Proof that \(\zeta(2)=\frac{{\pi}^{2}}{6}\)

Well here's another proof that \(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } }= \frac{{\pi}^{2}}{6}\)

We will start from the integral \(\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln(cos(x))dx } \)

now \(cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}\)

\(\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /2 }{ ln(\frac { { e }^{ ix }+{ e }^{ -ix } }{ 2 } )dx } \)

\( I = \displaystyle \int _{ 0 }^{ \pi /2 }{ ln(1+{ e }^{ 2ix })dx } - \int _{ 0 }^{ \pi /2 }{ (ln(2)+ix)dx } \)

Considering taylor expansion of \(ln(1+x)\) we have :

\(\displaystyle I = \int _{ 0 }^{ \pi /2 }{ \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } { e }^{ 2irx } } dx } -(\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8})\)

Changing the order of integration and summation we have :

\(\displaystyle I = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } \int _{ 0 }^{ \pi /2 }{ { e }^{ 2irx }dx } } -(\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8})\)

Integrating and putting limits we get :

\(\displaystyle I= \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ 2i{ r }^{ 2 } } ( } { e }^{ \pi ir }-1) - (\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8}) \)

Now \({e}^{\pi ir} = {(-1)}^{r} \)

Hence \(\displaystyle I = (\sum _{ r=1 }^{ \infty }{ \frac { 1-{ (-1) }^{ r } }{ 2{ r }^{ 2 } } } )i - (\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8})\)

We will be calculating the value of our summation.

\(S=(1+\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 5 }^{ 2 } } +......)\)

\(S=(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)-(\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +....)\)

\(S=(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)-\dfrac { 1 }{ 4 } (\dfrac { 1 }{ { 1 }^{ 2 } } +\dfrac { 1 }{ { 2 }^{ 2 } } +....)\)

\(S=(\dfrac { 3 }{ 4 } )(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)\)

\(S=(\dfrac { 3 }{ 4 } )\zeta (2)\)

Putting the value of \(S\) in our integral we have :

\(I=\dfrac { -\pi ln(2) }{ 2 } +(\dfrac { 3\zeta (2) }{ 4 } -\dfrac { { \pi }^{ 2 } }{ 8 } )i\)

Now since the integral is real hence imaginary part of our integral is \(0\) hence :

\(\dfrac { 3\zeta (2) }{ 4 } =\dfrac { { \pi }^{ 2 } }{ 8 } \)

Finally :

\(\zeta(2) = \dfrac{{\pi}^{2}}{6}\)

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Proof that \(\zeta(2)=\frac{{\pi}^{2}}{6}\)

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