Some integral representations of the Euler–Mascheroni constant

Question

What kind of substitution should I use to obtain the following integrals? $$\begin{align} \int_0^1 \ln \ln \left(\frac{1}{x}\right)\,dx &=\int_0^\infty e^{-x} \ln x\,dx\tag1\\ &=\int_0^\infty \left(\frac{1}{xe^x} - \frac{1}{e^x-1} \right)\,dx\tag2\\ &=-\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx\tag3\\ &=\int_0^\infty \left( e^{-x} - \frac{1}{1+x^k} \right)\,\frac{dx}{x},\qquad k>0\tag4\\ \end{align}$$

This is not homework problems and I know that the above integrals equal to $-\gamma$ (where $\gamma$ is the Euler-Mascheroni constant). I got these integrals while reading this Wikipedia page. According to Wikipedia, the Euler–Mascheroni constant is defined as the limiting difference between the harmonic series and the natural logarithm: $$\gamma=\lim_{N\to\infty} \left(\sum_{k=1}^N \frac{1}{k} - \ln N\right)$$ but I don't know why can this definition be associated to the above integrals?

I can obtain the equation $(1)$ using substitution $t=\ln \left(\frac{1}{x}\right)\rightarrow x=e^{-t} \rightarrow dx=-e^{-t}\,dt$ and I know that $$\int_0^\infty e^{-x} \ln x\,dx=\Gamma'(1)=\Gamma(1)\psi(1)=-\gamma$$ but I can't obtain the rest. Any idea? Any help would be appreciated. Thanks in advance.

Answer 1

I am not sure if you can obtain the other three from the first integral through some substitution.

The $(3)$ one can be obtained from the definition of constant..

Notice that $$\sum_{i=1}^N\frac{1}{i}=\int_0^1 \frac{1-t^N}{1-t}\,dt$$ and $$\ln N=\int_0^1 \frac{t^{N-1}-1}{\ln t}\,dt$$ The above can be proved using Frullani's integral. Therefore $$\begin{aligned} \lim_{N\rightarrow \infty} \left(\sum_{i=1}^N\frac{1}{i}-\ln N\right) &=\lim_{N\rightarrow \infty}\int_0^1\left(\frac{1-t^N}{1-t}-\frac{t^{N-1}-1}{\ln t}\,dt\right)\,dt \\ &=\int_0^1 \left(\frac{1}{1-t}+\frac{1}{\ln t}\right)\,dt \end{aligned}$$ Make the substitution $\ln x=-t$ to obtain the $(2)$.

Answer 2

For the fourth one, express the integral as an iterated integral and then switch the order of integration.

$$ \begin{align} \int_{0}^{\infty} \left( e^{-x}- \frac{1}{1+x^{k}} \right) \frac{dx}{x} &= \int_{0}^{\infty} \int_{0}^{\infty} \left(e^{-x} - \frac{1}{1+x^{k}} \right) e^{-xt} \ dt \ dx \\ &= \int_{0}^{\infty} \int_{0}^{\infty} \left(e^{-x} - \frac{1}{1+x^{k}} \right) e^{-tx} \ dx \ dt \\ &= \int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-(t+1)x} \ dx - \int_{0}^{\infty} \frac{e^{-tx}}{1+x^{k}} \ dx \right) \ dt \\ &= \int_{0}^{\infty} \left( \frac{1}{t+1} - \int_{0}^{\infty} \frac{e^{-u}}{1+ (\frac{u}{t})^{k}} \frac{du}{t} \right) \ dt \\ &= \int_{0}^{\infty} \left(\int_{0}^{\infty}\frac{e^{-u}}{t+1} \ du - \int_{0}^{\infty} \frac{t^{k-1}e^{-u}}{t^{k}+u^{k}} \ du \right) \ dt \tag{1} \\ &= \int_{0}^{\infty} \int_{0}^{\infty} \left(\frac{1}{t+1} - \frac{t^{k-1}}{t^{k}+u^{k}} \right)e^{-u} \ du \ dt \\ &= \int_{0}^{\infty} e^{-u} \int_{0}^{\infty} \left(\frac{1}{t+1} - \frac{t^{k-1}}{t^{k}+u^{k}} \right) \ dt \ du \\ &= \int_{0}^{\infty} e^{-u} \ln \left(\frac{1+t}{(t^{k}+u^{k})^{1/k}} \right) \Bigg|^{t=\infty}_{t=0} \ du \\ &= \int_{0}^{\infty} e^{-u} \ln u = -\gamma \end{align}$$

$(1)$ Since $\int_{0}^{\infty} e^{-u} \ du = 1$

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Since I'm pretty sure I saw sos440 use this approach on another site to show that $$ \int_{0}^{\infty} \left(\cos (x) - \frac{1}{1+x^{2}} \right) \frac{dx}{x} = - \gamma,$$ I'm going to post this answer as a community wiki.

