Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 / A / 2

summation by parts, with $S(x)={\sum }_{n\le x}\chi (n)$, gives for an integer $N$The boundary term tends to zero when $\mathrm{Re}s>0$ because $S$ is bounded. In the original region of absolute convergence this provesFor a compact subset of $\mathrm{Re}s>0$, choose $\delta >0$ below all its real parts and bound $|s|$ there. The integral tail is bounded by a constant times ${\int }_N^{\infty }{x}^{-1-\delta }dx$, uniformly on that compact set. Its truncated integrals are holomorphic function in $s$, so locally uniform convergence proves a analytic continuation throughout $\mathrm{Re}s>0$. This is holomorphy of a Dirichlet series from bounded partial sums.

The Euler product, valid for $\mathrm{Re}s>1$, isExpand the logarithms of these absolutely convergent Euler products. Writing $\theta =mt\log p$, the logarithm of the product in the desired inequality isFor odd $p$, $\chi (p{)}^m$ is $1$ or $-1$. In the first case the numerator is the given nonnegative trigonometric polynomial. In the second replace $\theta$ by $\theta +\pi$, which changes the sign of $\cos \theta$ and leaves $\cos 2\theta$ unchanged. At $p=2$ the numerator is $3+\cos 2\theta \ge 2$. Every summand is nonnegative. Exponentiation provesThis is the three-four-one inequality for Euler products applied to the real character modulo four.

Suppose $t\ne 0$ and $L(1+it)=0$. Holomorphy gives $L(\sigma +it)=O(\sigma -1)$ as $\sigma \downarrow 1$. The only pole of the Riemann zeta function is at one, so $\zeta (\sigma +2it)$ remains bounded, while $\zeta (\sigma {)}^3=O((\sigma -1{)}^{-3})$. The displayed product is then $O(\sigma -1)$, contradicting its lower bound. Hence $L(1+it)\ne 0$ for $t\ne 0$.

At $t=0$ the last zeta factor also has a pole, so that contradiction no longer follows from the inequality. Instead evaluate the convergent alternating series:For rigor, integrate the finite geometric identity for $(1+x^2{)}^{-1}$; the remainder has absolute integral at most $1/(2N+3)$ and tends to zero. summation by parts above identifies this value with the analytic continuation at one.