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We can use the function with and find its expansion into a trigonometric Fourier series

which is periodic and converges to in .

Observing that is even, it is enough to determine the coefficients

because

For we have

And for we get

because

Thus

Since , we obtain

Therefore

Second method (available on-line a few years ago) by Eric Rowland. From

and making the substitution one gets the series expansion

whose radius of convergence is 1. Now if we take the imaginary part of both sides, the RHS becomes

and the LHS

Since

the following expansion holds

Integrating the identity , we obtain

Setting , we get the relation between C and

And for , since

we deduce

Solving for C

we thus prove

Note: this 2nd method can generate all the zeta values by integrating repeatedly . This is the reason why I appreciate it. Unfortunately it does not work for .

Note also the

can be obtained by integrating and substitute

respectively.

Here is an other one which is more or less what Euler did in one of his proofs.

The function where is zero exactly at for each integer n. If we factorized it as an infinite product we get

We can also represent as a Taylor series at :

Multiplying the product and identifying the coefficient of we see that

or

Here are two interesting links:

• Euler's papers;

• Euler’s Solution of the Basel Problem – The Longer Story an essay on the subject written by Ed Sandifer.

Define the following series for

Now substitute to arrive at

if we find the roots of we find that

for and in the integers

With all of this in mind, recall that for a polynomial

with roots

Treating the above series for as polynomial we see that

then multiplying both sides by gives the desired series.

This method apparently was used by Tom Apostol in . I will outline the main ideas of the proof, the details can be found in here or this presentation (page )

Consider

You can verify that the left hand side is indeed by letting and

I have two favorite proofs. One is the last proof in Robin Chapman's collection; you really should take a look at it.

The other is a proof that generalizes to the evaluation of for all n, although I'll do it "Euler-style" to shorten the presentation. The basic idea is that meromorphic functions have infinite partial fraction decompositions that generalize the partial fraction decompositions of rational functions.

The particular function we're interested in is , the exponential generating function of the Bernoulli numbers . B is meromorphic with poles at , and at these poles it has residue . It follows that we can write, a la Euler,

Now we can expand each of the terms on the RHS as a geometric series, again a la Euler, to obtain

because, after rearranging terms, the sum over odd powers cancels out and the sum over even powers doesn't. (This is one indication of why there is no known closed form for .) Equating terms on both sides, it follows that

or

as desired. To compute it suffices to compute that , which then gives the usual answer.

Here is one more nice proof, I learned it from Grisha Mikhalkin:

Lemma: Let Z be a complex curve in . Let be the projection of Z onto its real parts and the projection onto its complex parts. If these projections are both one to one, then the area of is equal to the area of .

Proof: There is an obvious map from to , given by lifting to , and then projecting to . We must prove this map has Jacobian 1. WLOG, translate to and let Z obey near . To first order, we have and . So and . So the derivative of is and the Jacobian is 1. QED

Now, consider the curve , where and obey the following inequalities: , , and .

Given a point on , consider the triangle with vertices at 0, and . The inequalities on the y's states that the triangle should lie above the real axis; the inequalities on the x's state the horizontal base should be the longest side.

Projecting onto the x coordinates, we see that the triangle exists if and only if the triangle inequality is obeyed. So is the region under the curve . The area under this curve is

Now, project onto the y coordinates. Set for convenience, so the angles of the triangle are . The largest angle of a triangle is opposite the largest side, so we want , , plus the obvious inequalities , . So is the quadrilateral with vertices at , , and and, by elementary geometry, this has area .

I'll post the one I know since it is Euler's, and is quite easy and stays in . (I'm guessing Euler didn't have tools like residues back then).

Let

Then

But then

Since

We have

But

which yields

since all powers are odd.

This ultimately produces:

Let

Then

Which means

or

The most recent issue of The American Mathematical Monthly (August-September 2011, pp. 641-643) has a new proof by Luigi Pace based on elementary probability. Here's the argument.

Let and be independent, identically distributed standard half-Cauchy random variables. Thus their common pdf is for .

Let . Then the pdf of Y is, for ,

Since and are equally likely to be the larger of the two, we have . Thus

This is equivalent to

which, as others have pointed out, implies .