Answer 3

For equation $(2)$, make substitution $t=e^{-x}\,\Rightarrow\, x=-\ln t\,\Rightarrow\,dx=-\dfrac{dt}{t}$, then we get equation $(3)$ after $t\mapsto x$. \begin{equation} \int_0^\infty \left(\frac{1}{xe^x} - \frac{1}{e^x-1} \right)\,dx=\int_0^\infty \left(\frac{e^{-x}}{x} - \frac{e^{-x}}{1-e^{-x}} \right)\,dx=\int_0^1 \left(\frac{1}{1-x} \color{red}{+} \frac{1}{\ln x} \right)\,dx \end{equation} Indeed equation $(3)$ equals $-\gamma$ and here is the proof. We have \begin{equation} -\int_0^1 \left(\frac{1}{1-x} \color{red}{+} \frac{1}{\ln x} \right)\,dx=-\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx \end{equation} I check on the cited link (Wikipedia), there is a minor typo in $(3)$.

Proposition :

\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s \end{equation}

Proof :

Let \begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx \end{equation} then \begin{align*} I'(s)&=\int_0^1 \frac{1-x^s}{1-x}\,dx\\ I''(s)&=-\int_0^1 \frac{x^s\ln x}{1-x} \,dx\\ &=-\int_0^1\sum_{n=0}^\infty x^{n+s}\ln x\,dx\\ &=-\sum_{n=0}^\infty\partial_s\int_0^1 x^{n+s}\,dx\\ &=-\sum_{n=0}^\infty\partial_s\left[\frac{1}{n+s+1}\right]\\ &=\sum_{n=0}^\infty\frac{1}{(n+s+1)^2}\\ &=\psi_1(s+1)\\ I'(s)&=\int\psi_1(s+1)\,ds\\ I'(s)&=\int\frac{\partial}{\partial s}\bigg[\psi(s+1)\bigg]\,ds\\ I'(s)&=\psi(s+1)+C\\ \end{align*} For $s=0$, we have $I'(0)=0$. Implying $C=-\psi(1)=\gamma$, then \begin{align*} I'(s)&=\psi(s+1)+\gamma\\ I(s)&=\int \psi(s+1)\,ds +\gamma s+C\\ &=\int \frac{\partial}{\partial s}\bigg[\ln\Gamma(s+1)\bigg]\,ds +\gamma s+C\\ &=\ln\Gamma(s+1) +\gamma s+C\\ \end{align*} For $s=0$, we have $I(0)=0$. Implying $C=0$, then \begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s\qquad\qquad\square \end{equation}

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For $s=1$, we have

\begin{equation} -I(1)=-\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=-\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx=-\gamma \end{equation}

Answer 4

The easiest way to obtain this result is to use a regularization procedure. I will explain how to do the integral associated with a previous response, $$\int_{0}^{\infty}\left(\mathrm{cos}\space x-\frac{1}{1+x^2}\right)\frac{dx}{x}=-\gamma$$

This integral is correct. Easiest way to show is to regularize the integral by adding the small real number $\delta$ to the exponent of $x^{-1}$. This, after the $u$ substitution $u=ix$, gives $$\mathrm{Re}\left[\Gamma(\delta)\left(\frac{1}{i}\right)^\delta\right]-\frac{\pi}{2}\mathrm{csc}\left(\frac{\pi\delta}{2}\right).$$ Now, we just use the recursion identity $\delta\Gamma(\delta)=\Gamma(\delta+1)$ along with the Taylor series for $\Gamma(1+\delta)$ and sin($x$), to show that the integral is given by $-\gamma$. There are two $\frac{1}{\delta}$ terms that cancel, and the remaining contribution when $\delta$ is very small is $-\gamma$. This is a very neat integral for $\gamma$!
I'd like to see the reason why $$\int_{0}^{\infty}\frac{\mathrm{ln}(1+x)}{\mathrm{ln}^2(x)+\pi^2}\frac{dx}{x^2}=\gamma$$ I don't have any 'neat tricks' for this expression... would appreciate any!

Answer 5

The LHS of $(1)$ could be proven by using the fact that: $$\int_0^1(-\ln x)^sdx=\Gamma(s+1)$$Take the derivative of both sides: $$\int_0^1\ln\ln\left(\frac{1}{x}\right)(-\ln x)^sdx=\Gamma'(s+1)$$Let $s=0$. Then: $$\int_0^1\ln\ln\left(\frac{1}{x}\right)dx=\Gamma'(1)=-\gamma$$The derivative of $\Gamma$ is: $$\int_0^\infty e^{-t}t^{s-1}\ln tdt$$, and so we get the LHS of $(1)$. Take a look at this link for proofs on why $\Gamma'(1)=-\gamma$