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.

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

Can you product the definition of $\gamma$, which is $\displaystyle\lim_{n\rightarrow\infty}(H_n-\log n)$, from the integrals? Integration by parts might work, but not sure without evaluating... — pshmath0, Sep 06 '13 at 10:21

score 8 · Accepted Answer · answered Sep 06 '13 at 11:30

8

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

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

answered Sep 06 '13 at 11:30

Ron Gordon

141,538

Isn't this just using the "known" relation between $\Gamma'$ an $\gamma$, instead of the "known" relation between $Ei(x)$ and $\gamma$? – Nathaniel Bubis Sep 06 '13 at 11:41
@nbubis: Uhhh...yes. But this is what you asked for, no? If not, then you need to be more clear about what you want. – Ron Gordon Sep 06 '13 at 11:43
I guess I was looking for a "direct" proof that shows the integral sum is actually equivalent to the definition. Of course, there may not be one :) – Nathaniel Bubis Sep 06 '13 at 11:45
1

@nbubis: That the derivative of the gamma at $s=1$ is equal to $-\gamma$ is very well known and repeated enough within this site to be taken as fact. A proof of this uses the definition of the digamma as the derivative of the log of Gamma at $s=1$, which produces the sum of the reciprocal integers needed for the definition of $\gamma$. – Ron Gordon Sep 06 '13 at 11:50

score 8 · Answer 2 · answered Sep 06 '13 at 11:34

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,

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

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

2 Answers2

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