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I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"

Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$

I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.

I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.

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I would say that most of the standard proofs for computing $\zeta(2)$ applies to $\zeta(2n)$ too (here $n>0$ is an integer). – AD. Mar 21 '11 at 20:07
    
@AD: Yeah, that issue is part of what I was trying to get at with my question: Are there any proofs that $\zeta(4) = \pi^4/90$ that aren't just adaptions of a proof that $\zeta(2) = \pi^2/6$? I probably should have made that more explicit. – Mike Spivey Mar 21 '11 at 22:58
    
I wondering what does the $\zeta$ represent? Is that of any significance or just a variable? – night owl Feb 27 '12 at 2:40
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@nightowl: It refers to the Riemann zeta function. – Mike Spivey Feb 27 '12 at 3:22
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@AD. why does it not work for $\zeta(2n+1)$ (i.e., for odd $n$)? – Jessy Cat May 4 at 23:41

10 Answers 10

up vote 44 down vote accepted

In the same spirit of the 1st proof of this answer. If we substitute $x$ for $\pi $ in the Fourier trigonometric series expansion of $% f(x)=x^{4}$, with $-\pi \leq x\leq \pi $,

$$x^{4}=\frac{1}{5}\pi ^{4}+\sum_{n=1}^{\infty }\frac{8n^{2}\pi ^{2}-48}{n^{4}}\cos n\pi \cdot \cos nx,$$

we obtain

$$\begin{eqnarray*} \pi ^{4} &=&\frac{1}{5}\pi ^{4}+\sum_{n=1}^{\infty }\frac{8n^{2}\pi ^{2}-48}{% n^{4}}\cos ^{2}n\pi \\ &=&\frac{1}{5}\pi ^{4}+8\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}}% -48\sum_{n=1}^{\infty }\frac{1}{n^{4}}\text{.} \end{eqnarray*}$$

Hence

$$\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{\pi ^{4}}{48}\left( -1+\frac{1}{5}+ \frac{8}{6}\right) =\frac{\pi ^{4}}{48}\cdot \frac{8}{15}=\frac{1}{90}\pi ^{4}.$$

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Américo Tavares, can this proof be extended to calculate, recursively, values of $\zeta(2n)$ for larger values of $n$? – Mike Spivey Mar 21 '11 at 18:35
    
Mike Spivey, I think so, but I am not quite sure. One that can be extended for sure is the second proof in my answer math.stackexchange.com/questions/8337/…. – Américo Tavares Mar 21 '11 at 18:41
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Thanks! (extra characters) – Mike Spivey Mar 21 '11 at 18:49
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@Mike Spivey, It can. I posted the computation in this Portuguese post problemasteoremas.wordpress.com/2011/05/25/… . I got $$x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\left( \cos nx\cdot I_{2p}\right) ,$$ where $$I_{2p}=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x$$ satisfies $$I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p\left( 2p-1\right) }{% n^{2}}I_{2\left( p-1\right) }\qquad I_{0}=0$$ – Américo Tavares May 27 '11 at 10:39
    
"Substitute x for π", not the other way around. – marshaul Apr 10 '14 at 14:25

Consider the function $f(t):=t^2\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. Developping $f$ into a Fourier series we get $$t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).$$ If we put $t:=\pi$ here we easily find $\zeta(2)={\pi^2\over6}$. For $\zeta(4)$ we use Parseval's formula $$\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2\ .$$ Here $$\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}$$ and the $c_k$ are the complex Fourier coefficients of $f$. Therefore $c_0={\pi^2\over3}$ and $|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}$ $\ (k\geq1)$. Putting it all together gives $\zeta(4)={\pi^4\over 90}$.

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Thanks for your answer, Christian. – Mike Spivey Mar 22 '11 at 17:56
    
I believe this is exercise 14, of Chapter 8 of Rudin's Principles of Mathematical Analysis 3E. (Page 199) – Eric Naslund Mar 29 '11 at 16:21

If you are specially interested only in $\zeta(4)$, the following proof would work but this is an adaptation Euler's idea. The idea is just to mimic Euler's proof for the Basel problem. Euler looks at the function whose zeros are at $\pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$

To evaluate $\zeta(4)$, we can mimic Euler's idea and look at roots at $\pm \pi, \pm i \pi,\pm 2 \pi, \pm 2 i \pi,\pm 3 \pi, \pm 3 i \pi$.

