How to prove that?

$$\int_0^1 \frac{1 - e^{-t} - e^{-1/t}}{t}\ \text{d}t = \gamma$$

where $\gamma = 0.5772156649015328606065\ldots$ is the Euler-Mascheroni constant.

Additional question: is there a way to evaluate it via Residues Theorem too?


closed as off-topic by Carl Mummert, Pragabhava, Qwerty, Daniel W. Farlow, suomynonA Oct 24 '16 at 22:54

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  • $\begingroup$ OFF ToPIC? I believe you guys behave really weirdly. Off topic this question. HA HA HA HA HA HA!!!!! $\endgroup$ – Von Neumann Oct 28 '16 at 14:46
  • $\begingroup$ I agree. It is a good question, as can be seen by the quality and variety of solutions. Also, someone with your amount of reputation points should never have a question closed unless it is clearly absurd. Unfortunately, this type of behavior by moderators plagues all internet forums. $\endgroup$ – poweierstrass Nov 2 '16 at 14:23
  • $\begingroup$ @poweierstrass I completely agree with you! It's really a pity because moderators shall assume a very different behaviour. Well, still. I appreciated really much your answer and all the other ones here. So I cannot complain (too much) about :D $\endgroup$ – Von Neumann Nov 2 '16 at 14:41

Preliminary results.

Let us begin with a definition of the Euler-Mascheroni Constant \begin{equation} \gamma = \lim_{n \to \infty} H_{n} - \mathrm{ln}(n) \label{eq:1} \tag{1} \end{equation} where $H_{n}$ are the harmonic numbers defined as \begin{equation} H_{n} = \displaystyle\sum_{k=1}^{n} \frac{1}{k} \label{eq:2} \tag{2} \end{equation}

Let \begin{equation} \int\limits_{0}^{1} \frac{1-(1-x)^{n}}{x} \mathrm{d} x = H_{n} \label{eq:3} \tag{3} \end{equation}

Proof: \begin{align} \tag{a} \int\limits_{0}^{1} \frac{1-(1-x)^{n}}{x} \mathrm{d} x & = \int\limits_{0}^{1} \frac{1-y^{n}}{1-y} \mathrm{d} y \\ & = \int\limits_{0}^{1} \frac{1}{1-y} \mathrm{d} y \, - \int\limits_{0}^{1} \frac{y^{n}}{1-y} \mathrm{d} y \\ \tag{b} & = \int\limits_{0}^{1} \displaystyle\sum_{k=0}^{\infty} y^{k} \mathrm{d} y \,\, - \int\limits_{0}^{1} y^{n} \displaystyle\sum_{k=0}^{\infty} y^{k} \mathrm{d} y \\ & = \displaystyle\sum_{k=0}^{\infty} \frac{y^{k+1}}{k+1} \Big|_{0}^{1} \,\, - \displaystyle\sum_{k=0}^{\infty} \frac{y^{k+n+1}}{k+n+1} \Big|_{0}^{1} \\ & = (1 + \frac{1}{2} + \frac{1}{3} + \dots) \, - (\frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} + \dots) \\ & = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} = H_{n} \end{align}


a. Let $y=1-x$

b. Expand $$\frac{1}{1-y} = \sum\limits_{k=0}^{\infty} y^{k}$$

Main result. \begin{align} \gamma &= \int\limits_{0}^{1} \frac{1-\mathrm{e}^{-x}-\mathrm{e}^{-1/x}}{x} \mathrm{d}x \\ &= \int\limits_{0}^{1} \frac{1-\mathrm{e}^{-x}}{x} \mathrm{d}x - \int\limits_{0}^{1} \frac{\mathrm{e}^{-1/x}}{x} \mathrm{d}x \\ &= \int\limits_{0}^{1} \frac{1-\mathrm{e}^{-x}}{x} \mathrm{d} x \,\, - \int\limits_{1}^{\infty} \frac{\mathrm{e}^{-x}}{x} \mathrm{d} x \label{eq:4} \tag{4} \end{align}

We made the substitution $z=1/x$, then switched variables back to $x$.

Make the substitution $x = \frac{y}{n}$ in equation \eqref{eq:3} to obtain \begin{align} H_{n} & = \int\limits_{0}^{n} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\ & = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \int\limits_{1}^{n} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\ & = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \int\limits_{1}^{n} \frac{1}{y} \mathrm{d} y \,\, - \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\ & = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \mathrm{ln}(n) \, - \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \label{eq:5} \tag{5} \end{align}

Now we invoke the limit defnition of the exponential function \begin{equation} \mathrm{e}^{\pm x} = \lim_{n \to \infty} \left(1 \pm \frac{x}{n} \right)^{n} \label{eq:6} \tag{6} \end{equation} rearrange equation \eqref{eq:5} and take $\lim_{n \to \infty}$, we have

\begin{equation} \lim_{n \to \infty} \left(H_{n} - \mathrm{ln}(n)\right) = \lim_{n \to \infty} \left(\int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d}y - \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d}y\right) \end{equation}

The left hand side equals $\gamma$ by equation \eqref{eq:1} as does the right hand side by equations \eqref{eq:6}, \eqref{eq:5}, and \eqref{eq:4}.

  • $\begingroup$ You can prove the integral representation of H(n) much simpler by deriving the recursion relation h(n+1) = h(n) + 1/(n+1) directly from the integral. (The factor (1-x) leads to h(n) + trival integral giving 1/(1+n)). This avoids temporary use of divergent integrals. $\endgroup$ – Dr. Wolfgang Hintze Oct 22 '16 at 11:43

$$\int_{0}^{1}\frac{1-e^{-t}-e^{-1/t}}{t}dt=\int_{0}^{1}\frac{1-e^{-t}}{t}dt-\int_{0}^{1}\frac{e^{-1/t}}{t}dt $$ $$\stackrel{IBP}{=}\int_{0}^{1}\log\left(t\right)e^{-t}dt+\int_{0}^{1}\frac{\log\left(t\right)e^{-1/t}}{t^{2}}dt $$ $$\stackrel{1/t\rightarrow t}{=}\int_{0}^{1}\log\left(t\right)e^{-t}dt+\int_{1}^{\infty}\log\left(t\right)e^{-t}dt $$ $$=\int_{0}^{\infty}\log\left(t\right)e^{-t}dt=-\Gamma'\left(1\right)=\color{red}{\gamma}.$$

  • $\begingroup$ I just finished to prove that the very last integral you wrote is $\gamma$, hence this is awesome because now I can see a connection between that one and the integral in my question, thank you!! $\endgroup$ – Von Neumann Oct 22 '16 at 8:11
  • $\begingroup$ @AlanTuring You're welcome. $\endgroup$ – Marco Cantarini Oct 22 '16 at 8:12

$$\int_0^1 \frac{e^{-1/t}}{t}dt = \int_1^\infty \frac{e^{-t}}{t}dt$$ So your integral is $$\int_0^\infty \frac{1_{t < 1}-e^{-t}}{t}dt =\lim_{ s \to 0^+}\int_0^\infty t^{s-1}(1_{t < 1}-e^{-t})dt = \lim_{s \to 0^+} \frac{1}{s}-\Gamma(s)$$ $$ = \lim_{s \to 0^+} \frac{\Gamma(1)-\Gamma(s+1)}{s} = -\Gamma'(1) = \gamma$$

For proving that $\Gamma'(1) = -\gamma$ there are many ways, but I never remember the best one.

  • $\begingroup$ there is a derivation using $(1+x/n)^n \to e^x$ as poweierstrass, I'd like an easier way $\endgroup$ – reuns Oct 22 '16 at 0:00

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