Integral representation of Euler - Mascheroni Constant

[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.

 

[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}

 

[Yuqing]

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

 

[Dickfore]

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

<br /> \int_{0}^{\infty}{\ln{t} \, e^{-t} \, dt} = -\gamma<br />