# Integral representation of Euler's Gamma constant

Is there any repesentation of $\gamma$ (Euler-Mascheroni constant) of the form:

$$\int_2^\infty f(t) dt = \gamma ?$$

I have not yet found any (there are plenty of integral representations of this constant in Wolfram's Functions site but none starting by 2). I was just curious about it, since bot $\gamma$ and $2$ are important constants in Analytic Number Theory.

You could just change variables in, say, the familiar $$\gamma = \int_1^\infty \left(\frac{1}{\lfloor x\rfloor} - \frac1x\right)\;dx$$ giving $$\gamma = \int_2^\infty \left(\frac{1}{\lfloor t-1\rfloor} - \frac1{t-1}\right)\;dt$$

• Omg, thank you. I did not realise that simple solution – user3141592 Mar 7 '17 at 15:41

We can shift the number 1 in the very important representation $$\int\limits_0^1\frac{1-e^{-x}}{x}dx - \int\limits_1^\infty\frac{e^{-x}}{x}dx = \gamma$$ and get $$\int\limits_0^2\frac{1-e^{-x}}{x}dx - \int\limits_2^\infty\frac{e^{-x}}{x}dx = \gamma + \log2$$