# The bound behavior for an integral important in studying Chebyshev's prime counting function

I am working through a problem set in an analytic number theory course, and the following problem was included:

Let $$I(y,T) = \int_{c-iT}^{c+iT} \frac{y^s}{s} ds.$$

(1) If $y = 1$, $c>0$ and $T>0$, that $$|I(1,T)-\frac{1}{2}| \le \frac{c}{T}.$$

(2) If $y>1$, $c>0$ and $T>0$, that $$|I(y,T)-1| \le y^c \min (1, \frac{1}{T|\log y|}).$$ Also prove for $0<y<1$, $c>0$ and $T>0$, that $$|I(y,T)| \le y^c \min (1, \frac{1}{T|\log y|}).$$

(3) Comparing (1) and (2), where does the discontinuity come from when first $T \to \infty$, and then $y$ is moved? Change variables $y = e^u$ and interpret this result in terms of a Fourier transform. Explain the role of the requirement $c>0$ (versus, say, $c < 0$). Also, what would happen if for finite $T$ fixed we let $y \to \infty$ (resp., let $y \to 0$)? Compare the three answers

WARNING: The professor writing these problems has an unfortunate habit of TeX-ing problems up incorrectly! Hence part of the "fun" for students taking the course is to figure out if the statement of the problem itself is correct, and if not, to figure out how to modify the statement to make it workable.

I am wondering if anyone visiting would be able to tell whether the problem as stated is right (and if so, suggest a strategy for proving it); if the statement is false, I am curious to know if anyone could either suggest how to modify the statement to be workable, or even point me in the direction of a text that contains a correct statement.

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I'll take a stab at (1). I don't think it's written correctly.

Assume $c > 0$ and $T > 0$. We evaluate the integral and find that

$$\hspace{-1.625cm} I(1,T) = \log(c+iT) - \log(c-iT)$$ $$\hspace{1.65cm} = i\pi + \log\left(1-i\frac{c}{T}\right) - \log\left(1+i\frac{c}{T}\right).$$

Now assume that the ratio $c/T$ is small. Using the series expansion

$$\log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} + \cdots$$

and truncating at the term involving $(c/T)^3$ we get

$$I(1,T) = i\pi - 2i\frac{c}{T} + O\left(\frac{c}{T}\right)^3 \hspace{1cm} \left(\frac{c}{T} \to 0\right).$$

Thus if we take $c/T$ small enough we get, for example,

$$|I(1,T) - i\pi| \leq 3\frac{c}{T}.$$

Your professor must have meant something along these lines. Perhaps you could show that the the above inequality is true for all $c>0$ and $T>0$. Also, it is probably still valid if you replace the $3$ with a $2$.

Edit: I wanted to take a look at (2) just to set you in the right direction. Numerical experiments indicate that the integral converges to $i2\pi$ as $T \to \infty$, so the professor probably meant something like

$$|I(y,T)-i2\pi| \le y^c \min (1, \frac{1}{T|\log y|}).$$

You may be able to calculate the integral explicitly using the residue theorem with a Bromwich contour.

As for the case when $0 < y < 1$, it looks like

$$|I(y,T)| \le y^c \min (1, \frac{1}{T|\log y|})$$

is plausible.

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