \int_{-u}^{u}\frac{2}{\left(2-u^{2}\right) +v^{2}} dv =\displaystyle \int_{-u}^{u}\frac{2}{\left(2-u^{2}\right) \left(1+\frac{v^{2}}{2-u^{2}}\right)} dv$$

u is independent to v. $\displaystyle \frac{2 \arctan\left( \frac{v}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}\Bigr|_{-u}^{u}$

$= \displaystyle \frac{2 \arctan\left( \frac{u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}-\displaystyle \frac{2 \arctan\left( \frac{-u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}$

$\arctan -x =-\arctan x$

Therefore it is equal to $\displaystyle \frac{4 \arctan\left( \frac{u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}$.

Now next iteration $\displaystyle \int_{0}^{\frac{\sqrt{2}}{2}} \displaystyle \frac{4 \arctan\left( \frac{u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}} du$

$\displaystyle \int_{0}^{\frac{\sqrt{2}}{2}} \displaystyle \frac{4 \arctan\left( \frac{u}{\sqrt{2}\sqrt{1-\frac{u^{2}}{2}}}\right)}{\sqrt{2}\sqrt{1-\frac{u^{2}}{2}}} du$

Substitute $\frac{u}{\sqrt{2}}=\sin \theta$

$du=\sqrt{2}\cos \theta d\theta$

$\displaystyle \int_{0}^{\frac{\sqrt{2}}{2}} \displaystyle \frac{4 \arctan\left( \tan \theta\right)}{\sqrt{2}\cos \theta } \sqrt{2}\cos \theta d\theta$

$\arctan\left(\tan \theta\right)=\theta$

$\displaystyle \int 4\theta d\theta =2\theta^{2}$

Limits in terms of theta are: $\sin \theta =0$ to $\sin\theta =\frac{1}{2} , \theta =0$ to $\theta =\frac{\pi}{6}$

$\displaystyle 2\theta \Bigr|_{0}^{\frac{\pi}{6}} = \frac{\pi^{2}}{18}$ $\left( 1\right)$

Now we do the second part of the split integral :

$\displaystyle \int_{\frac{\sqrt{2}}{2}}^{\sqrt{2}} \int_{u-\sqrt{2}}^{\sqrt{2}-u} \frac{2}{2-u^{2} +v^{2}}dv du$

Again we iterate through v first and then u

Anti derivative for integration w.r.t v is same ,only the limits change.

$\displaystyle \frac{2 \arctan\left( \frac{v}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}\Bigr|_{u-\sqrt{2}}^{\sqrt{2}-u}$

$= \displaystyle \frac{2 \arctan\left( \frac{\sqrt{2}-u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}-\displaystyle \frac{2 \arctan\left( \frac{u-\sqrt{2}}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}$

$= \displaystyle \frac{2 \arctan\left( \frac{\sqrt{2}-u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}-\displaystyle \frac{2 \arctan\left( \frac{-\left(\sqrt{2} -u\right)}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}$

Again using the fact that $\arctan -x =-\arctan x$

$= \displaystyle \frac{4 \arctan\left( \frac{\sqrt{2}-u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}$

now the next iteration:

$\displaystyle \int_{\frac{\sqrt{2}}{2}}^{\sqrt{2}} \displaystyle \frac{4 \arctan\left( \frac{\sqrt{2}-u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}} du$

$=\displaystyle \int_{\frac{\sqrt{2}}{2}}^{\sqrt{2}} \displaystyle \frac{4 \arctan\left( \frac{\sqrt{2}\left(1-\frac{u}{\sqrt{2}}\right)}{\sqrt{2}\sqrt{1-\frac{u^{2}}{2}}}\right)}{\sqrt{2}\sqrt{1-\frac{u^{2}}{2}}} du$

$=\displaystyle \int_{\frac{\sqrt{2}}{2}}^{\sqrt{2}} \displaystyle \frac{4 \arctan\left( \frac{\left(1-\frac{u}{\sqrt{2}}\right)}{\sqrt{1-\frac{u^{2}}{2}}}\right)}{\sqrt{2}\sqrt{1-\frac{u^{2}}{2}}} du$

now substitute $\frac{u}{\sqrt{2}} =\sin\theta$

$du=\sqrt{2}\cos\theta d\theta$

integral becomes:

$\displaystyle \int \displaystyle 4 \arctan\left( \frac{\left(1-\sin\theta \right)}{\cos \theta}\right) d\theta$

$\displaystyle \int \displaystyle 4 \arctan\left(\sec \theta -\tan\theta \right) d\theta$

let $\sec \theta - \tan\theta =\tan m$

$\displaystyle \sec \theta\tan \theta - sec^{2} \theta=\sec^{2} m \frac{dm}{d\theta}$

