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Let $x>0$. Calculate:

$$\lim_{R\rightarrow\infty} \int_{\frac{1}{2}-iR}^{\frac{1}{2}+iR}\frac{x^s}{s}ds$$

for $x=1$ it is quite simple. Otherwise, I used the following contours:

enter image description here

The left one when $x>1$ and the right one when $0<x<1$. Let us discuss the case $x>1$.

The integral on $\Gamma_1$ is what we want. The integral on $\Gamma_3$ vanishes since:

$$\bigg| \int_{\Gamma_3}f(s)\bigg|\leq \int_{\Gamma_3}|f(s)| \leq 2R \bigg|\frac{x^s}{s}\bigg| \leq 2R \frac{e^{\ln x \Re s}}{\sqrt{2}R}=\sqrt{2}e^ {-R\ln x} \rightarrow 0$$

But I have no idea what to do with $\Gamma_2$ and $\Gamma_4$.

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  • $\begingroup$ The Maple command $$int((exp(1))^{k*(1/2+I*y)}/(1/2+I*y), y = -infinity .. infinity, CauchyPrincipalValue) assuming\,k > 0 $$ produces $2\pi$ and $$int((exp(1))^{k*(1/2+I*y)}/(1/2+I*y), y = -infinity .. infinity, CauchyPrincipalValue) assuming\,k <0 $$ produces $-2\pi$. The command $$int(1/(1/2+I*y), y = -infinity .. infinity, CauchyPrincipalValue) $$ gives $\pi .$ $\endgroup$
    – user64494
    Jul 31 '13 at 19:08
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    $\begingroup$ @user64494: Thanks, but I'm not intersted in the final solution - just how to get to it. $\endgroup$
    – catch22
    Jul 31 '13 at 19:52
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For the parts $\Gamma_2$ and $\Gamma_4$, you can estimate the integral by estimating the integrand

$$\left\lvert \frac{\exp \bigl((\sigma \pm iR)\log x\bigr)}{\sigma \pm iR} \right\rvert \leqslant \frac{1}{R} e^{\sigma\log x}$$

and then integrating

$$\left\lvert\int_{\Gamma_{2,\,4}} \frac{x^s}{s}\, ds\right\rvert \leqslant \frac{1}{R} \int_{-R}^{1/2} e^{\sigma\log x}\, d\sigma = \frac{1}{R} \frac{x^{1/2} - x^{-R}}{\log x} \leqslant \frac{C(x)}{R}.$$

For the $x < 1$ case, the estimate is similar.

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