Proving ∑n=1∞1n2=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=1∞1n2=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=1∞1n2\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 ∫01∫0111−xydydx\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 ∫01∫01∑n=0∞(xy)ndydx\displaystyle \int_{0}^{1} \int_{0}^{1} \displaystyle \sum_{n=0}^{\infty} \left(xy\right)^{n} dy dx

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

∫01∑n=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=1∞1n2\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=u−v2,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 det⁡∣12−121212∣=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−((u−v2)(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

∫∫22−u2+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=2−uv=\sqrt{2} -u and v=u−2v =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⁡(v2−u2)2−u2∣−uu \displaystyle \frac{2 \arctan\left( \frac{v}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}\Bigr|_{-u}^{u}

=2arctan⁡(u2−u2)2−u2−2arctan⁡(−u2−u2)2−u2= \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\arctan -x =-\arctan x

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

Now next iteration ∫0224arctan⁡(u2−u2)2−u2du\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⁡(u21−u22)21−u22du\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=sin⁡θ\frac{u}{\sqrt{2}}=\sin \theta

du=2cos⁡θdθdu=\sqrt{2}\cos \theta d\theta

∫0224arctan⁡(tan⁡θ)2cos⁡θ2cos⁡θdθ\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⁡(tan⁡θ)=θ\arctan\left(\tan \theta\right)=\theta

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

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

2θ∣0π6=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 :

∫222∫u−22−u22−u2+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⁡(v2−u2)2−u2∣u−22−u \displaystyle \frac{2 \arctan\left( \frac{v}{\sqrt{2-u^{2}}}\right)}{\sqrt{2-u^{2}}}\Bigr|_{u-\sqrt{2}}^{\sqrt{2}-u}

=2arctan⁡(2−u2−u2)2−u2−2arctan⁡(u−22−u2)2−u2= \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⁡(2−u2−u2)2−u2−2arctan⁡(−(2−u)2−u2)2−u2= \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\arctan -x =-\arctan x

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

now the next iteration:

∫2224arctan⁡(2−u2−u2)2−u2du\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(1−u2)21−u22)21−u22du=\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⁡((1−u2)1−u22)21−u22du=\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=sin⁡θ\frac{u}{\sqrt{2}} =\sin\theta

du=2cos⁡θdθdu=\sqrt{2}\cos\theta d\theta

integral becomes:

∫4arctan⁡((1−sin⁡θ)cos⁡θ)dθ\displaystyle \int \displaystyle 4 \arctan\left( \frac{\left(1-\sin\theta \right)}{\cos \theta}\right) d\theta

∫4arctan⁡(sec⁡θ−tan⁡θ)dθ\displaystyle \int \displaystyle 4 \arctan\left(\sec \theta -\tan\theta \right) d\theta

let sec⁡θ−tan⁡θ=tan⁡m\sec \theta - \tan\theta =\tan m

sec⁡θtan⁡θ−sec2θ=sec⁡2mdmdθ\displaystyle \sec \theta\tan \theta - sec^{2} \theta=\sec^{2} m \frac{dm}{d\theta}

sec⁡θ(tan⁡θ−secθ)=sec⁡2mdmdθ\displaystyle \sec \theta\left(\tan \theta - sec\theta\right)=\sec^{2} m \frac{dm}{d\theta}

−sec⁡θ(secθ−tan⁡θ)=sec⁡2mdmdθ\displaystyle -\sec \theta\left( sec\theta-\tan \theta\right)=\sec^{2} m \frac{dm}{d\theta}

−sec⁡θ(tan⁡m)=sec⁡2mdmdθ\displaystyle -\sec \theta\left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}

Now we know sec⁡2θ−tan⁡2θ=1\displaystyle \sec^{2} \theta -\tan^{2} \theta =1

(sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1\displaystyle \left(\sec\theta -\tan \theta\right)\left(\sec\theta +\tan \theta\right) =1

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

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

Now add this to sec⁡θ−tan⁡θ\displaystyle \sec \theta -\tan\theta which is equal to tan⁡m\tan m

sec⁡θ−tan⁡θ+sec⁡θ+tan⁡θ=1tan⁡m+tan⁡m\displaystyle \sec \theta -\tan\theta +\sec\theta +\tan \theta =\frac{1}{\tan m} +\tan m

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

2sec⁡θ=1+tan⁡2mtan⁡m=sec2mtan⁡m\displaystyle 2\sec \theta =\frac{1+\tan^{2} m}{\tan m} =\frac{sec^{2} m}{\tan m}

sec⁡θ=sec2m2tan⁡m\displaystyle \sec \theta=\frac{sec^{2} m}{2\tan m}

going back a few steps :

−sec⁡θ(tan⁡m)=sec⁡2mdmdθ\displaystyle -\sec \theta\left( \tan m\right)=\sec^{2} m \frac{dm}{d\theta}

=−sec2m2tan⁡m(tan⁡m)=sec⁡2mdmdθ=\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⁡(sec⁡θ−tan⁡θ)dθ\displaystyle \int \displaystyle 4 \arctan\left(\sec \theta -\tan\theta \right) d\theta

=∫4arctan⁡(tan⁡m)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 :

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

tan⁡m=sec⁡θ−tan⁡θ=1−sin⁡θcos⁡θ\tan m =\sec \theta -\tan \theta =\frac{1-\sin \theta}{\cos \theta}

So limits in terms of tan⁡m\tan m is equal to 1−sin⁡π6cos⁡π6\displaystyle \frac{1-\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}} to 1−sin⁡π2cos⁡π2\displaystyle\frac{1-\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}}

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

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

=−4m2∣pi60=\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} !!!

#Calculus

Note by Amal Hari
1 year, 6 months ago

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Comments

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, 2 months ago

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

A Former Brilliant Member - 1 year ago

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

Kushal Dey - 1 year, 2 months ago
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