2

Putting $ \displaystyle=4, \alpha=2, n=2 $ in my post,

$$\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}dx=\frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\right)$$

we can conclude that $$ \begin{aligned} I &=\frac{\pi}{4(2-1) !} \csc \frac{3 \pi}{4}\left(1-\frac{3}{4}\right) \\ &=\frac{\sqrt{2} \pi}{16} \end{aligned} $$

Question:

Is there any other method?

Your suggestion and alternative are warmly welcome.

Lai
  • 20,421
  • You can always use algorithms for rational integrands https://math.stackexchange.com/questions/3222937/algorithm-for-integration-of-a-rational-function :) – Maximilian Janisch Apr 03 '22 at 10:05
  • You can probably use the Residue Theorem. It might be more "simple" in terms of computations but you then assume a very powerful theorem. – Lelouch Apr 03 '22 at 10:08

3 Answers3

4

Integrate both sides of the equation $\left(\frac{x^3}{1+x^4}\right)’= \frac{4x^2}{(1+x^4)^2}-\frac{x^2}{1+x^4}$ to obtain $$\int_{0}^{\infty} \frac{x^{2}}{(1+x^{4})^{2}} d x = \frac14\int_{0}^{\infty} \frac{x^{2}}{1+x^{4}} d x=\frac14 \cdot \frac\pi{2\sqrt2}=\frac\pi{8\sqrt2} $$

Quanto
  • 97,352
2

You can use residue theorem. Start by extending the integral from $-\infty$ to $+\infty$, which you can do since the function is even. Then, use the semi-circle in the complex plane for integration: $\Gamma(\theta) = Re^{i\theta}$, with $\theta \in [0,\pi]$. You will get the desired result in the limit $R\rightarrow +\infty$. In that limit, the integral in the semi-circle vanishes due to Jordan Lemma. The residues inside the contour are $e^{i\pi/4}$ and $e^{i3\pi/4}$, with Res$\big(\frac{z^2}{(z^4+1)^2},z=e^{i\pi/4}\big)=\frac{e^{3i\pi/4}}{16}$ and Res$\big(\frac{z^2}{(z^4+1)^2},z=e^{i3\pi/4}\big)=\frac{e^{5i\pi/4}}{16}$. The result is then
$\int_{0}^{+\infty}dx \frac{x^2}{(x^4+1)^2} =\frac{1}{2} \int_{-\infty}^{+\infty}dx \frac{x^2}{(x^4+1)^2}=\pi i \sum \text{Res} = \frac{\pi i}{8}(e^{5i\pi/4} + e^{3i\pi/4} ) = \frac{\pi}{8\sqrt{2}} = \frac{\sqrt{2}\pi}{16} $

zuko1997
  • 61
2

Letting $x\mapsto \frac{1}{x} $ yields $$ I=\int_{0}^{\infty} \frac{x^{4}}{\left(x^{4}+1\right)^{2}} d x $$

Adding them together gives $$ \begin{aligned} 2 I &=\int_{0}^{\infty} \frac{x^{2}+x^{4}}{\left(x^{4}+1\right)^{2}} d x \\ &=\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}\right)^{2}} d x \\ &=\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^{2}+2\right]^{2}} \end{aligned} $$ Letting $x-\frac{1}{x}=\sqrt{2} \tan \theta$, we have

$$ I=\frac{\sqrt{2}}{8} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} \theta d \theta= \frac{\sqrt{2} \pi}{16} $$ We can conclude that

Lai
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