$$\int_{-\infty}^{\infty}\frac{dx}{(x^2+ax+a^2)(x^2+bx+b^2)}$$ a and b are real constants

How should I solve the above integral? Is there a nice, interesting method? or is partial fractions the only way to do it?

I did solve it using partial fractions, but it got really lengthy and cumbersome - wondering if there's a nicer way to do it.

Thanks a lot!

  • $\begingroup$ Are $a,b$ positive? $\endgroup$
    – Szeto
    Jul 2 '18 at 5:12
  • $\begingroup$ Nothing of that sort has been mentioned. $\endgroup$ Jul 2 '18 at 5:13
  • $\begingroup$ For what it's worth: WolframAlpha spits out an answer. wolframalpha.com/input/… $\endgroup$ Jul 2 '18 at 5:18
  • $\begingroup$ How's that gonna help? $\endgroup$ Jul 2 '18 at 5:19
  • $\begingroup$ Have you tried $y=x+(a+b)/4$ to make partial fractions less messy? $\endgroup$
    – J.G.
    Jul 2 '18 at 6:11

Wolfy’s answer is over-complicated, because it doesn’t know residue theorem:).

This integral is what exactly residue theorem can deal with easily.

The denominator can be factorized to $$(x-p|a|)(x-\overline p|a|)(x-q|b|)(x-\overline q|b|)$$

where $$p=\frac{-\text{sgn}(a)+i\sqrt 3}2$$ $$q=\frac{-\text{sgn}(b)+i\sqrt 3}2$$

Now, take a contour $C$ which is an infinitely large semicircle centered at the origin on the upper half of the complex plane.

By residue theorem, $$\oint_C\frac1{(x-p|a|)(x-\overline p|a|)(x-q|b|)(x-\overline q|b|)}dx=2\pi i\sum \text{residue included}$$

Also note that $$\oint_C=\underbrace{\int_{\text{arc}}}_{\to0}+\int^\infty_{-\infty}$$

The only residue included are at $A=p|a|$ and $B=q|b|$, and the residues at these two points are respectively $$\text{Res}_{p|a|}=\frac1{(p|a|-\overline p|a|)(p|a|-q|b|)(p|a|-\overline q|b|)}=\frac1{i\sqrt3|a| (p|a|-q|b|)(p|a|-\overline q|b|) }$$ $$\text{Res}_{q|b|}=\frac1{(q|b|-p|a|)(q|b|-\overline p|a|)(q|b|-\overline q|b|)}=\frac1{i\sqrt3|b| (q|b|-p|a|)(q|b|-\overline p|a|) }$$

The residues can be compactly written as

$$\text{Res}_{A}=\frac1{i\sqrt3|a| (A-B)(A-\overline B) }$$

$$\text{Res}_{B}=\frac1{i\sqrt3|b| (B-A)(B-\overline A) }$$

The computation is then complete:

$$\color{red}{\int^\infty_{-\infty}\frac1{(x^2+ax+a^2)(x^2+bx+b^2)}dx=\frac{2\pi}{\sqrt3(A-B)}\left(\frac1{|a|(A-\overline B)}+\frac1{|b|(\overline A-B)}\right)}$$

The residues might look complicated but do not be afraid to do some tedious algebra to add them up. There are surely some good simplifications/effective cancellations.

**Can you observe the hidden symmetry between $a$ and $b$ in the result? :)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.