How can one show that $\int_{0}^{\infty}\left({1\over 1+nx^n}-e^{-nx^n}\right)\cdot{\mathrm dx\over x^{1+n}}=1-\gamma?$

Question

Consider

$$\int_{0}^{\infty}\left({1\over 1+nx^n}-e^{-nx^n}\right)\cdot{\mathrm dx\over x^{1+n}}=1-\gamma\tag1$$ $n\ge1$;(integers)

n seem to be not involved in the closed form(why?)

How does one show that $(1)$ converges to $1-\gamma?$

An attempt:

$$(1+nx^n)^{-1}=1-nx^n+(nx^n)^2-(nx^n)^3+\cdots$$

$(1)$ becomes

$$\int_{0}^{\infty}\left(x^{-n-1}-nx^{-1}+n^2x^{n-1}-n^3x^{2n-1}+\cdots-{e^{-nx^n}\over x^{n+1}}\right)\mathrm dx\tag2$$

$(2)$ divgerges, how else can we tackle $(1)$?

Or do we have to differentiate m times w.r.t n

Answer 1

If you change variable $n x^n=y$, almost as suggested by tired in his comment, you end with $$\int_{0}^{\infty}\left({1\over 1+nx^n}-e^{-nx^n}\right)\cdot{\mathrm dx\over x^{1+n}}=\int_{0}^{\infty}\frac{\frac{1}{y+1}-e^{-y}}{y^2}\,dy$$ The antiderivative is given by $$\int\frac{\frac{1}{y+1}-e^{-y}}{y^2}\,dy=\text{Ei}(-y)+\frac{e^{-y}}{y}-\frac{1}{y}-\log (y)+\log (y+1)=f(y)$$ where appears the exponential integral.

Since $\lim_{y\to \infty } \, f(y)=0$, we are left with $\lim_{y\to 0 } \, f(y)$.

Using asymptotics , $$\text{Ei}(-y)=\gamma+\log (y) -y+\frac{y^2}{4}+O\left(y^3\right)$$ and Taylor series for small values of $y$

$$f(y)=(\gamma -1)+\frac{y}{2}-\frac{5 y^2}{12}+O\left(y^3\right)$$ which then leads to the result.

Answer 2

Claude Leibovicy already send a nice answer, while I was still typing.

Nevertheless, I joint my answer. Even if not smart, solving by brute force :

Answer 3

$$I= \int_{0}^{\infty}\frac{\frac{1}{x+1}-e^{-x}}{x^2}\,dx$$

Let

$$I(s)= \int_{0}^{\infty}\frac{x^{s-1}}{x+1}-x^{s-1}e^{-x}\,dx =\Gamma(s)\Gamma(1-s)-\Gamma(s)$$

Taking the limit

$$\lim_{s \to -1}\Gamma(s)\Gamma(1-s)-\Gamma(s) $$

Near $-1$ we have

$$\Gamma(s)\Gamma(1-s)-\Gamma(s) = -\frac{1}{s+1}+\frac{1}{s+1}-\gamma+1+O((s+1))$$

Hence

$$I= \int_{0}^{\infty}\frac{\frac{1}{x+1}-e^{-x}}{x^2}\,dx = 1-\gamma$$

Answer 4

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