unable to understand the contour integral while proving prime number theorem

Question

I am studying the proof on prime number theorem, but in a step I am getting confused.

They are integrating around the contour $R$, but after showing that the integral along the horizontal segments is zero, they concluded that $$\int\limits_{c- \infty i}^{c+ \infty i} x^{s-1}h(s)\mathop{ds}=\int\limits_{1- \infty i}^{1+ \infty i} x^{s-1}h(s)\mathop{ds}$$

After looking at the contour $R$ in figure 13.3(in the attached image), I think that instead of that they had written it should be $$\int\limits_{c- \infty i}^{c+ \infty i} x^{s-1}h(s)\mathop{ds}=\int\limits_{1+ \infty i}^{1- \infty i} x^{s-1}h(s)\mathop{ds}=$$.

Can anyone clear this doubt? The book I am following is Introduction to Analytic Number Theory, by Tom Apostol

Answer 1

No, the book is right. The integral around the entire contour is zero. Thus

$$ \int\limits_{c- \infty\mathrm i}^{c+ \infty\mathrm i} x^{s-1}h(s)\,\mathrm ds+\int\limits_{1+ \infty\mathrm i}^{1- \infty\mathrm i} x^{s-1}h(s)\,\mathrm ds=0\;, $$

since the two sides are traversed in opposite directions. Then when you bring one of the integrals over to the other side you get a minus sign, so you need to swap the limits on one of the integrals.

Answer 2

The integral shown in the picture could be summarised as $$\int_{c-iT}^{c+iT}+\int_{c+iT}^{1+iT}+\int_{1+iT}^{-1+iT}+\int_{-1+iT}^{c-iT}=0.$$ As $T\to\infty$ the horizontal integrals go to zero, so in the limit $$\int_{c-i\infty}^{c+i\infty}+\int_{1+i\infty}^{-1+i\infty}=0.$$ Theefore $$\int_{c-i\infty}^{c+i\infty}=-\int_{1+i\infty}^{-1+i\infty} =\int_{-1+i\infty}^{1+i\infty}$$ as $\int_a^b=-\int_b^a$ (as contour integrals along a straight line)