# Alternative approaches to showing that $\gamma=\int_0^\infty \left(\frac{1}{1+x^a}-\frac{1}{e^x}\right)\,\frac1x\,dx$, $a>0$

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 $(1)$?

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

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.

• I can't figure out the part $(\frac 1{1+x^a}-e^{-x})\log x{|}_0^{\infty}=0-0$ – polfosol Jan 7 '17 at 7:58
• Really nice answer. – Muse_China Jan 7 '17 at 10:54
• @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 $x \to \infty$. – Chappers Jan 7 '17 at 15:09
• I found my misunderstanding. Thanks – polfosol Jan 7 '17 at 15:13
• (+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. – Mark Viola Jan 7 '17 at 15:18

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

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*}

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

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

• 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 – Mark Viola Mar 9 '17 at 3:35