0

These notes I am reading claim the following:

$\int_{c-iT}^{c+iT}\frac{1}{s}ds \ll \log{T}$ where $c > 0$ and $T > 2$.

I think the following is true: $\int_{c-iT}^{c+iT}|\frac{1}{s}|ds \leq 2\int_{0}^{T}\frac{1}{|c + it|}dt$ but I am not sure how to proceed.

Am I able to say $\int_{0}^{T}\frac{1}{|c + it|}dt \leq \log(c+iT) - \log{c}.$ Where I am guessing the anti-derivative of $1/|z|$ is $\log{z}$. Is this correct? Is it true then that $ \log(c+iT) - \log{c} \ll \log T$?

trynalearn
  • 1,691
  • Taking the brach of logarithm $\mathbb{C}\setminus\ell_{-\pi}$, where $\ell_{-\pi}={te^{-\pi i}: t>0}$, the function $f(z)=\frac{1}{z}$ and has primitive $F(z)=\log z=\log|z| + i\operatorname{arg}(z)$ where $|\operatorname{arg}(z)|<\pi$. – Mittens Jul 06 '23 at 00:35
  • I see so I would immediately use FTC to say the integral equals log(c + iT) - log(c-iT) = iarg(c+iT) - iarg(c-iT) (since both complex numbers have the same absolute value). but then it would seem the integral is << $\pi$ which doesnt seem corect. – trynalearn Jul 06 '23 at 01:34
  • 1
    On $(1,T)$ you can use $\frac{1}{{\left| {c + {\rm i}t} \right|}} \le \frac{1}{{{\mathop{\rm Im}\nolimits} (c + {\rm i}t)}} = \frac{1}{t}$. The integral on $(0,1)$ is of order $1$. Also, after estimating you should write $|\mathrm{d}s|$. – Gary Jul 06 '23 at 06:13
  • @trynalearn: there is nothing wrong with the argument I made In fact it is to be expected by the inversion theorem of the Laplace transform. In this case $F(s)=\frac{1}{s}$ is the Laplace transform of $f(t)=\mathbb{1}{[0,\infty)}(t)$ and so $\lim{T\rightarrow\infty}\frac{1}{2\pi I}\int^{c+iT}_{c-iT}\frac{e^{-ts}}{s},ds=\frac{f(t+)+f(t-)}{2}$ for all $t\geq0$. – Mittens Jul 07 '23 at 21:45

0 Answers0