Let $s$ be a complex number with real part $>1$ and $x$ be a natural number.
I am given a result that $s\int_x^\infty \frac{w-x}{w^{s+1}}dw + O(|s|\int_x^\infty \frac{dw}{w^{\sigma+1}})=\frac{x^{1-s}}{s-1}+O(|s|x^{-\sigma})$ where $\sigma=Re(s)$.
I have tried to work the integrals but failed to achieve the result. This is my work so far.
$s\int_x^\infty \frac{w}{w^{s+1}}dw = s\int_x^\infty \frac{dw}{w^s}=-s\frac{w^{-s+1}}{s+1}|_x^\infty = s\frac{x^{1-s}}{s-1}.$ So I get the first part of the RHS multiplied by $s$.
For the second integral we have $s\int_x^\infty \frac{-x}{w^{s+1}}dw = -w^{-s}x|_x^\infty =x^{1-s}$.
So summing the two integrals do not give me $\frac{x^{1-s}}{s-1}$ as I wish..
Finally the integral in the big Oh term also gives $\int_x^\infty \frac{dw}{w^{\sigma+1}}=\frac{1}{\sigma}x^{-\sigma}.$ So I'm not getting $|s|x^{-\sigma}$.
What am I doing wrong here?
Edit: I have retried working the integrals and I think I got the answer. I would appreciate any comment about it.
First, $\int_x^\infty w^{-s}dw=\frac{1}{-s+1}w^{-s+1}|_x^\infty = \frac{x^{1-s}}{s-1}$.
Next, $\int_x^\infty \frac{x}{w^{s+1}}dw=x\int_x^\infty w^{-s-1}dw=x\frac{1}{-s}w^{-s}|_s^\infty = \frac{x^{1-s}}{s}=\frac{x^{1-s}}{s}.$
So $s\int_x^\infty \frac{w-x}{w^{s+1}}dw = \frac{sx^{1-s}}{s-1}-x^{1-s}=\frac{x^{1-s}}{s-1}, $ as required.
Finally, for the integral in the big oh term we have $\int_x^\infty \frac{dw}{w^{\sigma+1}}=-\frac{1}{\sigma}w^{-\sigma}|_x^\infty = \frac{1}{\sigma}x^{-\sigma}.$ Now since we assume $\sigma>1$, we have $\frac{|s|}{\sigma} <|s|$ and so $|s|\int_x^\infty \frac{dw}{w^{\sigma+1}} \ll |s|x^{-\sigma}$ as required.