As already mentioned on my title i need to estimate a contour integration on a straight line and have to show the value is $0$ when $R$ goes to $\infty$. Here is my contour:
I'm working on $\Psi^2$. The function itself is $$e^{zt}\operatorname{Log}\left(\frac{z+1}{z}\right),\quad t>0$$. And for your information, the straight vertical line goes from $\sigma-i\infty$ to $\sigma+i\infty$.
My parametrization: $$\Psi_2: z=-\xi+iR,\quad \xi\in[-\sigma,0]$$
The reason i wrote $-\xi$ is i don't want to write $\xi\in[\sigma,0]$ which the notation of the interval is wrong since $\sigma>0$
So, $$I_{\Psi_2}:=\int_{\Psi_2} e^{zt}\operatorname{Log}\left(\frac{z+1}{z}\right)\Bbb dz = -\int_{-\sigma}^0 e^{(-\xi+iR)t}\operatorname{Log}\left(\frac{-\xi+iR+1}{-\xi+iR}\right)\Bbb d\xi. $$
Then, for estimating the integral i need to put a modulus on it: $$\begin{align} \left|I_{\Psi_2}\right| &\leq \left| -\int_{-\sigma}^0 e^{(-\xi+iR)t}\operatorname{Log}\left(\frac{-\xi+iR+1}{-\xi+iR}\right)\Bbb d\xi\right|\\ &\leq \int_{-\sigma}^0 \left|e^{(-\xi+iR)t}\right| \left|\operatorname{Log}\left(\frac{-\xi+iR+1}{-\xi+iR}\right)\right|\Bbb d\xi \end{align}$$
Because $\left|e^z\right|\leq e^{|z|}$ and $|(-\xi+iR)t| = t|-\xi+iR| \leq t(|-\xi|+|iR|)\leq t|R-|\xi||$. Since $R$ is a big number then the last expression is $t(R-\xi)$ and $|e^{iRt}|=1$.
Then using the Laurent series at $z=\infty$ $$\operatorname{Log}\left(1+\frac 1z\right) = \frac 1z - \frac{1}{2z^2} + \mathcal{O}\left(\left(\frac 1z\right)^3\right)\approx \frac 1z$$
Then
$$\begin{align} \int_{-\sigma}^0 \left|e^{(-\xi+iR)t}\right| \left|\operatorname{Log}\left(\frac{-\xi+iR+1}{-\xi+iR}\right)\right|\Bbb d\xi &\lessapprox \int_{-\sigma}^0 e^{\xi t} \frac{1}{R-\xi}\Bbb d\xi\\ &\leq \frac{1}{R} \int_{-\sigma}^0 e^{\xi t} \Bbb d\xi \tag{$\sup\{\xi\in[-\sigma,0]\}=0$}\\ &= \frac{1}{Rt}(1-e^{-\sigma t}) \end{align}$$
Which approaches to $0$ as $R\to\infty$. But is this valid? When i tried to check the graph (real plane), seems like $\frac 1x > \ln(1+1/x)$ for $x>0$.
My second doubt is using $\sup\{\xi\in[-\sigma,0]\}=0$ for the denominator to get rid of the variable and makes it easy to calculate since if i include the variable, i will come to something involving exponential integral $\operatorname{E}_1(x)$
Please, if you have better approach to solve this ML inequality and show the thr value of the integral is $0$ as $R\to \infty$, kindly tell me. Thanks in advance!