7
$\begingroup$

Starting from the limit definition of the Euler-Mascheroni constant $\gamma$ as given by

$$\gamma=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n\frac1k\right)\tag 1$$

we can show that $\gamma$ has an integral representation

$$\gamma=\int_0^\infty\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx \tag 2$$


Proof of $(2)$: This is provided for completeness only and one can skip this part without losing context.

To show that the integral in $(2)$ is equivalent to $(1)$, we can proceed as follows.

$$\begin{align} \int_0^\infty\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx&= \int_0^\infty \frac{e^{-x}}{1-e^{-x}}\left(1-\frac{1-e^{-x}}{x}\right)\,dx\\\\ &=\sum_{k=1}^\infty \int_0^\infty\left(e^{-kx}-\frac{e^{-(kx}-e^{-(k+1)x}}{x}\right)\,dx\\\\ &=\sum_{k=1}^\infty \left(\frac{1}{k}-\log\left(\frac{k+1}{k}\right)\right)\\\\ &=\lim_{n\to \infty}\sum_{k=1}^n \left(\frac1k -\log\left(\frac{k+1}{k}\right)\right)\\\\ &=\lim_{n\to \infty}\left(-\log(n+1)-\sum_{k=1}^n\frac1k\right)\\\\ &=\lim_{n\to \infty}\left(-\log(n)-\sum_{k=1}^n\frac1k\right)\\\\ \end{align}$$


Another integral representation for $\gamma$ is given by

$$\gamma=\int_0^\infty \left(\frac{1}{x(1+x^a)}-\frac{1}{xe^x}\right) \,dx \tag 3$$

for $a>0$.

Equipped with $(2)$, we can show the equivalence of $(3)$ with $(1)$ by showing that

$$\int_0^\infty \left(\frac{1}{x(1+x^a)}-\frac{1}{e^x-1}\right)\,dx=0\tag 4$$


To prove $(4)$, I proceeded as follows.

$$\begin{align} \lim_{\epsilon\to 0}\int_{\epsilon}^\infty\left(\frac{1}{x(1+x^a)}-\frac{1}{e^x-1}\right)\,dx &=\lim_{\epsilon\to 0}\left.\left(-\frac1a \log(1+x^{-a})+\log(1-e^{-x})\right)\right|_{\epsilon}^{\infty}\\\\ &=\lim_{\epsilon\to 0}\left(\frac1a \log(1+\epsilon^{-a})+\log(1-e^{-\epsilon})\right)\\\\ &=0 \end{align}$$

And we are done!


This approach seemed a bit cumbersome and indirect.

QUESTION: So, what are alternative approaches to establishing equivalence of $(3)$ and $(1)$?

$\endgroup$
  • $\begingroup$ I deleted my answer. I misunderstood your question. $\endgroup$ – Zaid Alyafeai Jan 7 '17 at 17:42
  • $\begingroup$ Would the coward who down voted this question care to explain why? $\endgroup$ – Mark Viola Mar 12 '18 at 21:18
6
$\begingroup$

Integrating by parts, $$ \int_0^{\infty} \left( \frac{1}{1+x^a} - e^{-x} \right) \frac{dx}{x} = 0-0 + \int_0^{\infty} \left( \frac{ax^a}{x(1+x^a)^2} - e^{-x} \right) \log{x} \, dx $$

Of course, we recognise the second term as a familiar definition of/easy-to-derive formula for $\gamma$. The first term we need to show is zero. But $$ \int \frac{ax^a\log{x}}{x(1+x^a)^2} \, dx = \frac{x^a\log{x}}{1+x^a} - \frac{1}{a}\log{(1+x^a)}, $$ which is continuous and tends to zero at both endpoints since $a>0$.


The really interesting thing about this result in my opinion is that it shows the first term is a complete red herring: let $F$ be continuous and continuously differentiable on $(0,\infty)$ with the following properties:

  • $ F(x) = 1 + o(1/\log{x})$ as $x \downarrow 0 $,
  • $F(x) = o(x^{-\epsilon})$ as $x \uparrow \infty$ for some $\epsilon>0$,
  • $\int_0^{\infty} F'(x) \log{x} \, dx = 0$

Then $$ \gamma = \int_0^{\infty} \left( F(x) - e^{-x} \right) \frac{dx}{x}. $$ The proof is essentially identical to the above: $$ \int_0^{\infty} ( F(x) - e^{-x} ) \frac{dx}{x} = [(F(x) - e^{-x}) \log{x}]_0^{\infty} - \int_0^{\infty} ( F'(x) + e^{-x} ) \log{x} \, dx = \int_0^{\infty} e^{-x} \log{x} \, dx, $$ The integral on the left exists by the first two conditions on $F$, which are also enough to ensure the boundary terms from the integration by parts go to zero.

