Proving n=11n2=π26\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} =\frac{\pi^{2}}{6} using calculus

Most of us know of the fact that n=11n2=π26\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 n=11n2\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 010111xydydx\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dydx over the rectangular region bounded by points (0,0),(0,1),(1,1),(1,0)\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 n=0(xy)n \displaystyle \sum_{n=0}^{\infty} \left(xy\right)^{n} , so the integral becomes 0101n=0(xy)ndydx\displaystyle \int_{0}^{1} \int_{0}^{1} \displaystyle \sum_{n=0}^{\infty} \left(xy\right)^{n} dy dx

01n=0(xnyn+1n+1)dx01\displaystyle \int_{0}^{1} \displaystyle \sum_{n=0}^{\infty} \left(\frac{x^{n} y^{n+1}}{n+1}\right) dx \Bigr|_{0}^{1}

01n=0(xnn+1)dx\displaystyle \int_{0}^{1} \displaystyle \sum_{n=0}^{\infty} \left(\frac{x^{n}}{n+1}\right) dx

n=0(xn(n+1)2)01=n=11n2\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 u ,v space using the following equations x=uv2,y=u+v2\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 det12121212=1\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 : (0,0)in(x,y)=(0,0)in(u,v)\displaystyle\left( 0,0\right) in (x,y) =\left( 0,0\right) in (u,v)

(0,1)in(x,y)=(22,22)in(u,v)\displaystyle\left( 0,1\right) in (x,y) =\left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) in (u,v)

(1,0)in(x,y)=(22,22)in(u,v)\displaystyle\left( 1,0\right) in (x,y )=\left( \frac{\sqrt{2}}{2},\frac{-\sqrt{2}}{2}\right) in (u,v)

(1,1)in(x,y)=(2,0)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:

11((uv2)(u+v2))dudv\displaystyle \int\int \frac{1}{1-\left(\left(\frac{u-v}{\sqrt{2}}\right)\left(\frac{u+v}{\sqrt{2}}\right)\right)} du dv

22u2+v2dvdu\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=22u=\frac{\sqrt{2}}{2} is enclosed within lines v=uv=u and v=u v =-u and the region from this point to u=2u=\sqrt{2} is enclosed within lines v=2uv=\sqrt{2} -u and v=u2v =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. 2arctan(v2u2)2u2uu \displaystyle \frac{2 \arctan\left( \frac{v}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}\Bigr|_{-u}^{u}

=2arctan(u2u2)2u22arctan(u2u2)2u2= \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}}}

arctanx=arctanx\arctan -x =-\arctan x

Therefore it is equal to 4arctan(u2u2)2u2\displaystyle \frac{4 \arctan\left( \frac{u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}.

Now next iteration 0224arctan(u2u2)2u2du\displaystyle \int_{0}^{\frac{\sqrt{2}}{2}} \displaystyle \frac{4 \arctan\left( \frac{u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}} du

0224arctan(u21u22)21u22du\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 u2=sinθ\frac{u}{\sqrt{2}}=\sin \theta

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

0224arctan(tanθ)2cosθ2cosθdθ\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(tanθ)=θ\arctan\left(\tan \theta\right)=\theta

4θdθ=2θ2\displaystyle \int 4\theta d\theta =2\theta^{2}

Limits in terms of theta are: sinθ=0\sin \theta =0 to sinθ=12,θ=0\sin\theta =\frac{1}{2} , \theta =0 to θ=π6\theta =\frac{\pi}{6}

2θ0π6=π218\displaystyle 2\theta \Bigr|_{0}^{\frac{\pi}{6}} = \frac{\pi^{2}}{18} (1) \left( 1\right)

Now we do the second part of the split integral :

222u22u22u2+v2dvdu\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.

