$n$th Mathematics Blog

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Riemann Zeta Function: What Is $ζ(2)$

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First, start off with the integral: $$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \ln(\cos(x))\, dx } $$

Now, we know that $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$

$$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /2 }{ \ln\left(\frac { { e }^{ ix }+{ e }^{ -ix } }{ 2 } \right)\,dx } = \displaystyle \int _{ 0 }^{ \pi /2 }{ \ln\left(1+{ e }^{ 2ix }\right)\,dx } - \int _{ 0 }^{ \pi /2 }{ (\ln(2)+ix)\,dx } $$

Now, consider the taylor series expansion of $\ln(1+x)$:

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

Now, ${e}^{\pi ir} = {(-1)}^{r}$, so:

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

Now we calculate the summation:

$$S=\left(1+\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +\ldots\right)=\left(1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\ldots\right)-\left(\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\ldots\right)$$ $$=\left(1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\ldots\right)-\frac { 1 }{ 4 } \left(\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +\ldots\right)=\left(\frac { 3 }{ 4 } \right)\left(1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\ldots\right)$$ $$=\left(\frac { 3 }{ 4 } \right)\zeta (2)$$

After we substitute $S$ into our integral we get:

$$I=\frac { -\pi \ln(2) }{ 2 } +\left(\frac { 3\zeta (2) }{ 4 } -\frac {\pi^{ 2 }}{ 8 } \right)i$$

Since the integral is real, the imaginary part of our integral is $0$:

$$\frac { 3\zeta (2) }{ 4 } =\frac { { \pi }^{ 2 } }{ 8 }\Rightarrow\zeta(2) = \frac{\pi^{2}}{6}$$

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