Fdtxjr

Evaluate $$I=int_0^inftyfrac(x^2+x+1)dx(x^3+x+1)sqrtx$$

My try:

Letting $sqrtx=t$ we get

$$I=2int_0^infty fract^4+t^2+1(t^6+t^2+1),dt$$

Now

$$t^4+t^2+1=(t^2+1)^2-(t^2+1)+1$$

$$t^6+t^2+1=(t^2+1)^3-2(t^2+1)^2+(t^2+1)+1$$

Any clue then?

Let me attack the original problem, as stated in Brilliant.org.

Question. If

$$ alpha = frac1pi int_0^infty fracx^2 + x + 1x^3 + x + 1 fracmathrmdxsqrtx, $$

then what is the value of $31alpha^4 - 40alpha^2 - 24alpha$?

Remark. Once we identify the value of $31alpha^4 - 40alpha^2 - 24alpha$, the problem of determining $alpha$ reduces to solving the quartic equation. Since such formula is known, it suffices to answer to the question.

Step 1 - Reduction to algebraic sum. First, write $p(x) = x^2 + x + 1$ and $q(x) = x^3 + x + 1$ for simplicity. Then $alpha$ is written as

$$ alpha = frac1pi int_0^infty fracp(x)q(x)sqrtx , mathrmdx. $$

Let $operatornameLog$ denote the complex logarithm so that $ImoperatornameLog(z) in [0, 2pi)$. In other words, the branch cut is chosen as $[0, infty)$. Then by employing the standard keyhole contour, we may write

beginalign*
alpha
&= frac12pi left( int_infty-i0^+^-i0^+ fracp(z)q(z)e^-frac12operatornameLog(z) , mathrmdz + int_i0^+^infty+i0^+ fracp(z)q(z)e^-frac12operatornameLog(z) , mathrmdz right) \
&= i sum_q(x) = 0 frac p(x)q'(x) e^-frac12operatornameLog(x), tag1
endalign*

where the sum in the last line runs over all complex $x$'s solving $p(x)=0$, which itself is a consequence of the residue theorem.

Step 2 - Identifying the nature of the sum. Since $q(x)$ is a cubic polynomial with only simple zeros, the sum in $text(1)$ involves 3 terms, which we write as $z_1, z_2, z_3$. Then

$$alpha = z_1 + z_2 + z_3. $$

Also, consider

$$ y = left( fraci p(x)q'(x) e^-frac12operatornameLog(x) right)^2 = -fracp(x)^2xq'(x)^2 $$

and let $mathcalZ = z_1^2, z_2^2, z_3^2$ be the set of all possible values of $y$ given $q(x) = 0$. Rearranging this equality, we obtain $ yxq'(x)^2 + p(x)^2 = 0 $, and so, we observe that

• 
  $ q(x) $ and $yxq'(x)^2 + p(x)^2$ shares a root whenever $y in mathcalZ$.

On the other hand, if we regard both as polynomials with coefficients in $mathbbC[y]$, then the resultant between them is given by

$$ operatornameresleft(q(x), yxq'(x)^2 + p(x)^2 right) = 961y^3 - 620y^2 + 131y - 9. $$

Since the resultant between two complex polynomials is zero exactly when they share a root, it follows that $mathcalZ = y : 961y^3 - 620y^2 + 131y - 9 $, or equivalently,

$$ 961y^3 - 620y^2 + 131y - 9 = 961(y - z_1^2)(y - z_2^2)(y - z_3^2). $$

In particular, by Vieta's formulas, we check that

$$
z_1^2 + z_2^2 + z_3^2 = frac620961 = frac2031, qquad
z_1^2 z_2^2 + z_2^2 z_3^2 + z_3^2 z_1^2 = frac131961, \
z_1^2 z_2^2 z_3^2 = frac9961 = left(frac331right)^2.
$$

Moreover, although not immediate from the previous computation, we can in fact check that $z_1 z_2 z_3$ is not a negative real number, and so, $z_1 z_2 z_3 = frac331$.

Step 3 - Conclusion. From the conclusion of the previous step, we can identify the equation that $alpha$ satisfies. Indeed,

beginalign*
alpha^2
= left( z_1^2 + z_2^2 + z_3^2 right) + 2left( z_1z_2 + z_2z_3 + z_3z_1 right)
= frac2031+ 2left( z_1z_2 + z_2z_3 + z_3z_1 right)
endalign*

and

$$ left( alpha^2 - frac2031 right)^2
= 4 left( z_1^2 z_2^2 + z_2^2 z_3^2 + z_3^2 z_1^2 right) + 8 z_1 z_2 z_3 (z_1 + z_2 + z_3) = frac524961 + frac2431alpha.$$

Rearranging, we obtain

$$ 31 alpha^4 - 40alpha^2 - 24alpha = 4. $$

Moreover, it turns out that this equation has the unique positive zero, which is exactly $frac1pi$ times what WolframAlpha yields (as demonstrated in @Infiaria's answer).

According to Wolfram Alpha, the answer is

$$fracpisqrt186left(sqrt40-14sqrt[3]frac247-3sqrt93-sqrt[3]4left(47-3sqrt93right)+sqrt80+14sqrt[3]frac247-3sqrt93+sqrt[3]4left(47-3sqrt93right)+36sqrtfrac18640-14sqrt[3]frac247-3sqrt93-sqrt[3]4left(47-3sqrt93right)right)$$

which is approximately $4.34952$. Alternatively, it is simply the positive solution of

$$frac314pi^4x^4 - frac10pi^2x^2 - frac6pix - 1 = 0 $$