Let $$p(z) = \left(1 - \left(\frac{z}{i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{2 i \pi}\right)^4 \right) \times \left(1 - \left(\frac{z}{3 i \pi}\right)^4 \right) \times \cdots$$

It is not hard to guess that $p(z)$ is same as $$\frac{i \sin(z) \times \sin \left( \frac{z}{i} \right)}{z^2} = \left(1-\frac{z^2}{3!} + \frac{z^4}{5!} -\cdots \right) \times \left(1+\frac{z^2}{3!} + \frac{z^4}{5!} + \cdots \right)$$

Compare the coefficient of $z^4$ to get $$\zeta(4) = \frac{\pi^4}{90}$$

This proof could be extended for any even number to give that $$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $$

As expected for odd numbers, this doesn't work. For instance for $3$, if you try to work out by looking at $$p(z) = \left(1 - \left(\frac{z}{\omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{2 \omega \pi}\right)^3 \right) \times \left(1 - \left(\frac{z}{3 \omega \pi}\right)^3 \right) \times \cdots$$ where $\omega^3 = 1$ there is an asymmetry since $$\sin(z) \sin \left( \frac{z}{\omega}\right) \sin \left( \frac{z}{\omega^2}\right)$$ extends on both sides and the non-zero roots are at $$\pm \pi,\pm \omega \pi,\pm \omega^2 \pi,\pm 2 \pi,\pm 2 \omega \pi,\pm 2 \omega^2 \pi,\pm 3 \pi,\pm 3 \omega \pi,\pm 3 \omega^2 \pi,\ldots$$ and hence the $\zeta(3)$ terms nicely hides by canceling out and the resulting expression only gives $\zeta(6)$.

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Thanks, Sivaram. – Mike Spivey May 27 '11 at 14:02

Consider the contour integral $$ \oint_C\frac{\pi\cot\pi z}{z^4}\ dz $$ where $C$ is the counter-clockwise square contour centered at origin with vertices $\left(N+\frac12\right)(\pm1\pm i)$.


Lemma

Suppose that the function $\phi(z)$ is holomorphic at $z=n\in\mathbb{Z}$ with $\phi(n)\neq0$, then $\pi\phi(z)\cot\pi z$ has a simple pole at $n$ with residue $\phi(n)$.

Proof

Note that $\tan\pi z$ have simple zeros at $z=n$, hence $\pi\phi(z)\cot\pi z$ have simple poles there and $$ \text{Res}\left[\pi\phi(z)\cot\pi z\ ;\ n\right]=\text{Res}\left[\frac{\pi\phi(z)}{\tan\pi z}\ ;\ n\right]=\frac{\pi\phi(n)}{\pi\sec^2\pi n}=\phi(n). $$


Thus, by residue theorem for $z\neq0$ we obtain $$ \sum_{n=-N}^N\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=n\right]=\sum_{n=-N,\ n\neq0}^N\frac1{n^4}=2\sum_{n=1}^N\frac1{n^4}.\tag1 $$ From the Taylor series of $\cot\pi z$ at $z=0$ we obtain \begin{align} \frac{\pi\cot\pi z}{z^4}&=\frac\pi{z^4}\cos\pi z\csc\pi z\\ &=\frac\pi{z^4}\left(1-\frac{(\pi z)^2}{2!}+\frac{(\pi z)^4}{4!}-\frac{(\pi z)^6}{6!}+\cdots\right)\left(\frac1{\pi z}+\frac{\pi z}{6}+\frac{7(\pi z)^3}{360}+\cdots\right)\\ &=\frac1{z^5}\left(1-\frac{(\pi z)^2}{2!}+\frac{(\pi z)^4}{4!}-\frac{(\pi z)^6}{6!}+\cdots\right)\left(1+\frac{(\pi z)^2}{6}+\frac{7(\pi z)^4}{360}+\cdots\right)\\ \end{align} Expanding the series above, we see that $$ \text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=0\right]=-\frac{\pi^4}{2!\cdot6}+\frac{\pi^4}{4!}+\frac{7\pi^4}{360}=-\frac{\pi^4}{45}.\tag2 $$ Observe that at any point on the boundary, we have $$ \left|\frac{\pi\cot\pi z}{z^4}\right|\le\frac{\pi\coth\frac\pi2}{\left(N+\frac12\right)^4}.\tag3 $$


