Doubt in the proof of Prime Number Theorem

Question

Suppose we have that \begin{equation} \frac{\psi_1(x)}{x}-\frac{1}{2}\Big(1-\frac{1}{x}\Big)^2=\frac{1}{2 \pi i} \int\limits_{c-\infty i}^{c+\infty i} \frac{x^{z-1}}{z(z+1)}\Big(-\frac{\zeta'(z)}{\zeta(z)}-\frac{1}{z-1}\Big) \mathop{dz} \end{equation} If we sustitute $z=c+it$, then we have \begin{align} \frac{\psi_1(x)}{x}-\frac{1}{2}\Big(1-\frac{1}{x}\Big)^2&=\frac{1}{2 \pi i} \int\limits_{c-\infty i}^{c+\infty i} x^{c+it-1} f(c+it) \mathop{dz} \\ &=\frac{x^{c-1}}{2 \pi }\int\limits_{-\infty }^{\infty } e^{it \operatorname{log}(x)} f(c+it) \mathop{dt} \end{align}

where \begin{equation} f(z)= \frac{x^{z-1}}{z(z+1)}\Big(-\frac{\zeta'(z)}{\zeta(z)}-\frac{1}{z-1}\Big) \end{equation}

I couldn't understand what happens during the substitution. Can anyone make it clear for me? If we substitute $z=c+it$ then how will $dz$ and $dt$ be related? Will $dz=i dt$?

Answer 1

In the first line of your equation, you should have written \[ \cdots \equiv \frac{1}{2 \pi i } \int_{c - i \infty}^{c + i \infty} x^{z - 1} f(z)dz \] and then contour integral $z = c + it$, $dz = i dt$.

Please don't doubt the proof of the Prime Number Theorem. Once you come to this stage, you only need the facts that

1. the zeta function has no zero on the line $\text{Re}(s) = 1$,
2. some decreasing bound for $f(c + it)$ as $|t|$ goes to $\infty$, and
3. the Riemann-Lebesgue lemma.