The "inverse Mellin" transform is used in deriving the explicit Riemann/Von Mangoldt formula for the Chebyshev $\psi$ function. One comes across the following integrals which are puzzling to me. c>0. Can someone help me understand them?
$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\exp(k\cdot z)}{z}dz \;=1\;\;\;\forall\; k>0$$ $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\exp(k\cdot z)}{z}dz \;=0\;\;\;\forall\; k<0$$ $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\exp(k\cdot z)}{z}dz \;=0.5\;\;\; k=0$$
I'm reading H.M. Edwards' textbook as well as online resources, which call this a Fourier inversion formula. I'm not completely comfortable with the math for non-periodic Fourier functions, but apparently (I'm guessing) this is a sort of "low pass" filter like the Sinc function which sums up the $a(n)$ terms up to a given value of n for a Dirichlet series. There was a time in the past when I understood the Sinc function fourier transform. $$g(z)=\sum_{n=1}^{\infty}\frac{a(n)}{n^z}$$
Then the Dirichlet series is multiplied by $\frac{x^z}{z} $ and integrated to magically sum up the terms. The prime$'$ on the summation indicates that the last term of the sum must be multiplied by 1/2 when x is an integer, which matches the k=0 and k>0 formulas above.
$$A(x) = \sum_{n<=x}'a(n) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}g(z)\frac{x^z}{z}dz.$$