Estimation for integral I am trying to show that
$$\lim_{R\to\infty}\left | \int_{0}^{R}e^{-(R+iu)}(R+iu)^{s-1}\,\mathrm{d}u \right |=0$$
and
$$\lim_{R\to\infty}\left | \int_{0}^{R}e^{-(u+iR)}(u+iR)^{s-1}\,\mathrm{d}u \right |=0$$
for $s\in \mathbb{C}$ and $\Re s<1$ (I did it numerically), respectively. Here is $R>0$. It is obvious that $\left | e^{-(R+iu)} \right |=e^{-R}$ and $\left | e^{-(u+iR)} \right |=e^{-u}$. But it seems that I can't find some nice inequalities to $\left | (R+iu)^{s-1} \right |$ or $\left | (u+iR)^{s-1} \right |$.
 A: Let $z,w
 $, complex numbers and $\textrm{Re}\left(z\right)>0
 $. We have that $$z^{w}=\left|z\right|^{w}\exp\left(iw\arctan\left(\frac{\textrm{Im}\left(z\right)}{\textrm{Re}\left(z\right)}\right)\right)
 $$ $$=\left|z\right|^{\textrm{Re}\left(w\right)+i\textrm{Im}\left(w\right)}\exp\left(\left(i\textrm{Re}\left(w\right)-\textrm{Im}\left(w\right)\right)\arctan\left(\frac{\textrm{Im}\left(z\right)}{\textrm{Re}\left(z\right)}\right)\right)
 $$ so we have $$\left|z^{w}\right|=\left|z\right|^{\textrm{Re}\left(w\right)}\exp\left(-\textrm{Im}\left(w\right)\arctan\left(\frac{\textrm{Im}\left(z\right)}{\textrm{Re}\left(z\right)}\right)\right)
 $$ hence $$\left|\left(R+iu\right)^{s-1}\right|=\left(\sqrt{R^{2}+u^{2}}\right)^{\textrm{Re}\left(s\right)-1}\exp\left(-\textrm{Im}\left(s\right)\arctan\left(\frac{u}{R}\right)\right)\rightarrow0
 $$ as $R\rightarrow\infty
 $ since $\textrm{Re}\left(s\right)<1
 $. The other case is similar since $\arctan\left(x\right)
 $ is a bounded function.
A: You can estimate $ |(R+iu)^{s-1}|$ as follows. Let $s=r (\cos(\theta)+i\sin(\theta))$ and $R+iu=|R+iu|e^{i\phi} $; then
$$
(R+iu)^{s-1}=|R+iu|^{s-1} e^{i\phi(s-1)}=|R+iu|^{r \cos(\theta)-1+ir\sin(\theta)} e^{i\phi(r\cos\theta-1)-r\phi\sin\theta}
$$
hence
$$
|(R+iu)^{s-1}|=|R+iu|^{r \cos(\theta)-1}e^{-r\phi\sin\theta}\leq (R^2+u^2)^{\frac{1}{2}(r \cos(\theta)-1)}e^{2\pi r}\leq (R^2+u^2+1)^{r/2}e^{2\pi r}
$$
