I have been looking through this paper of Scourfield titled: "On the Divisiblity of $\sigma_\nu(n)$" (Link: https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/10/3/95439/on-the-divisibility-of-n). In Lemma 19, he estimates an integral of the form $\Sigma_1(x) := \int_{2-i\infty}^{2+i\infty} \frac{f(s)x^s}{s^2}ds$ using a contour of the form $\overline{A}ABCDE\overline{E}\overline{D} \overline{C}\overline{B}\overline{A}$ where $$A:=2+ix^2, B:=1-d_1(\log x^2)^{-9}+ix^2,$$ $$C:= 1-d_1(\log 3)^{-9}+3i, D:= 1-d_1(\log 3)^{-9}, E:= 1-\delta$$ and $\overline{A}$ denotes the conjugate of $A$ and so on. Here, $d_1$ has been chosen appropriately (c.f. Lemma $15$) and $0<\delta<<1$ (will tend to $0$ later), and all paths joining neighboring points are straight lines except for:
(i) $BC$, which is the curve $\Re(s)=1-d_1(\log \Im(s))^{-9}$ $(x^2 \geq \Im(s) \geq 3)$,
(ii) $\overline{B} \overline{C}$, which is the reflection of the previous curve about the real axis, and
(iii) $E \overline{E}$, which is the circle $|s-1|=\delta$.
In Lemma 19, he claims that the principal term of the integral $\Sigma_1(x)$ comes from the sum of the integrals of $\frac{f(s)x^s}{s^2}$ along $DE$ and $\overline{E} \overline{D}$. But aren't these two integrals of the same function along opposite paths (since $D$, $E$ clearly lie on the real axis), so shouldn't they just cancel? What is the (probably obvious) thing that I am missing? Thanks.