\[\large \lim_{x\to 0^{+}} \left( \text{Ei}(-x) - \ln(x)\right) = \gamma\]

EDIT: I figured out a solution; see my comment below.

To clarify:

\(\text{Ei}(y) = \displaystyle\int_{-y}^{\infty} \dfrac{-e^{-t}}{t} dt = \displaystyle\int_{-\infty}^{y} \dfrac{e^t}{t}dt\) denotes the exponential integral;

\( \gamma\) denotes the Euler-Mascheroni constant, \(\gamma \approx 0.5772 \).

**Notation**: \( H_n\) denotes the \(n^\text{th} \) harmonic number, \( H_n = 1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n\).

I came across this limit when I was trying to derive the derivative of the factorial function: \(\dfrac{d(n!)}{dn} = n!(H_n - \gamma)\) for all whole numbers \(n\). Let's denote the derivative as \(D(n)\) We know that \((n+1)! = (n+1)n!\). Therefore, differentiating gives us:

\[D(n+1) = (n+1)D(n) + n!\]

\[\dfrac{D(n+1)}{(n+1)!} = \dfrac{(n+1)D(n)}{(n+1)!} + \dfrac{n!}{(n+1)!} = \dfrac{D(n)}{n!} + \dfrac{1}{n+1}\]

It follows that \(\dfrac{D(n)}{n!} = H_n + a\) for some constant \(a\). What I can't figure out is how to prove that \(a = -\gamma\). I tried to find the value of \(D(0)\):

\[D(n) = \dfrac{d}{dn}\left(\int_{0}^{\infty} x^n e^{-x} dx \right) = \int_{0}^{\infty} \dfrac{d(x^n e^{-x})}{dn} dx \]

\[D(n) = \int_{0}^{\infty} x^n \ln(x) e^{-x} dx\]

Therefore, \[D(0) = \int_{0}^{\infty} \ln(x) e^{-x} dx\] Using integration by parts, this becomes:

\[ D(0) = -e^{-x} \ln(x) \rvert_{0}^{\infty} - \int_{0}^{\infty} \dfrac{-e^{-x}}{x}dx\]

\[=\lim_{y\to 0^{+}} \left[ -e^{-x} \ln(x) \rvert_{y}^{\infty} - \int_{y}^{\infty} \dfrac{-e^{-x}}{x}dx \right]\]

\[D(0) = \lim_{y\to 0^{+}} (e^{-y} \ln(y) - \text{Ei}(-y))\]

We can simplify this a little by observing the following: \[\lim_{y\to 0^{+}} (e^{-y} - 1) \ln(y) = \lim_{y\to 0^{+}} \dfrac{e^{-y} - 1}{-y} \lim_{y\to 0^{+}} (-y\ln(y)) = 1 * 0 = 0\]

Hence, the equation for \(D(0)\) can simplify to:

\[D(0) = \lim_{y\to 0^{+}} (\ln(y) - \text{Ei}(-y))\]

Now this is supposed to equal \(-\gamma\), but I cannot figure out how to prove it, and can't seem to find the proof anywhere either.

## Comments

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TopNewestBy definition, \(\displaystyle \text{Ei}(x) = -\int_{-x}^\infty \dfrac{e^{-t}}t \, dt \) and \(\displaystyle e^{-t} = \sum_{k=0}^\infty \dfrac{(-1)^{k+1} t^k }{k!} \). Integrating the Taylor series of \( \dfrac {e^{-t}}t \), we get

\[ \text{Ei}(x) = \gamma + \ln| x| + \sum_{k=1}^\infty \dfrac{x^k}{k \cdot k!} \qquad x\ne 0 .\]

Take the limit and the result follows. – Pi Han Goh · 4 months, 2 weeks ago

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– Ariel Gershon · 4 months, 2 weeks ago

Here's the thing though - how do you know the first term is \(\gamma\) ??Log in to reply

I've figured out the solution after a while. Here I will use the generalized harmonic number: \[H_x = \sum_{k=1}^{\infty} \left(\dfrac{1}{k} - \dfrac{1}{x+k}\right)\] Now let's start from the point where \(\dfrac{D(n)}{n!} = H_n + a\). Let's integrate this on both sides from \(0\) to \(1\): \[\int_{0}^{1} \dfrac{D(n)}{n!} dn = \int_{0}^{1} (H_n + a) dn\] \[\ln(n!) \large{\rvert_0^1} \normalsize{ = a + \int_{0}^{1} \sum_{k=1}^{\infty} \left(\dfrac{1}{k} - \dfrac{1}{n+k}\right) dn}\] \[0 = a + \sum_{k=1}^{\infty} \int_{0}^{1} \left(\dfrac{1}{k} - \dfrac{1}{n+k}\right) dn\] \[-a = \sum_{k=1}^{\infty} \left(\dfrac{n}{k} - \ln(n+k)\right) \large{\rvert_0^1}\] \[-a = \sum_{k=1}^{\infty} \left(\dfrac{1}{k} - \ln(1+k) + \ln(k)\right)\] \[a = -\lim_{m \to \infty} \left(\sum_{k = 1}^{m} \dfrac{1}{k} - \ln(1+m)\right)\] \[a = -\gamma\] – Ariel Gershon · 4 months, 1 week ago

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Hey Ariel, I see that you like to solve difficult calculus questions. We got a group on another social platform (online forum) that discusses all kinds of difficult calculus questions as well. Plus, most of the active members on Brilliant are on this online forum as well (even the staffs!). We discuss about everything related to math and science. Plus, it's free! Would you like to join us?

If yes, register here

See you! – Aditya Kumar · 4 months, 2 weeks ago

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– Ariel Gershon · 4 months, 1 week ago

Just joined! Thanks! :)Log in to reply

– Aditya Kumar · 4 months, 1 week ago

I've added you!Log in to reply