This is not really an answer, but rather a long comment prompted by David Speyer's answer. The proof that David gives seems to be the one in How to compute by solving triangles by Mikael Passare, although that paper uses a slightly different way of seeing that the area of the region (in Passare's notation) bounded by the positive axes and the curve ,

is equal to .

This brings me to what I really wanted to mention, namely another curious way to see why has that area; I learned this from Johan Wästlund. Consider the region illustrated below for :

Although it's not immediately obvious, the area of is . Proof: The area of is 1. To get from to one removes the boxes along the top diagonal, and adds a new leftmost column of rectangles of width and heights , plus a new bottom row which is the "transpose" of the new column, plus a square of side in the bottom left corner. The kth rectangle from the top in the new column and the kth rectangle from the left in the new row (not counting the square) have a combined area which exactly matches the kth box in the removed diagonal:

Thus the area added in the process is just that of the square, . Q.E.D.

(Apparently this shape somehow comes up in connection with the "random assignment problem", where there's an expected value of something which turns out to be .)

Now place in the first quadrant, with the lower left corner at the origin. Letting gives nothing but the region : for large N and for , the upper corner of column number in lies at

hence (in the limit) on the curve .

Note that

from complex analysis and that both sides are analytic everywhere except . Then one can obtain

Now the right hand side is analytic at and hence

Note

Thus

Just as a curiosity, a one-line-real-analytic-proof I found by combining different ideas from this thread and this question:

Update. By collecting pieces, I have another nice proof. By Euler's acceleration method or just an iterated trick like my here we get:

and the last series converges pretty fast. Then we may notice that the last series comes out from a squared arcsine. That just gives another proof of .

A proof of the identity

is also hidden in tired's answer here. For short, the integral

is clearly real, so the imaginary part of the sum of residues of the integrand function has to be zero.

Still another way (and a very efficient one) is to exploit the reflection formula for the trigamma function:

immediately leads to:

2018 update. We may consider that .
On the other hand, by Feynman's trick or Fubini's theorem

and since , by expanding as a geometric series we have

Here is a complex-analytic proof.

For { }, let

where the sum is taken over all branches of the logarithm. Each point in D has a neighbourhood on which the branches of are analytic. Since the series converges uniformly away from , is analytic on D.

Now a few observations:

(i) Each term of the series tends to 0 as . Thanks to the uniform convergence this implies that the singularity at is removable and we can set .

(ii) The only singularity of R is a double pole at due to the contribution of the principal branch of . Moreover, .

(iii) .

By (i) and (iii) R is meromorphic on the extended complex plane, therefore it is rational. By (ii) the denominator of is . Since , the numerator has the form . Then (ii) implies , so that

Now, setting yields

which implies that

and the identity follows.

The proof is due to T. Marshall (American Mathematical Monthly, Vol. 117(4), 2010, P. 352).

In response to a request here: Compute where the integral is taken around a square of side . Routine estimates show that the integral goes to 0 as .

Now, let's compute the integral by residues. At , the residue is , where q is some rational number coming from the power series for . For example, if , then we get .

At , for , the residue is . So

or

as desired. In particular, .

Common variants: We can replace with , with , or with similar formulas.

This is reminiscent of Qiaochu's proof but, rather than actually establishing the relation , one simply establishes that both sides contribute the same residues to a certain integral.

Another variation. We make use of the following identity (proved at the bottom of this note):

Now for and so

With summing the inequalities from to n we obtain

Hence

Taking the limit as we obtain

from which the result for follows easily.

To prove we note that

Therefore

And so setting we note that

has roots for from which follows since

A short way to get the sum is to use Fourier's expansion of in . Recall that Fourier's expansion of is

where

and

Easy calculation shows

Letting in both sides gives

Another way to get the sum is to use Parseval's Identity for Fourier's expansion of x in . Recall that Parseval's Identity is

Note

Using Parseval's Identity gives

or

Theorem: Let be a nonincreasing sequence of positive numbers such that converges. Then both series

converge. Morevere also converges, and we have the formula

Proof: Knopp. Konrad, Theory and Application of Infinite Series, page 323.

If we let in this theorem, then we have

Hence,

and now

At risk of contravening group etiquette w.r.t. old questions, I'm going to take this opportunity to post my own version. I don't see it in a transparent form in any of the other posts or in Robin Chapman's article, so I invite anyone to point out the correspondence if it's there. I like this argument because it's physical and can be followed without mathematical formalism.

