Generalisation of an already generalised integral

Question

Inspired by these two questions:

Closed form for $\int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$

Interesting integral formula

I ask whether the following integral has a closed form:

$$\int_0^\infty \frac{x^n}{(1+x^m)^\alpha} \text{d}x$$

For $\alpha \in \mathbb{N}$ and $0<n<\alpha m+1$

If that is too difficult, then I will also accept the standard restriction from the other questions

$0<n<m+1$

Other generalised forms of the integral are also acceptable, such as $$\int_0^\infty \frac{x^n}{(a+bx^m)^\alpha} \text{d}x$$

And an extension of this question for those who want something more:

Does a closed form exist for $\alpha \in \mathbb{R}^+$?

Answer 1

For all definite integrals of the form $\displaystyle\int_0^\infty \frac{x^n}{(1+x^m)^\alpha}~dx,~$ the substitution $t=\dfrac1{1+x^m}$

still applies. This will yield a result in terms of the beta function. If $a\in$ N, with the help

of the recurrence relation and reflection formula for the $\Gamma$ function, this can be then

further simplified to rational or algebraic multiples of $\pi.~$ As for $\displaystyle\int_0^\infty\frac{x^n}{(a+bx^m)^\alpha}~dx,~$

let $bx^m=au^m$ first, then factor a outside the parentheses, and apply the exact same

approach as the one presented above.

Answer 2

$\text {Let } y=\dfrac{1}{1+x^{m}}, \textrm{then } x=\left(\dfrac{1}{y}-1\right)^{\frac{1}{m}} \textrm{ and } d x=\dfrac{1}{m} \left(\dfrac{1}{y}-1\right)^{\frac{1}{m-1}}\left(\dfrac{d y}{-y^{2}}\right).$

Simplifying gives $$ \begin{aligned} \int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}} &=\frac{1}{m} \int_{0}^{1} y^{\left(\alpha-\frac{n+1}{m}\right)-1}(1-y)^{\frac{n+1}{m}-1} d y \\ &=\frac{1}{m} B\left(\alpha-\frac{n+1}{m}, \frac{n+1}{m}\right) \\ &=\frac{1}{m} \frac{\Gamma\left(\alpha-\frac{n+1}{m}\right) \Gamma\left(\frac{n+1}{m}\right)}{\Gamma(\alpha)} \end{aligned} $$ Using the property of Gamma Function: $\Gamma(z+1)=z \Gamma(z)$ repeatedly yields $$ \begin{aligned} \Gamma\left(\alpha-\frac{n+1}{m}\right) &=\left(\alpha-1-\frac{n+1}{m}\right) \Gamma\left(\alpha-1-\frac{n+1}{m}\right) \\ & \qquad\qquad \vdots \\ &=\prod_{k=1}^{\alpha -1}\left(\alpha-k-\frac{n+1}{m}\right) \Gamma\left(1-\frac{n+1}{m}\right) \end{aligned} $$ Using Euler’s Reflection Theorem: $ \quad \Gamma(z) \Gamma(1-z)=\pi \csc (\pi z) \textrm{ for } z\notin Z $ gives

Case 1: When $n=1$,

$$\int_{0}^{\infty} \frac{x^{n}}{1+x^{m}}= \frac{\pi}{m} \csc \frac{(n+1) \pi}{m} $$

Case 2: When $n\geq 2$, $$ \begin{aligned} \int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}&=\frac{1}{m(\alpha-1) !} \left[\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right) \right]\Gamma\left(1-\frac{n+1}{m}\right) \Gamma\left(\frac{n+1}{m}\right)\\&=\boxed{ \frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)} \end{aligned} $$

For the more general integral $\displaystyle \int_{0}^{\infty} \dfrac{x^{n}}{\left(a+b x^{m}\right)^{\alpha}} d x$, we let $\displaystyle u=\sqrt[m]{\frac{b}{a}} x$ and get $$ \begin{aligned} \int_{0}^{\infty} \dfrac{x^{n}}{\left(a+b x^{m}\right)^{\alpha}} d x &=\int_{0}^{\infty} \frac{\left(\frac{a}{b}\right)^{\frac{n}{m}} u^{n}}{\left(a+a u^{m}\right)^{\alpha}} \left(\frac{a}{b}\right)^{\frac{1}{m}} d u\\ &=\left(\frac{a}{b}\right)^{\frac{n+1}{m}} \cdot a^{-\alpha} \int_{0}^{\infty} \frac{u^{n}}{\left(1+u^{m}\right)^{\alpha}} d u \\ &=\boxed{\frac{a^{\frac{n+1}{m}-\alpha}\pi}{b^{\frac{n+1}{m}} m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)} \end{aligned} $$