Proof

Putting $z=x+iy$ and using the trigonometric sum formulas and basic identities, we have $$ |\cot\pi z|^2=\left|\frac{\cos\pi z}{\sin\pi z}\right|=\frac{\sinh^2\pi y+\cos^2\pi x}{\cosh^2\pi y-\cos^2\pi x}. $$ On the vertices sides of contour $C$, we have $x=\pm\left(N+\frac12\right)$ giving $\cos\left(N+\frac12\right)\pi=0$, hence $$ |\cot\pi z|=|\tanh\pi y|\le1. $$ On the horizontal sides we have $0\le\cos^2\pi x\le1$, hence $$ |\cot\pi z|^2\le\frac{\sinh^2\pi y+1}{\cosh^2\pi y-1}=\frac{\cosh^2\pi y}{\sinh^2\pi}=\coth^2\pi y. $$ Therefore $$ |\cot\pi z|\le\coth\pi y=\coth\left(N+\frac12\right)\pi\le\coth\frac\pi2 $$ Thus, on the boundary of contour $C$ we have $$ |\cot\pi z|\le\max\left[1,\coth\frac\pi2\right]=\coth\frac\pi2 $$


From $(3)$ and the property $$ \left|\int_C f(z)\ dz\right|\le ML, $$ where $M$ is $\max|f(z)|$ on C and $L$ is the length of $C$, we obtain $$ \oint_C\frac{\pi\cot\pi z}{z^4}\ dz\le\frac{\pi\coth\frac\pi2}{\left(N+\frac12\right)^4}\cdot8\left(N+\frac12\right)=\frac{8\pi\coth\frac\pi2}{\left(N+\frac12\right)^3}\to0 $$ as $N\to\infty$. Thus, using $(1)$ and $(2)$ also using residue theorem we obtain \begin{align} \lim_{N\to\infty}\frac{1}{2\pi i}\oint_C\frac{\pi\cot\pi z}{z^4}\ dz&=0\\ \sum_{n=-\infty}^\infty\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=n\right]+\text{Res}\left[\frac{\pi\cot\pi z}{z^4}\ ;\ z=0\right]&=0\\ 2\sum_{n=1}^\infty\frac1{n^4}-\frac{\pi^4}{45}&=0\\ \large\color{blue}{\sum_{n=1}^\infty\frac1{n^4}}&\large\color{blue}{=\frac{\pi^4}{90}}. \end{align}

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From How many ways to calculate: $\sum_{n=-\infty}^{+\infty}\frac{1}{(u+n)^2}$ where $u \not \in \Bbb{Z}$, we know that $$ \sum_{n=-\infty}^\infty \frac{1}{(z+n)^2}=\frac{\pi^2}{\sin^2(\pi z)}. $$ Differentiating this twice, we have $$ \sum_{n=-\infty}^\infty \frac{1}{(z+n)^4}=\frac{\pi^4(2+\cos(2\pi x))}{3\sin^4(\pi z)}. $$ So $$ \sum_{n=1}^\infty \left(\frac{1}{(z-n)^4}+\frac{1}{(z+n)^4}\right)=\frac{\pi^4(2+\cos(2\pi z))}{3\sin^4(\pi z)}-\frac{1}{z^4}. $$ Note that the LHS of the above is analytic $z=0$ and hence $$ \sum_{n=1}^\infty\frac{1}{n^4}=\lim_{z\to 0}\frac{1}{2}\left(\frac{\pi^4(2+\cos(2\pi z)}{3\sin^4(\pi z)}-\frac{1}{z^4}\right)=\frac{\pi^4}{90}. $$,

