Evaluating $I(n) = \int^{\infty}_{0} \frac{\ln(x)}{x^n(1+x)}\, dx$ for real $n$

Question

I am not sure how to handle the additional parameter $n$. I first need to find out for which real values of $n$ will the integral converge. Based on intuition and checking with mathematica, I believe it will converge only for $0 < n < 1$, although I am not sure and I would like to see exactly how to determine those values. For this integral, I would normally try using a branch cut, but again I am not sure how to apply it given my additional parameter $n$. $$I(n) = \int^{\infty}_{0} \dfrac{\ln(x)}{x^n(1+x)}\, dx$$

Answer 1

Let $y = \frac{x}{1+x}$, then

$$dy = \frac{dx}{(1+x)^2},\;1 - y = \frac{1}{1+x}\;\text{ and }\;x = \frac{y}{1-y}$$

For $0 < n < 1$, we have $$\begin{align} I_n = & \int_0^\infty \frac{\log x}{x^n(1+x)} dx\\ = & -\frac{\partial}{\partial n}\left[\int_0^\infty \frac{1}{x^n(1+x)} dx\right]\\ = & -\frac{\partial}{\partial n}\left[\int_0^\infty \left(\frac{1+x}{x}\right)^n\left(\frac{1}{1+x}\right)^{n-1} \frac{dx}{(1+x)^2}\right]\\ = & -\frac{\partial}{\partial n}\left[\int_0^1 y^{-n}(1-y)^{n-1} dy\right]\\ = & -\frac{\partial}{\partial n}\left[\frac{\Gamma(n)\Gamma(1-n)}{\Gamma(1)}\right] = -\frac{\partial}{\partial n}\left[\frac{\pi}{\sin(\pi n)}\right] = \frac{\pi^2 \cos(\pi n)}{\sin(\pi n)^2} \end{align} $$

Answer 2

I think the OP wanted to see some complex analysis. Here goes:

Consider

$$\oint_C dz \frac{z^{-n} \log{z}}{1+z}$$

where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$. Thus the contour integral has four pieces:

$$\int_0^R dx \frac{x^{-n} \log{x}}{1+x} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} R^{-n} e^{-i n \theta} \frac{\log{(R e^{i \theta})}}{1+R e^{i \theta}} \\ + e^{-i 2 \pi n} \int_R^0 dx \, x^{-n} \frac{\log{x}+i 2 \pi}{1+x} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \epsilon^{-n} e^{-i n \phi} \frac{\log{(\epsilon e^{i \phi})}}{1+\epsilon e^{i \phi}}$$

As $R \to\infty$, the second integral vanishes only when $n \gt 0$. As $\epsilon \to 0$, the fourth integral vanishes only when $n \lt 1$. Thus we restrict $n \in (0,1)$, and the integral is equal to

$$\left (1-e^{-i 2 \pi n} \right )\int_0^{\infty} dx \frac{x^{-n} \log{x}}{1+x} - i 2 \pi \, e^{-i 2 \pi n} \int_0^{\infty} dx \frac{x^{-n} }{1+x}$$

The contour integral is equal to, by the residue theorem, $i 2 \pi$ times the residue at the pole $z=-1 = e^{i \pi}$, or

$$i 2 \pi e^{-i n \pi} (i \pi) = -2 \pi^2 (\cos{n \pi} - i \sin{n \pi})$$

We find the integral sought after by equating real and imaginary parts. Let $A$ be the first integral (with the log) and $B$ the second (without the log). Then

$$(1-\cos{2 \pi n}) A - (2 \pi \sin{2 \pi n}) B = -2 \pi^2 \cos{\pi n}$$ $$(\sin{2 \pi n}) A - (2 \pi \cos{2 \pi n}) B = 2 \pi^2 \sin{\pi n}$$

Eliminating the $B$ pieces by multiplying by $\cos{2 \pi n}$ in the first equation and $-\sin{2 \pi n}$ in the second, we get

$$(\cos{2 \pi n}-1) A = -2 \pi^2 \cos{\pi n}$$

or

$$A = \int_0^{\infty} dx \frac{x^{-n} \log{x}}{1+x} = \pi^2 \frac{\cos{\pi n}}{\sin^2{\pi n}}$$

As a bonus, you can show that

$$B = \int_0^{\infty} dx \frac{x^{-n} }{1+x} = \frac{\pi}{\sin{\pi n}}$$

Answer 3

Note that if $$f(a) = \int_0^{\infty} \dfrac{dx}{x^a(1+x)}$$ then $I(n) = -f'(n)$. And from here, we have $$f(a) = \pi \csc(\pi a) \implies I(a) = \pi \csc(\pi a) \cot(\pi a)$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

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\begin{align} \on{I}\pars{n} & \equiv \bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{x} \over x^{n}\pars{1 + x}}\,\dd x\,\right\vert_{\,\Re\pars{n}\ <\ 1}} = \left.\partiald{}{\nu}\int_{0}^{\infty} {x^{\color{red}{\nu - n + 1} - 1} \over 1 + x}\,\dd x\,\right\vert_{\,\nu\ =\ 0} \end{align} Note that $\ds{{1 \over 1 + x} = \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}} \,{\pars{-x}^{k} \over k!}}$.

Then, \begin{align} \on{I}\pars{n} & \equiv \bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\ln\pars{x} \over x^{n}\pars{1 + x}}\,\dd x\,\right\vert_{\,\Re\pars{n}\ <\ 1}} \\[5mm] = &\ \left.\partiald{}{\nu}\, \Gamma\pars{\color{red}{\nu - n + 1}} \Gamma\pars{1 - \bracks{\color{red}{\nu - n + 1}}} \,\right\vert_{\,\nu\ =\ 0}\quad \pars{\substack{\ds{Ramanujan's}\\[0.5mm]\ds{Master} \\[0.5mm]\ds{Theorem}}} \\[5mm] = &\ \left.\partiald{}{\nu}\, {\pi \over \sin\pars{\pi\bracks{\nu - n + 1}}} \,\right\vert_{\,\nu\ =\ 0} = \left.-\pi\,\partiald{\csc\pars{\pi\bracks{\nu - n}}}{\nu} \,\right\vert_{\,\nu\ =\ 0} \\[5mm] = &\ \left.\pi^{2}\csc\pars{\pi\bracks{\nu - n}} \cot\pars{\pi\bracks{\nu - n}} \,\right\vert_{\,\nu\ =\ 0} \\[5mm] = &\ \bbx{\pi^{2}\csc\pars{n\pi}\cot\pars{n\pi}} \\ & \end{align}

Answer 5

Firstly, to determine the range of $n$ for which the integral converges, just look at the singularity at $0.$ To integrate, and put the pesky $n$ where it is easier to deal with, make the substitution $u=\log x.$ This will transform your integral to $\int_{-\infty}^{\infty} u \exp(-(n-1) u)/(1+\exp u) du,$ which should be easy by standard contour integration methods. To check your computation, the answer is $\pi^2 \cot(n \pi) \csc(n \pi),$ provided $0<n<1.$