Tag: Euler

One of the most famous identities in mathematics is

$$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots =\frac{{\pi }^2}{6}.$$

In summation notation,

$$\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}.$$

This is known as the Basel problem. Euler famously solved it in the eighteenth century. There are many proofs, but one particularly beautiful approach uses a double integral and an unexpected change of variables.

Start with the odd terms

Instead of attacking the entire series immediately, consider only the reciprocals of the odd squares:

$$S=1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots .$$

We will first prove that

$$S=\frac{{\pi }^2}{8}.$$

The full Basel sum will then follow almost immediately.

Turn the series into a double integral

Consider

$$I={\int }_0^1{\int }_0^1\frac{1}{1-x^2{y}^2}dxdy.$$

For points inside the unit square, the geometric-series identity gives

$$\frac{1}{1-x^2{y}^2}=\sum _{n=0}^{\infty }{xy}^{2n}.$$

Therefore,

$$I=\sum _{n=0}^{\infty }\left({\int }_0^1{x}^{2n}dx\right)\left({\int }_0^1{y}^{2n}dy\right).$$

Each one-dimensional integral is

$${\int }_0^1{x}^{2n}dx=\frac{1}{2n+1}.$$

Hence

$$I=\sum _{n=0}^{\infty }\frac{1}{{(2n+1)}^2}.$$

Thus the double integral is exactly the sum of the reciprocals of the odd squares:

$$I=1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots .$$

The key change of variables

Now comes the surprising part. Introduce new variables u and v by

$$x=\frac{\sin \left(u\right)}{\cos \left(v\right)},y=\frac{\sin \left(v\right)}{\cos \left(u\right)}.$$

The square

$$0\le x\le 1,0\le y\le 1$$

corresponds to the triangular region

$$u\ge 0,v\ge 0,u+v\le \frac{\pi }{2}.$$

To see where the last boundary comes from, notice that

$$x\le 1\iff \sin \left(u\right)\le \cos \left(v\right)\iff u+v\le \frac{\pi }{2},$$

and the condition on y gives the same inequality.

The Jacobian

We compute

$$\frac{\partial x}{\partial u}=\frac{\cos \left(u\right)}{\cos \left(v\right)},\frac{\partial x}{\partial v}=\frac{\sin \left(u\right)\sin \left(v\right)}{{\cos \left(v\right)}^2}.$$

Similarly,

$$\frac{\partial y}{\partial u}=\frac{\sin \left(u\right)\sin \left(v\right)}{{\cos \left(u\right)}^2},\frac{\partial y}{\partial v}=\frac{\cos \left(v\right)}{\cos \left(u\right)}.$$

After simplifying the determinant, the Jacobian is

$$\frac{\partial (x,y)}{\partial (u,v)}=1-x^2{y}^2.$$

This is exactly the expression that appears in the denominator of our original integral. Therefore,

$$\frac{dxdy}{1-x^2{y}^2}=dudv.$$

The complicated-looking integrand has completely disappeared.

The integral becomes an area

Our double integral is now simply

$$I={\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{\pi }{2}-u}dvdu.$$

Geometrically, this is the area of a right triangle whose two perpendicular sides both have length

$$\frac{\pi }{2}$$

Therefore,

$$I=\frac{1}{2}·\frac{\pi }{2}·\frac{\pi }{2}=\frac{{\pi }^2}{8}.$$

We have proved that

$$1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots =\frac{{\pi }^2}{8}.$$

Recovering the full series

Let

$$T=\sum _{n=1}^{\infty }\frac{1}{n^2}.$$

Split the series into its odd and even terms. The odd terms have sum

$$\frac{{\pi }^2}{8}$$

while the even terms have sum

$$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots =\frac{1}{4}T.$$

Consequently,

$$T=\frac{{\pi }^2}{8}+\frac{1}{4}T.$$

Thus,

$$\frac{3}{4}T=\frac{{\pi }^2}{8},$$

and finally,

$$T=\frac{{\pi }^2}{6}.$$

Why this proof is remarkable

We started with an infinite series involving nothing but reciprocals of squares. We then represented part of that series by a double integral over a square. A carefully chosen trigonometric change of variables transformed the square into a triangle and, at the same time, made the integrand disappear.

The infinite series was therefore reduced to the area of a triangle:

$$\frac{1}{2}·\frac{\pi }{2}·\frac{\pi }{2}=\frac{{\pi }^2}{8}.$$

From there, separating the odd and even terms gives the celebrated result

$$\sum _{n=1}\frac{1}{n^2}=\frac{{\pi }^2}{6}.$$

It is a striking example of how an infinite series, a double integral, a trigonometric substitution, and a simple geometric area can all describe the same number.

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Another Problem Where Two Dimensions Help

The double-integral proof of the Basel sum illustrates a powerful mathematical idea: sometimes a problem becomes easier when we move to a higher dimension.

One of the most beautiful examples is the Gaussian integral. A one-dimensional integral that resists ordinary antiderivative methods becomes accessible after it is squared and transformed into a two-dimensional integral.

Continue exploring: The Gaussian Integral and Beyond: From e^(-x²) to a Family of Integrals

Long before Euler solved the Basel problem, ancient Greek mathematicians had developed ingenious geometric methods for evaluating infinite sums. In particular, Archimedes used areas of triangles to establish a remarkable infinite series identity. Discover his method in How the Ancient Greeks Summed Infinite Series Without Calculus .