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Why that says LHS is analytic 0? – Diego Fonseca May 2 at 3:35
    
@Diego Fonseca, the LHS is analytic and so is the RHS. Hence one can take the limit. – xpaul May 2 at 19:46
    
Your cosines have xs in their arguments instead of zs. – J.G. May 22 at 7:11

Knowing a close form for $\zeta(2)$ there is an algebraic way to get $\zeta(4)$.

Consider $f(m,n):=\dfrac{2}{mn^3}+\dfrac{1}{m^2n^2}+\dfrac{2}{m^3n}$

There is the identity:

$f(m,n)-f(m,n+m)-f(m+n,n)=\dfrac{2}{m^2n^2}$

Therefore:

$\displaystyle \sum_{m,n>0} f(m,n)-\sum_{m,n>0} f(m,n+m)-\sum_{m,n>0} f(m+n,n)=\sum_{m,n>0} \dfrac{2}{m^2n^2}$

The right member is equal to:

$\displaystyle \sum_{m=1}^{+\infty}\Big(\sum_{n=1}^{+\infty}\dfrac{2}{m^2n^2}\Big)=2\sum_{m=1}^{+\infty}\dfrac{1}{n^2}\Big(\sum_{n=1}^{+\infty}\dfrac{1}{m^2}\Big)=2\Big(\sum_{n=1}^{+\infty}\dfrac{1}{n^2}\Big)\Big(\sum_{n=1}^{+\infty}\dfrac{1}{m^2}\Big)=2\zeta(2)^2$

The second sum in the left member is equal to:

$\displaystyle \sum_{n>m>0} f(m,n)$

The third one is equal to:

$\displaystyle \sum_{m>n>0} f(m,n)$

Therefore the left member is equal to:

$\displaystyle \sum_{n=1}^{+\infty}f(n,n)=\sum_{n=1}^{+\infty}\dfrac{5}{n^4}=5\zeta(4)$

(proof found in "Quelques conséquences surprenantes de la cohomologie de $SL_2(\mathbb{Z})$, Don Zagier) http://people.mpim-bonn.mpg.de/zagier/files/tex/ConsequencesCohomologySL/fulltext.pdf

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By induction we can easily prove that for any nonnegative real numbers $a_k$ $$1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j-\sum_{1\le i<j<k\le n}a_ia_ja_k\le\prod_{k=1}^n(1-a_k)\le1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j$$ Taking $a_k=\frac{x^2}{k^2\pi^2}$,we get $$1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2-\frac{x^6}{\pi^6}\frac{\zeta_n(2)^3-\zeta_n(6)}{6}\le\prod_{k=1}^n(1-\frac{x^2}{k^2\pi^2})\le1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2$$ Since $\prod_{k=1}^{\infty}(1-\frac{x^2}{k^2\pi^2})=\frac{\sin(x)}x$(proof here), by taking $n\to\infty$ $$1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$ subtraciting $1-\frac{x^2}{\pi^2}\zeta_n(2)$ $$\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$ dividing by $x^4$ and putting $x=0$ we get $$\frac1{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2=\frac1{5!}$$ and this follows that $\zeta(4)=\frac{\pi^4}{90}$.

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This is just a sketch of one of many possible proofs.

