Question

Prove that :-


`\quad \leadsto \quad \bf \zeta ( z ) = \displaystyle \bf \sum_(\bf n=1)^(\infty) \bf (1)/(n^x) \cdot e^(-iy \cdot log ( n ))`

Where , z = x + yi

Riemann Hypothesis :) ​

Answer 1

We have

`e^(-iy \log(n)) = e^{\log(n^(-iy))} = n^(-iy)`

and

`\frac1{n^x} \cdot e^(-iy \log(n)) = (n^(-iy))/(n^x) = \frac1{n^(x+iy)} = \frac1{n^z}`

The sum

`\displaystyle \sum_(n=1)^\infty \frac1{n^z}`

is exactly the definition of the Riemann zeta function
`\zeta(z)`.