Improper integral of $\int_{0}^{\infty} \frac{x^p}{1+x}dx$ with$ -1<p<0$

Question

I have a question, I need to solve the following equation:

$\int \limits_0^\infty \frac{x^p}{1+x}dx$

I can see that $x^p$ can be rewritten using $q\equiv -p$:

$\int \limits_0^\infty \frac{1}{x^q(1+x)}dx, 0<q<1$

There are two singularities, at $z=0$ and $z=-1$. Only $z=0$ is on the path, but what would be the best way to tackle this problem? I am stuck from here on...

Answer 1

Here is a very effective approach using integration in the complex plane. Let choose the branch cut of the function $z^{p}$ along the positive real semi-axis and consider the following integration contour $\Gamma$: $$\begin{align} &1)\;z=x,& x:0\to R\\ &2)\;z=Re^{i\phi},&\phi:0\to 2\pi\\ &3)\;z=x,& x:R\to0 \end{align}$$

The integral of the function $$ f(z)=\frac{z^{p}}{1+z} $$ over the circle vanishes as $R\to\infty$ due to Jordan's lemma. While on the top of the branch cut the function evaluates to $$\frac{x^{p}}{1+x}$$ on the bottom of the cut it is $$\frac{(xe^{i2\pi})^{p}}{1+x}=e^{i2\pi p}\frac{x^{p}}{1+x}.$$

Thus we have $$\operatorname{Res}_{z=-1}f(z)=e^{i\pi p}=\frac1{2\pi i} \int_{\Gamma}f(z)dz=\frac{1-e^{i2\pi p}}{2\pi i}\int_{0}^\infty\frac{x^{p}}{1+x}dx $$ or $$\int_{0}^\infty\frac{x^{p}}{1+x}dx=\frac{2\pi i\; e^{i\pi p} }{1-e^{i2\pi p}} =-\frac\pi{\sin\pi p}. $$

Answer 2

Let rewrite the integral as $$ \int_0^\infty\frac {x^{-p}}{1+x}dx, $$ where $p$ is a real number. The integral converges if and only if $0<p<1$.

Now consider the following transformations: $$\begin{align} \int_0^\infty\frac {x^{-p}}{1+x}dx&=\int_0^1+\int_1^\infty\frac {x^{-p}}{1+x}dx\tag1\\ &=\int_0^1\frac {x^{-p}+x^{p-1}}{1+x}dx\tag2\\ &=\int_0^1(x^{-p}+x^{p-1})\left(\sum_{n\ge0}(-x)^n\right)dx\tag3 \\ &=\sum_{n\ge0}(-1)^n\int_0^1(x^{-p+n}+x^{p-1+n})dx\tag4\\ &=\sum_{n\ge0}(-1)^n\left[\frac1{-p+n+1}+\frac1{p+n}\right]\tag5\\ &=\sum_{n\ge0}\left[\frac{(-1)^{-n-1}}{p-n-1}+\frac{(-1)^n}{p+n}\right]\tag6\\ &=\sum_{n=-\infty}^\infty\frac{(-1)^n}{p+n}.\tag7 \end{align}$$

It can be proved in various ways that: $$ \sum_{n=-\infty}^\infty\frac{(-1)^n}{p+n}=\frac\pi{\sin\pi p}.\tag8 $$

The simplest way is probably to use the Mittag-Leffler's theorem. Indeed both expressions in $(8)$ considered as functions of $p$ have the same set of simple poles at $p=n$ with residues $(-1)^n$.

Explanations:

$(1)\to(2)$: substitution $x\mapsto\frac1x$ into the second integral.

$(2)\to(3)$: geometric series expansion of $\frac1{1+x}$ $(0<x<1)$.

$(3)\to(4)$: the order of integration and summation are interchanged.

$(4)\to(5)$: powers are integrated (observe that both $-p+n+1$ and $p+n$ are positive for $n\ge0$).

$(5)\to(6)$: the factor $(-1)^n$ is moved to fractions. the numerator and denominator of first fraction are then negated.

$(6)\to(7)$: reindexing.