$\endgroup$
  • 1
    $\begingroup$ I can't figure out the part $(\frac 1{1+x^a}-e^{-x})\log x{|}_0^{\infty}=0-0$ $\endgroup$ – polfosol Jan 7 '17 at 7:58
  • $\begingroup$ Really nice answer. $\endgroup$ – Muse_China Jan 7 '17 at 10:54
  • $\begingroup$ @polfosol $\log{x} = o(x^{\alpha})$ at zero and $\infty$, for any $\alpha>0$. The bracket is $O(\max{\{ x,x^a \}})$ at zero, and $O(x^{-a})$ as $x \to \infty$. $\endgroup$ – Chappers Jan 7 '17 at 15:09
  • $\begingroup$ I found my misunderstanding. Thanks $\endgroup$ – polfosol Jan 7 '17 at 15:13
  • $\begingroup$ (+1) Thia works. I woke up in the middle of the night and realuzed IBP was viable, but thought I ought to prove for completeness that the resulting integrral representation was equivalent to $(1)$ in the OP. I'll post an answer that leverages yours. $\endgroup$ – Mark Viola Jan 7 '17 at 15:18
2
$\begingroup$

I thought it might be instructive to present a supplement to the nice solution posted by @Chappers in order to have a self-contained way forward. To that end, we proceed.


In THIS ANSWER, I showed using the integral representation of the Gamma function

$$\Gamma(x)=\int_0^\infty s^{x-1}e^{-s}\,ds \tag 1$$

that Gamma can be expressed as the limit

$$\Gamma(x)=\lim_{n\to \infty}\frac{n^x\,n!}{x(x+1)(x+2)\cdots (x+n)} \tag 2$$

Now, note that $(2)$ can be rewritten as

$$\begin{align} \Gamma(x)&=\lim_{n\to \infty}\frac{e^{x(\log(n)-1-1/2-\cdots -1/n)}\, e^{x}e^{x/2}\cdots e^{x/n}}{x(1+x)(1+x/2)\cdots (1+x/n)} \\\\&=\frac{e^{-\gamma x}}{x}\prod_{n=1}^\infty e^{x/n}\left(1+\frac xn\right)^{-1}\tag 3 \end{align}$$

where $(3)$ gives the well-known Weierstrass product for Gamma.

Differentiating the logarithm of $(3)$ and setting $x=1$ reveals

$$\Gamma'(1)=\Gamma(1)\left(-\gamma -1+\sum_{n=1}^\infty \left(\frac1n-\frac{1}{n+1}\right)\right)=-\gamma \tag 4$$

Differentiating $(1)$ and setting $x=1$ yields

$$\Gamma'(1)=\int_0^\infty \log(x)e^{-x}\,dx \tag 5$$

whence comparing $(4)$ and $(5)$ we obtain the coveted result

$$\int_0^\infty \log(x)e^{-x}\,dx =-\gamma$$


In THIS ANSWER, I showed that that $\gamma$ as given by $\gamma=-\int_0^\infty e^{-x}\,\log(x)\,dx$ is equal to $\gamma$ as expressed by the limit $\gamma=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n\frac1k\right)$.

$\endgroup$
2
$\begingroup$

The same idea as in my previous answer applies. Notice that

$$ f(x) = \frac{1}{1+x} \quad \Rightarrow \quad \int_{\epsilon}^{\infty} \frac{f(x)}{x} \, dx = -\log\epsilon + \log(1+\epsilon) = -\log\epsilon + o(1) $$

as $\epsilon \to 0^+$. Now from the linked answer above, we find that

\begin{align*} f(x) = e^{-x} &\quad \Rightarrow \quad c(f) = \lim_{R\to\infty} \left( \int_{0}^{R} \frac{ds}{1+s} - \log R \right) - \gamma = -\gamma \\ &\quad \Rightarrow \quad \int_{\epsilon}^{\infty} \frac{e^{-x}}{x} \, dx = -\log\epsilon - \gamma + o(1). \end{align*}

