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Where $a,b$ and $c$ are consecutive arithmetic terms.

We wish to evaluate this integral,

$$\int_{-\infty}^{\infty}\frac{x(x+a)(x+b)(x+c)}{(x^3+2x^2-x-1)^2+(x^2+x-1)^2}\mathrm dx$$

I don't even really know how to make an attempt.

If I expanded the denominator it is very messy.

I can't factorise $x^3+2x^2-x-1$ or $x^2+x-1.$

  • The denominator of your integrand $${x}^{6}+4,{x}^{5}+3,{x}^{4}-4,{x}^{3}-4,{x}^{2}+2$$ has only complex roots. – Dr. Sonnhard Graubner Jul 09 '19 at 10:02
  • Maple says your integral is infinity – Dr. Sonnhard Graubner Jul 09 '19 at 10:11
  • It cannot be infinity. The denominator is always positive and acts like $x^6$ when $x\to\pm\infty$, numerator goes like $x^4$, integral of $1/x^2$ converges on $x\to\pm\infty$ – Vasily Mitch Jul 09 '19 at 10:18
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    Integral is $(3 - a - b - c + a b + a c + b c) \pi$, but I need some time to explain why – Vasily Mitch Jul 09 '19 at 10:37
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    Is this some kind of contest problem? Can you share the source of it? – Zacky Jul 09 '19 at 12:08
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    @VasilyMitch It's just that $$x(x+a)(x+b)(x+c)=(abc)x + (ab+bc+ac)x^2 +(a+b+c)x^3+x^4$$ And the information about the consecutive arithmetic terms is useless as it holds for any numbers. Let $p(x)$ to be the denominator, then: $$\int_{-\infty}^\infty \frac{x}{p(x)}dx=0,\quad \int_{-\infty}^\infty \frac{x^2}{p(x)}dx=\pi,\quad \int_{-\infty}^\infty \frac{x^3}{p(x)}dx=-\pi,\quad \int_{-\infty}^\infty \frac{x^4}{p(x)}dx=3\pi $$ My belief is that it's just another case of this: https://math.stackexchange.com/a/2821162/515527. Finding what substitution was made, is quite hard though. – Zacky Jul 09 '19 at 12:49
  • @Zacky, Yeah, I took it with residues. But I can't explain how to take those without Mathematica. They seem magical. I will think about your link. Thanks – Vasily Mitch Jul 09 '19 at 12:55

2 Answers2

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This is an adaption of my answer here, many similar evaluations are also given there.

Two reasons enable the integral to evaluate nicely:

  • Firstly reason: the denominator $(x^3+2x^2-x-1)^2+(x^2+x-1)^2$ factors as $$\underbrace{\left(x^3+(2-i) x^2-(1+i) x+(-1+i)\right)}_{h_1(x)} \underbrace{\left(x^3+(2+i) x^2-(1-i) x+(-1-i)\right)}_{h_2(x)}$$

  • Second reason: roots of $h_1$ all lie in upper plane, denote them by $\alpha,\beta,\gamma$.

Now residue theorem implies $$\frac{1}{{2\pi i}}\int_{ - \infty }^\infty {\frac{{p(x)}}{{{h_1}(x){h_2}(x)}}dx} = \frac{{p(\alpha )}}{{{h_1}'(\alpha ){h_2}(\alpha )}} + \frac{{p(\beta )}}{{{h_1}'(\beta ){h_2}(\beta )}} + \frac{{p(\gamma )}}{{{h_1}'(\gamma ){h_2}(\gamma )}}$$

The RHS is a symmetric function in $\alpha,\beta,\gamma$, roots of $h_1 \in \mathbb{Q}(i)[x]$. If $p(x)\in \mathbb{Q}[x]$, then without any computation, we know that $$\frac{1}{{2\pi i}}\int_{ - \infty }^\infty {\frac{{p(x)}}{{{h_1}(x){h_2}(x)}}dx} \in \mathbb{Q}(i)$$ since the integral is real, $\frac{1}{\pi }\int_{ - \infty }^\infty {\frac{{p(x)}}{{{h_1}(x){h_2}(x)}}dx} \in \mathbb{Q}$. This explains the nice result. The rational number can be explicitly calculated via elementary symmetric polynomials, a cumbersome but mechanical process, the results for $p(x) = x,x^2,x^3,x^4$ are already pointed out in comment.

pisco
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I will show the calculation of (On upper plane) $$\sum_r Res\left(\frac{x}{h(x)},r\right)=0,\quad \sum_r Res\left(\frac{x^2}{h(x)},r\right)=\frac{1}{2i},\quad \sum_r Res\left(\frac{x^3}{h(x)},r\right)= -\frac{1}{2i},\quad \sum_r Res\left(\frac{x^4}{h(x)},r\right)=\frac{3}{2i} $$ using formula $Res(f,r)=\frac{g(r)}{h'(r)}$

Denominator $$h(x)=(x^3+2x^2-x-1)^2+(x^2+x-1)^2 \\ =((x^3 + 2 x^2 - x - 1) - i (x^2 + x - 1))((x^3 + 2 x^2 - x - 1) + i (x^2 + x - 1))$$

Numeric calculation show that all root(all roots are simple) in upper plane is given by $$(x^3 + 2 x^2 - x - 1) - i (x^2 + x - 1)=0$$ So if $r$ is a root of $$(x^3 + 2 x^2 - x - 1) - i (x^2 + x - 1)$$ Then $$r^3 = -(2-i) r^2+(1+i) r+(1-i)$$ Let $S=-(2-i),J=-(1+i),P=(1-i)$, Vieta's formulas will be applied later.

