2
$\begingroup$

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?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Pragabhava, Qwerty, Daniel W. Farlow, suomynonA
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
5
$\begingroup$

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}

Notes:

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}.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

$$\int_0^1 \frac{e^{-1/t}}{t}dt = \int_1^\infty \frac{e^{-t}}{t}dt$$ 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.

$\endgroup$
  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.