# Integral representation of Euler - Mascheroni Constant

1. Aug 21, 2011

### Yuqing

The definition of the Euler - Mascheroni constant, $\gamma$, is given as
$$\gamma = \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k} - \ln(n)$$
or equivalently in integral form as $$\gamma = \int_{1}^{\infty}\frac{1}{\left\lfloor x\right\rfloor} - \frac{1}{x}\ dx$$

I saw a seeming related integral representation
$$\gamma = 1 - \int_{1}^{\infty} \frac{x - \left\lfloor x\right\rfloor}{x^2}\ dx$$
but I can't seem to derive it. I was wondering if anyone can shed some light on this.

2. Aug 21, 2011

### Citan Uzuki

The easiest way to do this is to just expand out the integral, like so:

\begin{align}&1 - \int_{1}^{\infty} \frac{x-\lfloor x \rfloor}{x^2}\ dx \\ &= 1 - \lim_{n \rightarrow \infty} \int_{1}^{n} \frac{x-\lfloor x \rfloor}{x^2}\ dx \\ &= \lim_{n \rightarrow \infty} 1 - \sum_{k=1}^{n-1} \int_{k}^{k+1} \frac{x - k}{x^2}\ dx \\ &= \lim_{n \rightarrow \infty} 1 - \sum_{k=1}^{n-1} \left( \ln x + \frac{k}{x}\right) \Big|_{k}^{k+1} \\ &=\lim_{n \rightarrow \infty} 1 - \sum_{k=1}^{n-1} \left( \ln (k+1) - \ln k + \frac{k}{k+1} - 1\right) \\ &=\lim_{n \rightarrow \infty} 1 - \sum_{k=1}^{n-1} \left( \ln (k+1) - \ln k - \frac{1}{k+1}\right) \\ &=\lim_{n \rightarrow \infty} 1 - \sum_{k=1}^{n-1} (\ln (k+1) - \ln k) + \sum_{k=1}^{n-1} \frac{1}{k+1} \\ &= \lim_{n \rightarrow \infty} 1 - \ln n + \sum_{k=2}^{n} \frac{1}{k} \\ &=\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n = \gamma \end{align}

3. Aug 21, 2011

### Yuqing

Hmm, that's quite a novel approach. Thank you.

4. Dec 8, 2011

### Dickfore

I think this integral is related to the Euler-Mascheroni Constant as well:

$$\int_{0}^{\infty}{\ln{t} \, e^{-t} \, dt} = -\gamma$$