Step1. Prove that over the interval $[0,2\pi]$, the function: $$f(x)=\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n^2}$$ is a second degree-polynomial whose graph goes through the points: $$(0,\pi^2/6),\quad (\pi,-\pi^2/12),\quad (2\pi,\pi^2/6).$$

Step2. Deduce from Lagrange interpolation that: $$ f(x) = \frac{\pi^2}{6}-\frac{x(2\pi-x)}{4}.$$

Step3. Apply Parseval's identity to $f(x)$: $$\int_{0}^{2\pi}f(x)^2\,dx = \pi\sum_{n=1}^{+\infty}\frac{1}{n^4}.$$

Step4. Prove, through the second step, that: $$\int_{0}^{2\pi}f(x)^2\, dx = \frac{\pi^5}{90}.$$

Conclusion:

$$\zeta(4)=\sum_{n=1}^{+\infty}\frac{1}{n^4} = \frac{\pi^4}{90}.$$

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3  
Duplicate of this answer. – robjohn Aug 7 '14 at 0:26
    
@robjohn: yes, I know, I just copied it from that closed answer to here for visibility, I think it is interesting. – Jack D'Aurizio Aug 7 '14 at 1:09

Here is proof I discovered involving Stokes Theorem.

Consider the quadruple integral:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{1-x_1^2x_2^2x_3^2x_4^2}dx_4dx_3dx_2dx_1.$$

Since $0<x_1,...,x_4<1$, we can expand the integrand into a geometric series:

$$\frac{1}{1-x_1^2x_2^2x_3^2x_4^2}=\sum_{n=0}^{\infty}(x_1x_2x_3x_4)^{2n}.$$

We compute: $$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}(x_1x_2x_3x_4)^{2n}dx_4dx_3dx_2dx_1$$ by noticing it the same as computing: $$\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(x_1x_2x_3x_4)^{2n}dx_4dx_3dx_2dx_1,$$ and it follows that the result of the quadruple integration is the sum: $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4}.$$ We now make a surprising change of variables, first discovered by Beukers, Calabi, and Kolk. Put:

$$x_i=\frac{\sin(u_i)}{\cos(u_{i+1})},x_4=\frac{\sin(u_4)}{\cos(u_1)},1\leq i \leq 3$$

The Jacobian determinant, worked out in the paper, http://www.staff.science.uu.nl/~kolk0101/Publications/calabi.pdf, is: $$\det \frac{\partial(x_1,...,x_4)}{\partial(u_1,...,u_4)}=1-x_1^2x_2^2x_3^2x_4^2,$$ which cancels with the integrand, and the region of integration, also worked out in the paper, is the open polytope described by the inequalities: $$0<u_i+u_{i+1}<\frac{\pi}{2},0<u_1+u_4<\frac{\pi}{2},1\leq i\leq 3$$ and here, $$0<u_1,...,u_4<\frac{\pi}{2}.$$ Denote this polytope as $U$. The volume of $U$ is equal to $\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4}$. But consider the scaled polytope $P$, defined by the inequalities: $$0<v_i+v_{i+1}<1,0<v_1+v_4<1,1\leq i\leq 3$$ and $$0<v_1,...,v_4<1.$$

It can be shown through the change of variables $u_i=\frac{\pi}{2}v_i, 1\leq i\leq 4$ that $$\text{Vol}(U)=\left(\frac{\pi}{2}\right)^4 \text{Vol}(P),$$ where $\text{Vol}$ means volume.

Now the goal is to find $\text{Vol}(P)$. Now we use Stokes theorem. Stokes theorem states that $$\int_{\partial P} \omega=\int_{P} d\omega$$ provided that $\omega$ is a smooth 3-form and $P$ is both compact and orientable manifold (which it is if you replace $<$ with $\leq$ in all the inequalities defining $P$).

For simplicity let $\omega=v_1\wedge dv_2\wedge dv_3\wedge dv_4$ because $d\omega=dv_1\wedge dv_2\wedge dv_3\wedge dv_4$. We will compute the left hand side of Stokes theorem with the defined $\omega$ to get $\text{Vol}(P)$. To do this, we need to find the faces of $P$ and explicitly parameterize them. Given the inequalities defining $P$, the faces of $P$ are the coordinate hyperplanes: $v_i=0,1\leq i\leq 4$, the faces where only $v_i+v_{i+1}=1,1\leq i \leq 3$ (and the rest of the inequalities holds) and the face where only $v_1+v_4=1$ (where the rest of the inequalities holds). This might seem like a herculean task integrating over every face, but the good news is that all of these faces produce surface integrals whose values are $0$, except for the faces where $v_1+v_2=1$ and $v_1+v_4=1$.

