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Integral $$I=\int_{-\infty}^{\infty}\left(\frac{x^{2}-x+\pi}{x^{4}-x^{2}+1}\right)^{2}dx$$ I have solved this integral using partial decomposition but it is very lengthy and I had to use wolfram to do the heavy computation. Is their any other way of solving this integral. The answer I got is $\pi+\pi^{2}+\pi^{3}$. Since the answer is so clean I think there seems to be a trick which can be utilized to solve this.

jayesh kumar gupta
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  • True, partial fraction decomposition will result with a lengthy antiderivate – I was not able to write it here as a comment. It will be interesting to see does someone else have alternative approach which will yield to a simplest antiderivate. – Anton Vrdoljak Nov 18 '23 at 17:05
  • If you know a bit of complex analysis you can try to use the residue theorem – Marco Nov 18 '23 at 17:44
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    Note that $\pi$ is a red herring. In fact $$\int_{-\infty}^{\infty}\left(\frac{x^{2}-x+a}{x^{4}-x^{2}+1}\right)^{2}dx = \pi a^2+\pi a + \pi$$ – GEdgar Nov 18 '23 at 18:00

1 Answers1

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The four roots of the denominator are $x=\pm i \pm \sqrt 3$.

The integrand goes to zero as $r^{-4}$ for $r \to\infty$.

The integral can be closed by a half circle of radius $r> |i + \sqrt 3|$ in the upper or lower complex half plane, divided by $2 \pi $, evaluating the residuum, the coefficient of $1/(z-z0_k)$ for the different series expansions around the poles to the left of the countour. This yields

$$2 \pi \ i \ \text{res}\left(\left(\frac{x^2-x+\pi }{x^4-x^2+1}\right)^2,\frac{1}{2} \left(\sqrt{3}-i\right)\right)+2 \pi \ i \ \text{res}\left(\left(\frac{x^2-x+\pi }{x^4-x^2+1}\right)^2,-\frac{1}{2} \left(\sqrt{3}+i\right)\right)=- \pi (1 + \pi + \pi^2)$$

$$2 \pi \ i \ \text{res}\left(\left(\frac{x^2-x+\pi }{x^4-x^2+1}\right)^2,-\frac{1}{2} \left(\sqrt{3}+i\right)\right)+2 \pi \ i \ \text{res}\left(\left(\frac{x^2-x+\pi }{x^4-x^2+1}\right)^2,\frac{1}{2} \left(\sqrt{3}-i\right)\right)= \pi (1 + \pi + \pi^2)$$

eg by

$$\begin{align}&\text{Coefficient}\left[\text{Series}\left[\left(\frac{x^2-x+\pi }{x^4-x^2+1}\text{/.}\, \left\{x\to y+\frac{1}{2} \left(-\sqrt{3}+i\right)\right\}\right)^2,\{y,0,0\}\right],\frac{1}{y}\right]+\\ & \text{Coefficient}\left[\text{Series}\left[\left(\frac{x^2-x+\pi }{x^4-x^2+1}\text{/.}\, \left\{x\to y+\frac{1}{2} \left(\sqrt{3}+i\right)\right\}\right)^2,\{y,0,0\}\right],\frac{1}{y}\right] =-\frac{1}{2} i \ \left(1+\pi +\pi ^2\right) \end{align} $$

Roland F
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