# 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.

• If you only need it numerically for various $\alpha$ then how about splitting it into two parts, making the substitution $x \to 1/x$ in the second piece, and then expanding each as a powerseries and integrating termwise ( maybe more pieces if you ned faster convergence). Maybe the resulting series looks familiar. Mar 28, 2022 at 16:44
• Is $\alpha \geq 0$? Mar 28, 2022 at 16:48
• @jwimberley, must be bigger than $0$! I'll edit it now. Mar 28, 2022 at 16:52
• @Testcase, can you give me a bit more detail on what you're suggesting? Not sure I follow (might be my physics background). Mar 28, 2022 at 16:52
• The change of variables $v=\frac{1}{1+\lambda x^\alpha}$, assuming $\lambda>0$ will produce simple expression in terms of the Beta function. Mar 28, 2022 at 17:39

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 $$I = \frac{\pi(1-\frac3 a)(1-\frac3{2a})(1-\frac3{3a})(1-\frac3{4a})(1-\frac3{5a})(1-\frac3{6a})(1-\frac3{7a})}{a\lambda^{\frac{3-a}{a}}\sin\frac{3\pi}{a} }$$

• Thanks! Just to double check, 'a' in the final line is $\alpha$, right? Mar 28, 2022 at 19:44
• @FairyLiquid - correct, easier to type Mar 28, 2022 at 20:11

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.

• If $\alpha > 1/2$, then writing the second integrand as $u^{8\alpha - 4}/(u^\alpha + \lambda)^8$ keeps all the calculations finite. $3/8 < \alpha < 1/2$ is more annoying as you have an integrable divergence at the origin. Mar 28, 2022 at 17:36
• Thank you! I've followed this through to something that looks nice and manageable! Mar 28, 2022 at 19:45
• @fairyliquid You are welcome, but both Oliver Diaz remark and Quanto s solution are a lot smarter and cleaner than working with series. Mar 28, 2022 at 20:46

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}