I've come across the following useful device from complex analysis:
$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}{\frac{y^z}{z}}{dz} = \left\{\begin{array}{lll} 0 & \text{if} & 0<y<1 \\ \frac{1}{2} & \text{if} &y=1 \\ 1 & \text{if} & y > 1\end{array}\right.,$$ where $\text{Re}(c)>0$ is conveniently chosen.
I understand, amongst other applications, this can be used (is essential?) in the accepted proof of the Prime Number Theorem.
I'd just like some clarification I'm on the right track with how to use this device. For example, suppose we were to have the following:
$$f(z;y) = \frac{y^z}{z},$$ and I could see it would be beneficial to use the above device. Then I can write
$$\int_{c-i\infty}^{c+i\infty}f(z;y)dz = \int_{c-i\infty}^{c+i\infty}\frac{y^z}{z}dz.$$
However, am I correct in thinking I can't just go ahead and use the information in the curly brace. Instead, loosely speaking, I have to create a (probably square) contour, two corners of which are $c+i\infty$ and $c-i\infty$. Then I would need to check the top, bottom, and left line integrals go to zero (or else compute their values), and furthermore account for all residues at poles within the contour of the meromorphic function. Only then would I be able to use the information on the line integral given in the curly brace?
EDIT
Now I know the name of this device, I have found a related post. After reading this post, Perron's formula may come from the Mellin transform?...
EDIT
As mentioned by @FelixMarin in a comment to @robjohn's answer, it is possible to view this function as the Heaviside step function. See comment for further information.
The contour used to compute this integral may be important in verifying its applicability. Let's rewrite $y=e^a$ so that the integral is $$ \int_{c-i\infty}^{c+i\infty}\frac{e^{az}}{z}\mathrm{d}z\tag{1} $$
If $a\gt0$ (corresponding to $y\gt1$), the contour would be $$ \overbrace{c+iR[-1,1]}^{\gamma_0}\cup\overbrace{[c,-R]+iR}^{\gamma_1}\cup\overbrace{-R+iR[1,-1]}^{\gamma_2}\cup\overbrace{[-R,c]-iR}^{\gamma_3}\tag{2} $$ where, for $k\in\{1,3\}$, we can control $$ \begin{align} \left|\,\int_{\gamma_k}\frac{e^{az}}{z}\mathrm{d}z\,\right| &\le\frac1R\int_{-\infty}^ce^{at}\mathrm{d}t\\ &=\frac{e^{ac}}{|a|R}\tag{3} \end{align} $$ and we can control $$ \begin{align} \left|\,\int_{\gamma_2}\frac{e^{az}}{z}\mathrm{d}z\,\right| &\le\frac1R\int_{-R}^Re^{-|a|R}\mathrm{d}t\\ &=2e^{-|a|R}\tag{4} \end{align} $$ Thus, the integral over $\gamma_k$ vanishes for $k\in\{1,2,3\}$. This contour circles $z=0$ counterclockwise, so the integral over $\gamma_0$ is $2\pi i$.
If $a\lt0$ (corresponding to $y\lt1$), the contour would be $$ \overbrace{c+iR[-1,1]}^{\gamma_0}\cup\overbrace{[c,R]+iR}^{\gamma_1}\cup\overbrace{R+iR[1,-1]}^{\gamma_2}\cup\overbrace{[R,c]-iR}^{\gamma_3}\tag{5} $$ where, for $k\in\{1,3\}$, we can control $$ \begin{align} \left|\,\int_{\gamma_k}\frac{e^{az}}{z}\mathrm{d}z\,\right| &\le\frac1R\int_c^{\infty}e^{at}\mathrm{d}t\\ &=\frac{e^{ac}}{|a|R}\tag{6} \end{align} $$ and we can control $$ \begin{align} \left|\,\int_{\gamma_2}\frac{e^{az}}{z}\mathrm{d}z\,\right| &\le\frac1R\int_{-R}^Re^{-|a|R}\mathrm{d}t\\ &=2e^{-|a|R}\tag{7} \end{align} $$ Thus, the integral over $\gamma_k$ vanishes for $k\in\{1,2,3\}$. This contour does not circle $z=0$, so the integral over $\gamma_0$ is $0$.
If $a=0$ (corresponding to $y=1$), then the integral is simply the logarithm. In this case, we need to make sure the top and bottom limits of integration are conjugates of each other, otherwise, the integral will have an unwanted real part. If they are conjugates, $(1)$ is $\arg(i)-\arg(-i)=\pi i$.
This case converges in the principal value sense; that is, neither the top half nor the bottom half of the integral converges, but they cancel if we integrate over conjugate halves. The other two cases converge because of oscillation, but this case relies solely on the cancellation that comes from adding conjugates: $$ \begin{align} \int_0^\infty\left(\frac1{c+it}+\frac1{c-it}\right)i\,\mathrm{d}t &=i\int_0^\infty\frac{2c}{c^2+t^2}\,\mathrm{d}t\\ &=i\pi\tag{8} \end{align} $$ whereas neither $$ \int_0^\infty\frac1{c+it}i\,\mathrm{d}t \quad\text{nor}\quad \int_0^\infty\frac1{c-it}i\,\mathrm{d}t\tag{9} $$ converge.
However, the integral in $(1)$ does converge without including the contours added to compute its value via contour integration. So I think to answer your question about whether you need to worry about these contours for a particular application of the formula, we would need to know more about the way this formula is being used.