We start by assuming the well-known series for in alternating odd fractions. We can recognize it as the sum of the Fourier series of the square wave, evaluated at the origin:

It is easily argued on physical grounds that this adds up to a square wave; and that the height of the wave is pi/4 follows from the alternating sequence already mentioned. Now we are going to interpret this wave as an electric current flowing through a resistor. There are two ways of calculating the power and they must agree. First, we can just take square of the amplitude; in the case of this square wave, this is obviously a constant and it is just . The other way is to add up the power of the sinusoidal components. These are the squares of the individual amplitudes:

No, not quite; I've been a little sloppy and neglected to mention that when calculating the power of a sine wave, you use its RMS amplitude and not its peak amplitude. This introduces a factor of two; so in fact the series as written adds up to This isn't quite what we want; remember we've just added up the odd fractions. But the even fractions contribute in a rather picturesque way; it's easy to group them by powers of two into a geometric sum leading to the desired result of

I like this one:

Let , where is the space of Lipschitz functions on . So its well defined the number for (called Fourier series of f)

By the inversion formula, we have

Now take , . Note that

We have

Using the inversion formula, we have on that

Then,

\begin{eqnarray} 0 &=& \frac{\pi}{2}-\sum_{k\in\mathbb{Z}\ |k|\ odd}\frac{2}{k^{2}\pi} \nonumber \\ &=& \frac{\pi}{2}-\sum_{k\in\mathbb{N}\ |k|\ odd}\frac{4}{k^{2}\pi} \nonumber \\ \end{eqnarray}

This implies

If we multiply the last equation by with ,we get

Now

The sum in the left is equal to:

The sum in the right is equal to :

So we conclude:

Note: This is problem 9, Page 208 from the boof of Michael Eugene Taylor - Partial Differential Equation Volume 1.

Here's a proof based upon periods and the fact that and are periods forming an accessible identity.

The definition of periods below and the proof is from the fascinating introductory survey paper about periods by M. Kontsevich and D. Zagier.

Periods are defined as complex numbers whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coefficient over domains in given by polynomial inequalities with rational coefficients.

The set of periods is therefore a countable subset of the complex numbers. It contains the algebraic numbers, but also many of famous transcendental constants.

In order to show the equality we have to show that both are periods and that and form a so-called accessible identity.

First step of the proof: and are periods

There are a lot of different proper representations of showing that this constant is a period. In the referred paper above following expressions (besides others) of are stated:

showing that is a period. The known representation

shows that is also a period.

Second step: and form an accessible identity.

An accessible identity between two periods A and B is given, if we can transform the integral representation of period A by application of the three rules: Additivity (integrand and domain), Change of variables and Newton-Leibniz formula to the integral representation of period B.

This implies equality of the periods and the job is done.

In order to show that and are accessible identities we start with the integral I

Expanding as a geometric series and integrating term-by-term,

we find that

providing another period representation of .

Changing variables:

with Jacobian , we find

the last equality being obtained by considering the involution and comparing this with the last integral representation of above we obtain:

So, we have shown that and are accessible identities and equality follows.

As taken from my upcoming textbook:

There is yet another solution to the Basel problem as proposed by Ritelli (2013). His approach is similar to the one by Apostol (1983), where he arrives at

by evaluating the double integral

Ritelli evaluates in this case the definite integral shown in . The starting point comes from realizing that is equivalent to

To evaluate the above sum we consider the definite integral

We evaluate first with respect to x and then to y

where we used the substitution in the last step. If we reverse the order of integration one gets

Hence since and are the same, we have

Furthermore

where we used the substitution . Combining and yields

By expanding the denominator of the integrand in into a geometric series and using the Monotone Convergence Theorem,

Using integration by parts one can see that

Hence from , and

which finishes the proof.

References:

Daniele Ritelli (2013), Another Proof of Using Double Integrals, The American Mathematical Monthly, Vol. 120, No. 7, pp. 642-645

T. Apostol (1983), A proof that Euler missed: Evaluating the easy way, Math. Intelligencer 5, pp. 59–60, available at http://dx.doi.org/10.1007/BF03026576.

