Why is $-\gamma = \int_0^1 \frac{e^{-z}-1}{z}dz+\int_1^\infty \frac{e^{-z}}{z}dz$

Question

It seems like the sum of the two RHS integrals is "well known"$^\dagger$ to be Euler's constant: $$\gamma \equiv \int_1^\infty \frac{1}{\lfloor z\rfloor} - \frac{1}{z}dz \quad\stackrel{?}{=}\quad -\int_0^1 \frac{e^{-z}-1}{z}dz-\int_1^\infty \frac{e^{-z}}{z}dz$$

How can I prove this is so?

Edit:

I can prove this converges to a constant by showing that this is equivalent to: $$\int_0^1 \frac{e^{-1/z}+e^{-z}-1}{z}dz$$ And that the limit as $z\to0$ exists, so the integral converges. This however doesn't bring me closer to proving what this constant is.

$\Tiny^\dagger\text{ From the collected papers of L. Landau. }$

Answer 1

Integration by parts on the first integral produces

$$[(1-e^{-z}) \log{z}]_0^{1} -\int_0^1 dz \, e^{-z} \, \log{z} = -\int_0^1 dz \, e^{-z} \, \log{z}$$

Integration by parts on the second integral produces

$$-[e^{-z} \log{z}]_1^{\infty} - \int_1^{\infty}dz \, e^{-z} \, \log{z} = - \int_1^{\infty}dz \, e^{-z} \, \log{z}$$

Putting this together, these integrals sum to

$$-\int_0^{\infty} dz \, e^{-z} \, \log{z} = -\left [\frac{d}{ds} \int_0^{\infty} dz \, e^{-z} \, z^s\right ]_{s=1} = -\left [\frac{d}{ds} \Gamma(s)\right]_{s=1} = -\gamma $$

Answer 2

For $\alpha > 0$, write

$$\begin{align} F(\alpha) &:= \int_0^1 \frac{e^{-z}-1}{z^{1-\alpha}}\,dz + \int_1^\infty \frac{e^{-z}}{z^{1-\alpha}}\,dz\\ &= \int_0^\infty z^{\alpha-1}e^{-z}\,dz - \int_0^1 z^{\alpha-1}\,dz\\ &= \Gamma(\alpha) - \frac{1}{\alpha}\\ &= \frac{\alpha\Gamma(\alpha) - 1}{\alpha}\\ &= \frac{\Gamma(1+\alpha) - \Gamma(1)}{\alpha} \xrightarrow{\alpha\to 0} \Gamma'(1) = -\gamma. \end{align}$$

On the other hand,

$$\lim_{\alpha\to 0} F(\alpha) = \int_0^1 \frac{e^{-z}-1}{z}\,dz + \int_1^\infty \frac{e^{-z}}{z}\,dz.$$