We first compute $\int_{F_1}\omega,$ where $F_1$ is the face of $P$ where $v_1+v_2=1$. Define the parameterization for $F_1$: $$r_1(t_1,t_2,t_3)=(v_1(t_1,t_2,t_3),v_2(t_1,t_2,t_3),v_3(t_1,t_2,t_3),v_4(t_1,t_2,t_3))=(t_1,1-t_1,t_2,t_3).$$ Notice, in this parameterization $v_1+v_2=t_1+1-t_1=1$, which is the equality constraint $v_1+v_2=1$ of $F_1$. Given $\omega=v_1\wedge dv_2 \wedge dv_3\wedge dv_4$ and substituting the parameterization into $\omega$, we get:

$$\int_{F_1}\omega=\int_{F_1}\omega(r_1)=\int_{F_1} t_1 \det \left(\frac{\partial(v_2,v_3,v_4)}{\partial(t_1,t_2,t_3)}\right)dt_1 dt_2 dt_3=\int_{F_1} -t_1 dt_1 dt_2 dt_3$$ Now we need to find the bounds of $F_1$. If we let $0<t_1<1$, since one of the inequalities of $P$ is that $0<v_1+v_4<1$ and using the definition of $r_1$, it must follow that $0<t_3<1-t_1$. We also keep in mind that another inequality of $P$ is that $0<v_2+v_3<1$, so based on this and the parameterization of $r_1$, $0<t_2<t_1$. Thus, we see:

$$\int_{F_1}\omega=\int_{0}^{1}\int_{0}^{t_1}\int_{0}^{1-t_1}-t_1 dt_3 dt_2 dt_1=\frac{-1}{12}.$$

Now define the face $F_2$ of $P$ where $x_1+x_4=1$. Let $$r_2(t_1,t_2,t_3)=(t_1,t_2,t_3,1-t_1).$$

$$\int_{F_2}\omega=\int_{F_2}\omega(r_2)=\int_{F_2} t_1 \det \left(\frac{\partial(v_2,v_3,v_4)}{\partial(t_1,t_2,t_3)}\right)dt_1 dt_2 dt_3=\int_{F_2} -t_1 dt_1 dt_2 dt_3.$$ The bounds of $F_2$ are $0<t_1<1,0<t_2<1-t_1,0<t_3<t_1$ which result from the inequalities of $P$. We see now:

$$\int_{F_2}\omega=\int_{0}^{1}\int_{0}^{1-t_1}\int_{0}^{t_1}-t_1 dt_3 dt_2 dt_1=\frac{-1}{12}.$$

Now, we need to orient $F_1$ and $F_2$ using positive orientation, which affects the signs of both integrals. That is, if for every $p \in F_1$ obtained by $r_1$, the normal vector to $F_1$ points away from the polytope $P$, then we keep the sign of $\int_{F_1}\omega$; otherwise we change the sign. Similarly, if for every $p \in F_2$ obtained by $r_2$, the normal vector to $F_2$ points away from the polytope $P$, then we keep the sign of $\int_{F_2}\omega$; otherwise we change the sign.

However, based on the cross products: $$\frac{\partial r_1}{\partial t_1}\times\frac{\partial r_1}{\partial t_2}\times\frac{\partial r_1}{\partial t_3}=(-1,-1,0,0)$$ and $$\frac{\partial r_2}{\partial t_1}\times\frac{\partial r_2}{\partial t_2}\times\frac{\partial r_2}{\partial t_3}=(-1,0,0,-1)$$

the normal vector to every $p\in F_1$ given by $r_1$ is inward pointing towards $P$, not outward pointing, and the same issue exists with every $p \in F_2$ given by $r_2$. So both integrals change sign, and we add them to get that $$\text{Vol}(P)=\int_{\partial P}\omega=\frac{1}{6}.$$ So this tells us that: $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4}=\text{Vol}(U)=\left(\frac{\pi}{2}\right)^4\frac{1}{6}=\frac{\pi^4}{96}.$$ By seeing: $$\sum_{n=1}^{\infty}\frac{1}{n^4}=\sum_{n=1}^{\infty}\frac{1}{(2n)^4}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4},$$ the final result can be obtained: $$\zeta(4)=\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{2^4}{2^4-1}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4}=\left(\frac{16}{15}\right)\frac{\pi^4}{96}=\frac{\pi^4}{90}.$$