2arctan(v2u2)2u2u22u \displaystyle \frac{2 \arctan\left( \frac{v}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}\Bigr|_{u-\sqrt{2}}^{\sqrt{2}-u}

=2arctan(2u2u2)2u22arctan(u22u2)2u2= \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}}}

=2arctan(2u2u2)2u22arctan((2u)2u2)2u2= \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 arctanx=arctanx\arctan -x =-\arctan x

=4arctan(2u2u2)2u2= \displaystyle \frac{4 \arctan\left( \frac{\sqrt{2}-u}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}

now the next iteration:

2224arctan(2u2u2)2u2du\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

=2224arctan(2(1u2)21u22)21u22du=\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

=2224arctan((1u2)1u22)21u22du=\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 u2=sinθ\frac{u}{\sqrt{2}} =\sin\theta

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

integral becomes:

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

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

let secθtanθ=tanm\sec \theta - \tan\theta =\tan m

secθtanθsec2θ=sec2mdmdθ\displaystyle \sec \theta\tan \theta - sec^{2} \theta=\sec^{2} m \frac{dm}{d\theta}

secθ(tanθsecθ)=sec2mdmdθ\displaystyle \sec \theta\left(\tan \theta - sec\theta\right)=\sec^{2} m \frac{dm}{d\theta}

secθ(secθtanθ)=sec2mdmdθ\displaystyle -\sec \theta\left( sec\theta-\tan \theta\right)=\sec^{2} m \frac{dm}{d\theta}

secθ(tanm)=sec2mdmdθ\displaystyle -\sec \theta\left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}

Now we know sec2θtan2θ=1\displaystyle \sec^{2} \theta -\tan^{2} \theta =1

(secθtanθ)(secθ+tanθ)=1\displaystyle \left(\sec\theta -\tan \theta\right)\left(\sec\theta +\tan \theta\right) =1

(tanm)(secθ+tanθ)=1\displaystyle \left(\tan m\right)\left(\sec\theta +\tan \theta\right) =1

secθ+tanθ=1tanm\displaystyle \sec\theta +\tan \theta =\frac{1}{\tan m}

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

secθtanθ+secθ+tanθ=1tanm+tanm\displaystyle \sec \theta -\tan\theta +\sec\theta +\tan \theta =\frac{1}{\tan m} +\tan m

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

2secθ=1+tan2mtanm=sec2mtanm\displaystyle 2\sec \theta =\frac{1+\tan^{2} m}{\tan m} =\frac{sec^{2} m}{\tan m}

secθ=sec2m2tanm\displaystyle \sec \theta=\frac{sec^{2} m}{2\tan m}

going back a few steps :

secθ(tanm)=sec2mdmdθ\displaystyle -\sec \theta\left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}

=sec2m2tanm(tanm)=sec2mdmdθ=\displaystyle -\frac{sec^{2} m}{2\tan m} \left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}

dθ=2dmd\theta =-2 dm

going back to our integral :

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

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

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

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

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

Now to express limits in terms of m :

sinθ=u2,\sin \theta =\frac{u}{\sqrt{2}} , from u=22u=\frac{\sqrt{2}}{2} to 2\sqrt{2} is equal to limit from sinθ=12\sin\theta =\frac{1}{2} to sinθ=1 \sin\theta =1 θ=π6 \theta =\frac{\pi}{6} to θ=π2 \theta =\frac{\pi}{2}

tanm=secθtanθ=1sinθcosθ\tan m =\sec \theta -\tan \theta =\frac{1-\sin \theta}{\cos \theta}

So limits in terms of tanm\tan m is equal to 1sinπ6cosπ6\displaystyle \frac{1-\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}} to 1sinπ2cosπ2\displaystyle\frac{1-\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}}

tanm=13\tan m =\frac{1}{\sqrt{3}} to tanm=0\tan m=0

in terms of m : m=π6 m =\frac{\pi}{6} to m=0 m=0

=4m2pi60=\displaystyle-4m^{2} \Bigr|_{\frac{pi}{6}}^{0}

=4×02(4(π6)2)=\displaystyle-4\times 0^{2} - \left( - 4\left(\frac{\pi}{6}\right)^{2}\right)

π29 \frac{\pi^{2}}{9} .

Now finally adding this to the result from our first split integral , π218+π29\displaystyle \frac{\pi^{2}}{18} +\frac{\pi^{2}}{9}

=3π218=π26!!!\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.....

Kushal Dey - 1 year, 3 months ago

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@Amal Hari, does this mean n=16n2=π\displaystyle\sum_{n=1}^{\infty} \frac{6}{n^2} = \pi?

A Former Brilliant Member - 1 year, 1 month ago

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

Kushal Dey - 1 year, 3 months ago

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