Editing Basel problem (section)

==Proof using differentiation under the integral sign==

It's possible to prove the result using elementary calculus by applying the [[Leibniz integral rule|differentiation under the integral sign]] technique to an integral due to Freitas:<ref>{{cite arXiv|last1=Freitas|first1=F. L.|title=Solution of the Basel problem using the Feynman integral trick|year=2023|class=math.CA|eprint=2312.04608|mode=cs2}}</ref>
<math display=block>I(\alpha) = \int_0^\infty \ln\left(1+\alpha e^{-x}+e^{-2x}\right)dx.</math>

While the [[primitive function]] of the integrand cannot be expressed in terms of elementary functions, by differentiating with respect to <math>\alpha</math> we arrive at

<math display=block>\frac{dI}{d\alpha} = \int_0^\infty \frac{e^{-x}}{1+\alpha e^{-x}+e^{-2x}}dx,</math>
which can be integrated by [[Integration by substitution|substituting]] <math>u=e^{-x}</math> and decomposing into [[partial fractions]]. In the range <math>-2\leq\alpha\leq 2</math> the definite integral reduces to

<math display=block>\frac{dI}{d\alpha} = \frac{2}{\sqrt{4-\alpha^2}}\left[\arctan\left(\frac{\alpha+2}{\sqrt{4-\alpha^2}}\right)-\arctan\left(\frac{\alpha}{\sqrt{4-\alpha^2}}\right)\right].</math>

The expression can be simplified using the [[Inverse trigonometric functions#Arctangent addition formula|arctangent addition formula]] and integrated with respect to <math>\alpha</math> by means of [[trigonometric substitution]], resulting in

<math display=block>I(\alpha) = -\frac{1}{2}\arccos\left(\frac{\alpha}{2}\right)^2 + c.</math>

The [[integration constant]] <math>c</math> can be determined by noticing that two distinct values of <math>I(\alpha)</math> are related by

<math display=block>I(2) = 4I(0),</math>
because when calculating <math>I(2)</math> we can [[Factorization|factor]] <math>1+2e^{-x}+e^{-2x} = (1+e^{-x})^2</math> and express it in terms of <math>I(0)</math> using the [[List of logarithmic identities#Logarithm of a power|logarithm of a power identity]] and the [[Integration by substitution|substitution]] <math>u=x/2</math>. This makes it possible to determine <math>c = \frac{\pi^2}{6}</math>, and it follows that

<math display=block>I(-2) = 2\int_0^\infty \ln(1-e^{-x})dx = -\frac{\pi^2}{3}.</math>

This final integral can be evaluated by expanding the natural logarithm into its [[Mercator series|Taylor series]]:

<math display=block>\int_0^\infty \ln(1-e^{-x})dx = - \sum_{n=1}^\infty \int_0^\infty \frac{e^{-nx}}{n}dx = -\sum_{n=1}^\infty\frac{1}{n^2}.</math>

The last two identities imply

<math display=block>\sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}.</math>