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# analytic continuation of Riemann zeta (using integral)

The Riemann zeta function can be analytically continued to the whole complex plane minus the point 1 by means of an integral representation. Remember that the zeta functon is defined by the series

$\zeta(s)=\sum_{{n=1}}^{\infty}{1\over n^{s}}.$ |

When $\Re s>1$, this series converges; furthermore, this convergence is uniform on compact subsets of this half-plane, hence the series converges to an analytic function on this half plane. However, the series diverges when we have $\Re s<1$, so this series cannot be used to define the zeta function in the whole complex plane, which is why we must make an analytic continuation.

To make this continuation, we start by changing the variable in an integration:

$n^{s}\int_{0}^{{\infty}}e^{{-nx}}x^{{s-1}}\,dx=\int_{0}^{{\infty}}e^{{-y}}y^{{% s-1}}\,dy=\Gamma(s)$ |

This provides us with an integral representation of our summand. Substituting this into the series, we find that

$\zeta(s)=\sum_{{n=1}}^{\infty}{1\over\Gamma(s)}\int_{0}^{{\infty}}e^{{-nx}}x^{% {s-1}}\,dx={1\over\Gamma(s)}\sum_{{n=1}}^{\infty}\int_{0}^{{\infty}}e^{{-nx}}x% ^{{s-1}}\,dx.$ |

We note that

$\sum_{{n=1}}^{\infty}\int_{0}^{{\infty}}\left|e^{{-nx}}x^{{s-1}}\right|\,dx=% \sum_{{n=1}}^{\infty}\int_{0}^{{\infty}}e^{{-nx}}x^{{|s-1|}}\,dx=\sum_{{n=1}}^% {\infty}{\Gamma(|s-1|+1)\over n^{s}};$ |

because the series converges, hence it is possible to interchange integration and summation and subsequently sum a geometric series.

$\zeta(s)={1\over\Gamma(s)}\sum_{{n=1}}^{\infty}\int_{0}^{{\infty}}e^{{-nx}}x^{% {s-1}}\,dx={1\over\Gamma(s)}\int_{0}^{{\infty}}\sum_{{n=1}}^{\infty}e^{{-nx}}x% ^{{s-1}}\,dx={1\over\Gamma(s)}\int_{0}^{{\infty}}{x^{{s-1}}\over e^{x}-1}\,dx$ |

As it stands, the integral representation we have is not of much use for analytically continuing the zeta function because the integral diverges when $\Re s<1$ on account of the fact that the integrand behaves like $x^{{-s}}$ when $x$ is close to zero. However, it is possible to make use of the theorem of Cauchy to move the path of integration away from zero.

Given a real number $r>0$, define the contour $C_{r}$ on the Riemann surface of $z^{{s-1}}$ as follows: $C_{r}$ passes from $+\infty$ to $r$ along a lift of the real axis, then continues along the circle of radius $r$ clockwise, and finally goes from $r$ to $+\infty$.

We now examine the integral over such a contour by breaking it into three pieces.

$\int_{{C_{r}}}{x^{{s-1}}\over e^{x}-1}\,dx=\int_{r}^{\infty}{x^{{s-1}}\over e^% {x}-1}\,dx+\int_{{|x|=r}}{x^{{s-1}}\over e^{x}-1}\,dx-\int_{r}^{\infty}{(e^{{-% 2\pi i}}x)^{{s-1}}\over e^{x}-1}\,dx.$ |

We may estimate the third integral in absolute value like so:

$\left|\int_{{|x|=r}}{x^{{s-1}}\over e^{x}-1}\,dx\right|\leq 2\pi r\sup_{{|x|=r% }}\left|{x^{{s-1}}\over e^{x}-1}\right|$ |

The expression $x/(e^{x}-1)$ represents an analytic function of $x$, and hence a bounded function of $x$ in a neighborhood of $0$. When $\Re s>1$, it happens that $\lim_{{x\to 0}}|x|x^{{s-2}}=0$, so

$\lim_{{r\to 0}}\left|\int_{{|x|=r}}{x^{{s-1}}\over e^{x}-1}\,dx\right|=0.$ |

The third integral differs from the first integral by a phase, so they may be combined by pulling out this common factor. When $\Re s>0$, we may take the limit as $r$ approaches $0$ after doing so to obtain the following:

$\lim_{{r\to 0}}\int_{{C_{r}}}{x^{{s-1}}\over e^{x}-1}\,dx=\left(1+e^{{2\pi i(1% -s)}}\right)\int_{0}^{\infty}{x^{{s-1}}\over e^{x}-1}\,dx$ |

Since, aside from the branch point at $0$, the only singularities of our integrand occur at multiples of $2\pi i$, it follows from Cauchy’s theorem that

$\int_{{C_{a}}}{x^{{s-1}}\over e^{x}-1}\,dx=\int_{{C_{b}}}{x^{{s-1}}\over e^{x}% -1}\,dx$ |

whenever $0<a<2\pi$ and $0<b<2\pi$, which trivially implies that

$\lim_{{r\to 0}}\int_{{C_{r}}}{x^{{s-1}}\over e^{x}-1}\,dx=\int_{{C_{r}}}{x^{{s% -1}}\over e^{x}-1}\,dx$ |

for any $r$ between $0$ and $2\pi$. Therefore,

$\zeta(s)={1\over\left(1+e^{{2\pi i(1-s)}}\right)\Gamma(s)}\int_{{C_{\pi}}}{x^{% {s-1}}\over e^{x}-1}\,dx$ |

when $\Re z>1$. This integral converges for all complex $s$ because the exponential grows more rapidly than the power. Furthermore, this integral defines an analytic function of $s$, so we have an analytic continuation of the zeta function to the whole complex plane minus the point 1.

## Mathematics Subject Classification

30B40*no label found*30A99

*no label found*

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