Estimating complex integral

Question

I am reading chapter 12 of "Lectures on the Riemann Zeta function" by H. Iwaniec. I am stuck at understanding the computation done at the beginning of page 45. We want to estimate the following integral $$ I_1(s)=\frac{1}{2\pi i}\int_{(-1/4)}\Gamma(w)y^{-w}\zeta(1-s-w) \,dw $$ where $(-1/4)$ should denote the vertical line at $\text{Re}(w)=-1/4$ and $s=1/2+it$. He says that to compute this we have to move the integration to the line $\text{Re}(w)=-3/4$ and then expand $\zeta(1-s-w)$ into its Dirichlet series, interchange the summation and integration to eventually get $$ I_1(s)=-\sum_n n^{s-1}e^{-n/y} $$ Here are my doubts

1. When we move the line of integration from $\text{Re}(w)=-1/4$ to $\text{Re}(w)=-3/4$ don't we pass through the pole of $\zeta$ at $w=-s$? What about the residue?
2. Once we expand $\zeta$ into its Dirichlet series and interchange the summation and integral we are left with computing $$ \sum_n n^{s-1}\frac{1}{2\pi i}\int_{(-3/4)}\Gamma(w)\left(\frac{y}{n}\right)^{-w}\,dw $$ I know that $$ e^{-y}=\frac{1}{2\pi i}\int_{(c)}\Gamma(w)\left(y\right)^{-w}\,dw $$ for all c>0. So the difference between the two should be given by the residue of $\Gamma(w)y^{-w}$ at $w=1$, that is 1. But what they claim is that we pick a minus sign.

I am surely overlooking something but I cannot figure out what. I really appreciate any advice. Thank you.

Answer 1

• For $c > 0,y > 0$ $$e^{-y}=\frac{1}{2\pi i}\int_{(c)}\Gamma(w)y^{-w}\,dw$$ you can see it from the residue theorem (summing over the poles at negative integers and showing $\lim_{a \to -\infty}\frac{1}{2\pi i}\int_{(a)}\Gamma(w)\left(y\right)^{-w}\,dw = 0$)

  or from the inverse Fourier/Laplace/Mellin transform of $\Gamma(w) = \int_0^\infty y^{w-1} e^{-y}dy=\int_{-\infty}^\infty e^{-uw} e^{-e^{-u}}du$ given $e^{-uc} e^{-e^{-u}}$ is Schwartz thus so is $\Gamma(c+it)$.

• Thus for $\Re(s) < 0,-\Re(s) > c > 0$ as everything converges absolutely $$\sum_{n=1}^\infty n^{s-1} e^{-y/n}=\frac{1}{2\pi i}\int_{(c)}\Gamma(w)\sum_{n=1}^\infty n^{s-1} \left(y/n\right)^{-w}\,dw=\frac{1}{2\pi i}\int_{(c)}\Gamma(w)\zeta(1-s-w)y^{-w}\,dw$$

• When $\Re(s) < 0$ and shifting the contour to the left, $\zeta(1-s-w)$ stays bounded and $\Gamma(w)$ stays fast decreasing (because $\Gamma(w+1) =w \Gamma(w)$ and $\Gamma(w)$ is Schwartz-fast decreasing on vertical lines for $\Re(w)>0$)

  so there is no fear to say for $\Re(s) < 0,-\Re(s) > c > 0$

$$\frac{1}{2\pi i}\int_{(-1/4)}\Gamma(w)\zeta(1-s-w)y^{-w}\,dw\\= \frac{1}{2\pi i}\int_{(c)}\Gamma(w)\zeta(1-s-w)y^{-w}\,dw-\text{Res}_{w=0}(\Gamma(w)\zeta(1-s-w)y^{-w})\\=\sum_{n=1}^\infty n^{s-1} e^{-y/n}-\zeta(1-s)$$

• What happens for $\Re(s) < 0, c > -\Re(s)$ ? Even if $\zeta(1-s-w)$ isn't bounded $\Gamma(w)\zeta(1-s-w)$ stays fast decreasing. So there is just to add the residue at $w=-s$ giving

$$\frac{1}{2\pi i}\int_{(-1/4)}\Gamma(w)\zeta(1-s-w)y^{-w}\,dw\\=\frac{1}{2\pi i}\int_{(c)}\Gamma(w)\zeta(1-s-w)y^{-w}\,dw-\text{Res}_{w=0}(\Gamma(w)\zeta(1-s-w)y^{-w})-\text{Res}_{w=-s}(\Gamma(w)\zeta(1-s-w)y^{-w})$$

• What happens for $\Re(s) > 0$ ? Something similar holds except that you'll need to regularize $F_y(s)=\sum_{n=1}^\infty n^{s-1} e^{-y/n}$, for example $\sum_{n=1}^\infty n^{s-1} (e^{-y/n}-1) = F_y(s)-\zeta(1-s)$ for $\Re(s)\in (0,1),y > 0$ and hence $$\frac{1}{2\pi i}\int_{(-1/4)}\Gamma(w)\zeta(1-s-w)y^{-w}\,dw=\sum_{n=1}^\infty n^{s-1} (e^{-y/n}-1)$$