Solution to nasty power-law integral $ \int^\infty_0\frac{x^2}{(1+\lambda x^\alpha)^8} dx$ I have the following integral
$$
\int^\infty_0\frac{x^2}{(1+\lambda x^\alpha)^8} dx
$$
with $ \alpha \gt 0$. Everything is real, and $\lambda$ is a numerical constant.
There is a nice easy solution for the case $\alpha = 2$. But I need a general solution as a function of $\alpha$. I would solve numerically, but unfortunately I require the $\alpha$ dependency for my final result.
Any pointers or advice welcome.
 A: Let $t= \lambda x^\alpha$
$$
I= 
\int^\infty_0\frac{x^2}{(1+\lambda x^\alpha)^8} dx=
\frac1{\alpha\lambda^{\frac{3-\alpha}{\alpha}} }\int^\infty_0\frac{t^{\frac{3-\alpha}{\alpha}} }{(1+t)^8} dt
$$
Then, apply the recursion
$$
\int^\infty_0\frac{t^b}{(1+t)^n} dt=\frac b{n-1}
\int^\infty_0\frac{t^{b-1}}{(1+t)^{n-1}} dt
$$
along with $\int^\infty_0\frac{t^{s-1}}{1+t} dt=\frac\pi{\sin{\pi s}}$ to obtain
$$I = \frac{\pi(1-\frac3 a)(1-\frac3{2a})(1-\frac3{3a})(1-\frac3{4a})(1-\frac3{5a})(1-\frac3{6a})(1-\frac3{7a})}{a\lambda^{\frac{3-a}{a}}\sin\frac{3\pi}{a} }
$$
A: Hint: One approach might be to write the integral as
$$ \int_0^{\infty}\frac{x^2}{(1+\lambda x^{\alpha})^8} dx =  \int_0^1\frac{x^2}{(1+\lambda x^{\alpha})^8 dx} +  \int_1^{\infty}\frac{x^2}{(1+\lambda x^{\alpha})^8} dx$$
In the second of the integrals on the right make the substitution $u=1/x$ giving us
$$ \int_0^{\infty}\frac{x^2}{(1+\lambda x^{\alpha})^8} dx =  \int_0^1\frac{x^2}{(1+\lambda x^{\alpha})^8} dx -  \int_0^1\frac{1}{u^2(1+\lambda u^{-\alpha})^8} \frac{-1}{u^2}du$$
(this caluculation should probably be checked carefully...).
Now we can use the expansion
$$\frac{1}{(1+\lambda x^{\alpha})^8} = \sum_{n=0}^{\infty} (-1)^{n}\frac{n(n+1)(n+2)\ldots (n+6)}{2 \cdots 7} \lambda^n x^{\alpha n}$$
which essientially arise from differentiating the usual geometric sum 7 times and inserting $(-\lambda x^\alpha)$ as the variable (again, you should check the details, this was litterally done on the back of an envelope).Using this the first of the integrals above can be written as power-series and integrated termwise, and a similar trick will apply ot the second one after some rewriting.
Note that if $\lambda$ is larger than $1$ or $\alpha$ is sufficiently small (less than 1/2) you will run into issues of convergence which you should probably give some thought to (the first one doesnt seem that serious, as for the second one I am not sure your original integral is even well-defined for $\alpha<1/2$). Anyway this could be a start.
A: First of all, transform the integral by letting $y=\lambda x^{\alpha}$, then
$$
I=\frac{1}{\alpha \lambda^{\frac{3}{\alpha}}} \int_{0}^{\infty} \frac{y^{\frac{3}{\alpha}-1}}{(1+y)^{8}} d y
$$
Then convert $I$ into a Beta Function by letting $z=\frac{1}{1+y}$,
$$
\begin{aligned}
I = \int_{0}^{1}z^{\left(8-\frac{3}{\alpha} \right)-1}(1-z)^{\frac{3}{\alpha}-1} d z =\frac{1}{\alpha \lambda^{\frac{3}{\alpha}}} B\left(8-\frac{3}{\alpha}, \frac{3}{\alpha}\right) =\frac{\Gamma\left(8-\frac{3}{\alpha}\right) \Gamma\left(\frac{3}{\alpha}\right)}{ \alpha \lambda^{\frac{3}{\alpha}}\Gamma(8)}
\end{aligned}
$$
Using the property of Gamma Function
$$
\Gamma(z+1)=z \Gamma(z),
$$
we can simplify the numerator as
$$
\begin{array}{l} \Gamma\left(8-\frac{3}{\alpha}\right) \Gamma\left(\frac{3}{\alpha}\right)=
\left(7-\frac{3}{\alpha}\right)\left(6-\frac{3}{\alpha}\right)\left(5-\frac{3}{\alpha}\right)\left(4-\frac{3}{\alpha}\right)\left(3-\frac{3}{\alpha}\right)
\left(2-\frac{3}{\alpha}\right)\left(1-\frac{3}{\alpha}\right) \Gamma\left(1-\frac{3}{\alpha}\right) \Gamma\left(\frac{3}{\alpha}\right)
\end{array}
$$
Using the Reflection Property of Gamma Function
$$
\Gamma(z) \Gamma(1-z)=\pi \csc (\pi z), \textrm{ where } z\not\in Z,
$$
we can now conclude that
$$
\begin{aligned}I&=\frac{\left(7-\frac{3}{\alpha}\right)\left(6-\frac{3}{\alpha}\right) \left(5-\frac{3}{\alpha}\right)\left(4-\frac{3}{\alpha}\right)\left(3-\frac{3}{\alpha}\right)
\left(2-\frac{3}{\alpha}\right)\left(1-\frac{3}{\alpha}\right) \pi}{7!\alpha \lambda^{\frac{3}{\alpha}} \sin \left(\frac{3 \pi}{\alpha}\right)}\\&=\frac{\pi \displaystyle  \prod_{k=1}^{7}\left(1-\frac{3}{k \alpha}\right)}{\alpha \lambda^{\frac{3}{\alpha}} \sin \left(\frac{3 \pi}{\alpha}\right)}\end{aligned}
$$
