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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Testcase
    Mar 28, 2022 at 16:44
  • $\begingroup$ Is $\alpha \geq 0$? $\endgroup$
    – jwimberley
    Mar 28, 2022 at 16:48
  • $\begingroup$ @jwimberley, must be bigger than $0$! I'll edit it now. $\endgroup$ Mar 28, 2022 at 16:52
  • $\begingroup$ @Testcase, can you give me a bit more detail on what you're suggesting? Not sure I follow (might be my physics background). $\endgroup$ Mar 28, 2022 at 16:52
  • 2
    $\begingroup$ The change of variables $v=\frac{1}{1+\lambda x^\alpha}$, assuming $\lambda>0$ will produce simple expression in terms of the Beta function. $\endgroup$
    – Mittens
    Mar 28, 2022 at 17:39

3 Answers 3


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

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

  • $\begingroup$ 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. $\endgroup$ Mar 28, 2022 at 17:36
  • $\begingroup$ Thank you! I've followed this through to something that looks nice and manageable! $\endgroup$ Mar 28, 2022 at 19:45
  • $\begingroup$ @fairyliquid You are welcome, but both Oliver Diaz remark and Quanto s solution are a lot smarter and cleaner than working with series. $\endgroup$
    – Testcase
    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 $$ \begin{aligned}I&=\frac{\left(7-\frac{3}{\alpha}\right)\left(6-\frac{3}{\alpha}\right) \left(5-\frac{3}{\alpha}\right)\left(4-\frac{3}{\alpha}\right)\left(3-\frac{3}{\alpha}\right) \left(2-\frac{3}{\alpha}\right)\left(1-\frac{3}{\alpha}\right) \pi}{7!\alpha \lambda^{\frac{3}{\alpha}} \sin \left(\frac{3 \pi}{\alpha}\right)}\\&=\frac{\pi \displaystyle \prod_{k=1}^{7}\left(1-\frac{3}{k \alpha}\right)}{\alpha \lambda^{\frac{3}{\alpha}} \sin \left(\frac{3 \pi}{\alpha}\right)}\end{aligned} $$


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