Here is Euler's Other Proof by Gerald Kimble

I saw this proof in an extract of the College Mathematics Journal.

Consider the Integeral :

From , we have:

The Taylor series expansion of

Thus , , then for I :

By evaluating we get something like this..

Hence

So now we have a real integral equal to an imaginary number, thus the value of the integral should be zero.

Thus,

But let .We get

And as a result

This popped up in some reading I'm doing for my research, so I thought I'd contribute! It's a more general twist on the usual pointwise-convergent Fourier series argument.

Consider the eigenvalue problem for the negative Laplacian on with Dirichlet boundary conditions; that is, with . Through inspection we can find that the admissible eigenvalues are for

One can verify that the integral operator

where

inverts the negative Laplacian, in the sense that on the admissible class of functions (twice weakly differentiable, satisfying the boundary conditions). That is, G is the Green's function for the Dirichlet Laplacian. Because is a self-adjoint, compact operator, we can form an orthonormal basis for from its eigenfunctions, and so may express its trace in two ways:

and

The latter quantity is

Hence, we have that

Consider the function which has poles at where n is an integer. Using the L'hopital rule you can see that the residue at these poles is 1.

Now consider the integral where the contour is the rectangle with corners given by ±(N + 1/2) ± i(N + 1/2) so that the contour avoids the poles of . The integral is bouond in the following way:

It can easily be shown that on the contour that where M is some constant. Then we have

where (8N+4) is the lenght of the contour and is half the diagonal of . In the limit that N goes to infinity the integral is bound by 0 so we have

by the cauchy residue theorem we have 2πiRes(z = 0) + 2πi Residues(z 0) = 0. At z=0 we have Res(z=0)= , and so we have

Where the 2 in front of the residue at n is because they occur twice at +/- n.

We now have the desired result .

I have another method as well. From skimming the previous solutions, I don't think it is a duplicate of any of them

In Complex analysis, we learn that which is an entire function with simple zer0s at the integers. We can differentiate term wise by uniform convergence. So by logarithmic differentiation we obtain a series for .

Therefore,

We can expand as

Thus,

I would like to present you a method I found recently here.

Let and .

The first integral is well known, by per partes we get the recurecnce relation :

By per partes for the second integral:

First two terms vanish, so we are left only with the integral and since we have :

for . Rearranging and substituing from we get :

Summing from 1 to some k natural we get by telescoping property

Next, using the inequality on and by :

so in the limit the last term vanishes by the sqeeze theorem, so we are left with

That concludes the result.

Applying the usual trick ¹ transforming a series to an integral, we obtain

where we use the Monotone Convergence Theorem to integrate term-wise.

Then there's this ingenious change of variables ², which I learned from Don Zagier during a lecture, and which he in turn got from a colleague:

One verifies that it is bijective between the rectangle and the triangle , and that its Jacobian determinant is precisely , which means would be a neater integrand. For the moment, we have found

(the area of the triangular domain in the plane).

There are two ways to transform into something ish:

• Manipulate : We have so . Applying the series-integral transformation, we get so

  

• Manipulate : Substituting we have so

  

  whence

  

(It may be seen that they are essentially the same methods.)

After looking at the comments it seems that this looks a lot like Proof 2 in the article by R. Chapman.

See also: Multiple Integral

¹ See e.g. Proof 1 in Chapman's article.
² It may have been a different one; maybe as in the above article. Either way, the idea to do something trigonometric was not mine.

This is, by no measure, the best nor the simplest approach, but I think the approach is pretty peculiar.

We estimate the number of integer solutions to as . On one hand, this is the number of lattice points inside the the 4-ball of radius , which has volume , hence .

On the other hand, let be the number of solutions to . Following the derivation in the book by Iwaniec-Kowalski, by Jacobi's four-square identity we can write

(I have copied the steps as they were in the book, it's a neat exercise to justify every transition). In particular, we have

See evaluations of Riemann Zeta Function in mathworld.wolfram.com and a solution by in D. P. Giesy in Mathematics Magazine:

D. P. Giesy, Still another elementary proof that , Math. Mag. 45 (1972) 148–149.

Unfortunately I did not get a link to this article. But there is a link to a note from Robin Chapman seems to me a variation of proof's Giesy.