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I also want to make a remark that the parameterizations I used on both those faces aren't the only ones that work. It turns out, for example you can parameterize $v_1+v_2=1$ by considering the hyperplane between $(1,0,0,0),(0,1,0,0),(1,0,1,0),(0,1,0,1),$ which has parameterization $(1-u-w,u+w,v,w)$. Now let $0<u<1,0<u+v<1,0<u+v+w<1$ and integrate $1-u-w$ over that region (with positive orientation). – Vivek Kaushik May 26 at 16:04

Old thread, but one of my favorite methods is to construct a Fourier series that looks more or less like $$f(x)=\sum_{n=1}^\infty \frac{\cos nx}{n^4}$$ Functions which have a finite number of discontinuities in their periodic extension have Fourier coefficients that are $O(n^{-1})$, so functions continuous through their second derivative should have coefficients $O(n^{-4})$. For an even function of $x$, the even derivatives take care of themselves so only the odd derivatives need be forced to $0$ at the endpoints, and the only odd derivative the matters for $\zeta(4)$ is the first. If we consider the fundamental interval to be $[-\pi,\pi]$, such a function would be $$f(x)=\left(x^2-\pi^2\right)^2$$ A slight improvement would be to find the average of $f(x)$ $$\langle f(x)\rangle=\frac1{2\pi}\int_{-\pi}^{\pi}f(x)dx=\frac1{2\pi}\int_{-\pi}^{\pi}\left(x^4-2\pi^2x^2+\pi^4\right)dx=\frac1{2\pi}(2\pi^5)\frac8{15}=\frac{8\pi^4}{15}$$ And subtract it so that $$g(x)=\left(x^2-\pi^2\right)^2-\frac{8\pi^4}{15}$$ has the right kind of continuity and zero average. It's an even function with period $2\pi$ so we can represent it as $$g(x)=\sum_{n=1}^{\infty}a_n\cos nx$$ Then $$\begin{align}\int_{-\pi}^{\pi}g(x)\cos nx\,dx&=\int_{-\pi}^{\pi}\left[\left(x^2-\pi^2\right)^2-\frac{8\pi^4}{15}\right]\cos nx\,dx\\ &=\left[\left(\frac{\left(x^2-\pi^2\right)^2}{n}-\frac{8\pi^2}{15n}-\frac{12x^2-4\pi^2}{n^3}+\frac{24}{n^5}\right)\sin nx\right.\\ &\left.+\left(\frac{4x\left(x^2-\pi^2\right)}{n^2}-\frac{24x}{n^4}\right)\cos nx\right]_{-\pi}^{\pi}\\ &=-\frac{48\pi}{n^4}(-1)^n\\ &=\sum_{k=1}^{\infty}a_k\int_{-\pi}^{\pi}\cos kx\cos nx\,dx\\ &=\sum_{k=1}^{\infty}a_k\pi\delta_{kn}=\pi a_n\end{align}$$ Where we have used tabular integration to accelerate integration by parts, the Sturm-Liouville properties of the Fourier series to evaluate the orthogonality integrals, and the average value of $\cos^2nx$ of $\frac12$ over the interval of width $2\pi$ to evaluate the normalization integrals. Then for $x\in[-\pi,\pi]$, $$\left(x^2-\pi^2\right)^2-\frac{8\pi^4}{15}=-48\sum_{n=1}^{\infty}\frac{(-1)^n}{n^4}\cos nx$$ When $x=\pi$ for example, $$-\frac{8\pi^4}{15}=-48\zeta(4)$$ Which is equivalent to the desired result. With Parseval's theorem we could get $\zeta(8)$ out of this as well.

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