Derivative of denominator is given by $$6 x^5+20 x^4+12 x^3-12 x^2-8 x$$ Note that $$Res(f,r)=\frac{r^n}{6 r^5+20 r^4+12 r^3-12 r^2-8 r}\\ =\frac{r^{n-1}}{6 r^4+20 r^3+12 r^2-12 r-8}$$ Using $r^3 = -(2-i) r^2+(1+i) r+(1-i)$ we can always reduces polynomial of r to degree not greater than $2$, i.e.: $$ 6 r^4+20 r^3+12 r^2-12 r-8=(6 - 2 i) - (4 - 8 i) r - (4 - 2 i) r^2 $$ The denominator of $$\sum_{r=A,B,C} \frac{g(r)}{(6 - 2 i) - (4 - 8 i) r - (4 - 2 i) r^2 }$$ is given by $$ (52 + 36 i) \\ -(88-16 i) A-(88-16 i) B-(88-16 i) C \\ -(40 + 20 i) A^2 - (40 + 20 i) B^2 - (40 + 20 i) C^2 \\ +(72 - 104 i) A B + (72 - 104 i) A C + (72 - 104 i) B C \\ +(60 - 20 I) A^2 B + (60 - 20 I) A B^2 + (60 - 20 i) A C^2 + (60 - 20 i) B C^2 + (60 - 20 i) A^2 C + (60 - 20 i) B^2 C\\ +(30 + 10 i) A^2 B^2 + (30 + 10 i) A^2 C^2 + (30 + 10 i) B^2 C^2 \\ -(40 - 80 i) A^2 B C - (40 - 80 i) A B^2 C - (40 - 80 i) A B C^2 \\ -(40 - 20 i) A^2 B^2 C - (40 - 20 i) A B^2 C^2 - (40 - 20 i) A^2 B C^2\\ +(32 + 176 i) A B C \\ -(22 + 4 i) A^2 B^2 C^2 $$ which is equal to $$ (52 + 36 i) - (88 - 16 i) S - (40 + 20 i) (S^2 - 2 J) + (72 - 104 i) J + (60 - 20 i) (S J - 3 P) + (30 + 10 i) (J^2 - 2 P S) - (40 - 80 i) P S - (40 - 20 i) P J + (32 + 176 i) P - (22 + 4 i) P^2 \\ =56 + 16 i $$ When $g(r)=1$, the numerator is given by $$ (18 + 24 i)\\ -(28 + 4 i) A - (28 + 4 i) B - (28 + 4 i) C \\ -(10 + 10 i) A^2 - (10 + 10 i) B^2 - (10 + 10 i) C^2 \\ +(16 - 12 i) A B + (16 - 12 i) B C + (16 - 12 i) A C \\ +(4 + 3 i) B^2 C^2 + (4 + 3 i) A^2 C^2 + (4 + 3 i) A^2 B^2 \\ +10 A^2 B + 10 A B^2 + 10 A^2 C + 10 B^2 C + 10 A C^2 + 10 B C^2 $$ which is equal to $$ (18 + 24 i) - (28 + 4 i) S - (10 + 10 i) (S^2 - 2 J) + (16 - 12 i) J + (4 + 3 i) (J^2 - 2 P S) + 10 (S J - 3 P) \\ =0 $$ When $g(r)=r$, the numerator is given by $$ (6 + 8 i) A + (6 + 8 i) B + (6 + 8 i) C\\ -(28 + 4 i) A B - (28 + 4 i) A C - (28 + 4 i) B C\\ -(5 + 5 i) A^2 B - (5 + 5 i) A B^2 - (5 + 5 i) A^2 C - (5 + 5 i) B^2 C - (5 + 5 i) A C^2 - (5 + 5 i) B C^2\\ +(48 - 36 i) A B C\\ +20 A^2 B C + 20 A B^2 C + 20 A B C^2\\ +(4 + 3 i) A^2 B^2 C + (4 + 3 i) A^2 B C^2 + (4 + 3 i) A B^2 C^2 $$ which is equal to $$ (6 + 8 i) S - (28 + 4 i) J - (5 + 5 i) (S J - 3 P) + (48 - 36 i) P + 20 S P + (4 + 3 i) P J\\ =8 - 28 i $$ When $g(r)=r^2$, the numerator is given by $$ (6 + 8 i) A^2 + (6 + 8 i) B^2 + (6 + 8 i) C^2 \\ -(14 + 2 i) A^2 B - (14 + 2 i) A B^2 - (14 + 2 i) A^2 C -(14 + 2 i) B^2 C - (14 + 2 i) A C^2 - (14 + 2 i) B C^2\\ -(10 + 10 i) A^2 B^2 - (10 + 10 i) A^2 C^2 - (10 + 10 i) B^2 C^2 \\ +(16 - 12 i) A^2 B C + (16 - 12 i) A B^2 C + (16 - 12 i) A B C^2 \\ +20 A^2 B^2 C + 20 A^2 B C^2 + 20 A B^2 C^2\\ +(12 + 9 i) A^2 B^2 C^2\\ $$ which is equal to $$ (6 + 8 i) (S^2 - 2 J) - (14 + 2 i) (S J - 3 P) - (10 + 10 i) (J^2 - 2 P S) + (16 - 12 i) S P + 20 P J + (12 + 9 i) P^2\\ =-8 + 28 i $$ The result followed by togethering all the things.