I am studying the proof of the prime number theorem in Stein & Shakarchi's Complex Analysis, and am not sure about the justification of a particular equality.
The equality I'm interested in is equation (7) in the screenshots below. In particular, I don't understand the sentence "Since $F$ is regular in that region, and its decrease at infinity is rapid enough, the assertion (7) is established." Are the authors referring to some uncited result about the growth of a holomorphic function and its relation to integrals around contours?
I really appreciate any help establishing this result!
So as I said in the comments what you need to do is make a box, apply the Residue Theorem, and push a side of the box to infinity. Precisely, and in maybe too much detail (but hey maybe it's useful for someone sometime and that's why we're here), this means the following:
As you say, $c>1$. For $R>T$ denote by $\mathcal B_R$ the box with sides \[ \overbrace {(c+iT,c+Ri)}^{\text {right side}}\cup \overbrace {(c+Ri,1+Ri)}^{\text {top side}}\cup \overbrace {(1+Ri,1+iT)}^{\text {left side}}\cup \overbrace {(1+iT,c+iT)}^{\text {bottom side}},\] and it's best to think of $R$ as large. Since $F(s)$ is holomorphic in a region strictly containing this box (probably not terribly unwise to note this uses $\zeta (s)\not =0$ on the line $\sigma =1$) then the Residue Theorem says \[ \int _{\mathcal B_R}F(s)ds=0\] or said differently \[ \int _{c+iT}^{c+Ri}F(s)ds=\left (\int _{1+Ri}^{c+Ri}+\int _{1+iT}^{1+Ri}+\int _{c+iT}^{1+iT}\right )F(s)ds.\] Now we show the first integral on the RHS -the one over the top side of the box- is small. Since on $(1+Ri,c+Ri)$ certainly $F(s)\ll _x1/t=1/R$, we have \[ \int _{1+Ri}^{c+Ri}F(s)ds\ll _x\frac {1}{R}\int _1^cd\sigma \ll _c\frac {1}{R}\] i.e. \[ \lim _{R\rightarrow \infty }\int _{1+Ri}^{c+Ri}F(s)ds=0.\] Therefore \[ \int _{c+iT}^{c+i\infty }F(s)ds=\lim _{R\rightarrow \infty }\int _{c+iT}^{c+Ri}F(s)ds=\lim _{R\rightarrow \infty }\left (\int _{1+Ri}^{c+Ri}+\int _{1+iT}^{1+Ri}+\int _{c+iT}^{1+iT}\right )F(s)ds=\left (\int _{1+iT}^{1+i\infty }+\int _{c+iT}^{1+iT}\right )F(s)ds.\] Doing a similar argument in the lower half of the plane should give us \[ \int _{c-i\infty }^{c-iT}F(s)ds=\left (\int _{1-i\infty }^{1-iT}+\int _{1-iT}^{c-iT}\right )F(s)ds\] which allows us to conclude \[ \left (\int _{c-i\infty }^{c-iT}+\int _{c-Ti}^{c+iT}+\int _{c+iT}^{c+i\infty }\right )F(s)ds=\left (\int _{1-i\infty }^{1-iT}+\int _{1-iT}^{c-iT}+\int _{c-iT}^{c+iT}+\int _{c+iT}^{1+iT}+\int _{1+iT}^{1+i\infty }\right )F(s)ds\] i.e. \[ \int _{c-i\infty }^{c+i\infty }F(s)ds=\int _{\gamma (T)}F(s)ds.\]