It is worth to remark that the $\gamma$ term above is computed from the identity $\gamma = -\int_{0}^{\infty} e^{-x}\log x \, dx$, which you are already aware of. From this,

\begin{align*} \int_{\epsilon}^{\infty} \frac{1}{x^a + 1} \, \frac{dx}{x} - \int_{\epsilon}^{\infty} e^{-x} \, \frac{dx}{x} &= \frac{1}{a}\int_{\epsilon^a}^{\infty} \frac{1}{x + 1} \, \frac{dx}{x} - \int_{\epsilon}^{\infty} e^{-x} \, \frac{dx}{x}\\ &= \gamma + o(1) \end{align*}

and taking $\epsilon \to 0^+$ gives the result.


Using the quantity $c(f)$, you can compute various integrals (including all the integrals you have asked) together with some tabulated results for $c(f)$:

\begin{align*} c\left\{\frac{1}{(1+x)^\alpha}\right\} &= -H_{\alpha-1}, & c\{e^{-x}\} &= -\gamma, \\ c\left\{\frac{x}{e^x-1}\right\} &= 0, & c\{\cos x\} &= -\gamma, \end{align*}

where $H_n$ is the harmonic numbers. For instance, if $a > 0$ then

\begin{align*} \int_{0}^{\infty} \left( \frac{1}{\sqrt{1+a x}} - e^{-x^2} \right) \frac{dx}{x} &= c\left\{ \frac{1}{\sqrt{1+a x}} - e^{-x^2} \right\} \\ &= c\left\{ \frac{1}{\sqrt{1+a x}} \right\} - c\{e^{-x^2}\} \\ &= c\left\{ \frac{1}{\sqrt{1+x}} \right\} - \log a - \frac{1}{2}c\{e^{-x}\} \\ &= H_{-1/2} - \log a + \frac{\gamma}{2} \\ &= \frac{\gamma}{2} - \log(4a). \end{align*}

$\endgroup$
  • 1
    $\begingroup$ Thank you, Sangchul, for adding this. (+1) of course. And I like the idea of linking related questions/answers. -Mark $\endgroup$ – Mark Viola Feb 22 '17 at 19:26
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\left.\int_{0}^{\infty}\pars{{1 \over 1 + x^{a}} - {1 \over \expo{x}}} \,{\dd x \over x}\,\right\vert_{\ a\ >\ 0} = \lim_{\epsilon \to 0^{+}}\pars{% \int_{\epsilon}^{\infty}{\dd x \over \pars{1 + x^{a}}x} - \int_{\epsilon}^{\infty}{\expo{-x} \over x}\,\dd x} \\[5mm] = &\ \lim_{\epsilon \to 0^{+}}\braces{% {1 \over a}\int_{\epsilon^{a}}^{\infty}{\dd x \over \pars{1 + x}x} - \bracks{-\ln\pars{\epsilon}\expo{-\epsilon} + \int_{\epsilon}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}} \\[5mm] = &\ \gamma + \lim_{\epsilon \to 0^{+}}\bracks{% -{1 \over a}\ln\pars{\epsilon^{a} \over 1 + \epsilon^{a}} + \ln\pars{\epsilon}\expo{-\epsilon}}\label{1}\tag{1} \end{align}

because $\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \left.\totald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x \,\right\vert_{\ \mu\ =\ 0} = \left.\totald{\Gamma\pars{\mu + 1}}{\mu} \,\right\vert_{\ \mu\ =\ 0} =\Psi\pars{1} = -\gamma}$.

Expression \eqref{1} becomes: \begin{align} &\left.\int_{0}^{\infty}\pars{{1 \over 1 + x^{a}} - {1 \over \expo{x}}} \,{\dd x \over x}\,\right\vert_{\ a\ >\ 0} = \gamma\ +\ \underbrace{\lim_{\epsilon \to 0^{+}}\bracks{% {1 \over a}\ln\pars{1 + \epsilon^{a}} - \ln\pars{\epsilon}\pars{1 - \expo{-\epsilon}}}}_{\ds{=\ 0}}\ =\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{\gamma}} \end{align}

$\endgroup$
  • $\begingroup$ Felix, I appreciate your posting this. But if you have a look at the section that follows Equation $(4)$ in the OP, I was seeking alternative approaches to this. -Mark $\endgroup$ – Mark Viola Mar 9 '17 at 3:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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