$\displaystyle \sec \theta\left(\tan \theta - sec\theta\right)=\sec^{2} m \frac{dm}{d\theta}$

$\displaystyle -\sec \theta\left( sec\theta-\tan \theta\right)=\sec^{2} m \frac{dm}{d\theta}$

$\displaystyle -\sec \theta\left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}$

Now we know $\displaystyle \sec^{2} \theta -\tan^{2} \theta =1$

$\displaystyle \left(\sec\theta -\tan \theta\right)\left(\sec\theta +\tan \theta\right) =1$

$\displaystyle \left(\tan m\right)\left(\sec\theta +\tan \theta\right) =1$

$\displaystyle \sec\theta +\tan \theta =\frac{1}{\tan m}$

Now add this to $\displaystyle \sec \theta -\tan\theta$ which is equal to $\tan m$

$\displaystyle \sec \theta -\tan\theta +\sec\theta +\tan \theta =\frac{1}{\tan m} +\tan m$

$\displaystyle 2\sec \theta =\frac{1}{\tan m} +\tan m$

$\displaystyle 2\sec \theta =\frac{1+\tan^{2} m}{\tan m} =\frac{sec^{2} m}{\tan m}$

$\displaystyle \sec \theta=\frac{sec^{2} m}{2\tan m}$

going back a few steps :

$\displaystyle -\sec \theta\left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}$

$=\displaystyle -\frac{sec^{2} m}{2\tan m} \left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}$

$d\theta =-2 dm$

going back to our integral :

$\displaystyle \int \displaystyle 4 \arctan\left(\sec \theta -\tan\theta \right) d\theta$

$=\displaystyle \int \displaystyle 4 \arctan\left(\tan m \right) d\theta$

$=\displaystyle \int \displaystyle 4 m d\theta$

$=\displaystyle \int \displaystyle 4 m\times\left( -2\right) dm$

$=\displaystyle \int \displaystyle-8m dm$

Now to express limits in terms of m :

$\sin \theta =\frac{u}{\sqrt{2}} ,$ from $u=\frac{\sqrt{2}}{2}$ to $\sqrt{2}$ is equal to limit from $\sin\theta =\frac{1}{2}$ to $\sin\theta =1$ $\theta =\frac{\pi}{6}$ to $\theta =\frac{\pi}{2}$

$\tan m =\sec \theta -\tan \theta =\frac{1-\sin \theta}{\cos \theta}$

So limits in terms of $\tan m$ is equal to $\displaystyle \frac{1-\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}}$ to $\displaystyle\frac{1-\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}}$

$\tan m =\frac{1}{\sqrt{3}}$ to $\tan m=0$

in terms of m : $m =\frac{\pi}{6}$ to $m=0$

$=\displaystyle-4m^{2} \Bigr|_{\frac{pi}{6}}^{0}$

$=\displaystyle-4\times 0^{2} - \left( - 4\left(\frac{\pi}{6}\right)^{2}\right)$

$\frac{\pi^{2}}{9}$ .

Now finally adding this to the result from our first split integral , $\displaystyle \frac{\pi^{2}}{18} +\frac{\pi^{2}}{9}$

$\displaystyle=\frac{3\pi^{2}}{18} =\frac{\pi^{2}}{6} !!!$

Note by Amal Hari
1 year, 8 months ago

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Now, I will tell you another method of proving the above which I did in my own way. The Taylor series expansion, sin(x)=x-x^3/3!+x^5/5!........ creates an algebraic series of this trigonometric function in the form of a polynomial. In other words, I am trying to visualise sin(x) as a polynomial through this series. Thus I say that sin(x)=Lt n-> infinity ((-1)^nx((x^2/pi^2)-1)((x^2/4pi^2)-1).........((x^2/(npi)^2)-1)) so that this polynomial vanishes at x=npi and has relative extrema at odd multiples of pi/2(through logarithmic differentiation) just like sin(x). Since the Taylor series and my assumed polynomial are an identity, thus you can just compare the coefficient of x^3 of both series and see that : -1/6 = -(1/pi^2)*(1+1/4+1/9+.....) =>(pi^2)/6=1+1/4+1/9.....

- 1 year, 3 months ago

@Amal Hari, does this mean $\displaystyle\sum_{n=1}^{\infty} \frac{6}{n^2} = \pi$?

- 1 year, 1 month ago

Really cool way of proving the series. Using the same integral 2 different ways is impressive.

- 1 year, 3 months ago