Cfrgtkky

I have no idea how to do this question.

I'm given $int^{infty}_{-infty}frac{1}{x^2+2ax+b^2}dx=frac{pi}{sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $int^{infty}_{-infty}frac{1}{(x^2+x+1)^3}dx=frac{4pi}{3sqrt3}$.

What I've tried:

$int^{infty}_{-infty}frac{1}{(x^2+x+1)}dx=frac{2pi}{sqrt3}$. I've tried setting a variable as the power, ie: $I(r)=int^{infty}_{-infty}frac{1}{(x^2+x+1)^r}dx$ but differenciating the inside w.r.t to $r$ doesn't seem to form any differential equation that can be solved.

Answers and hints appreciated!

Differentiate with respect to $b$ gives
$$dfrac{d}{db}int^{infty}_{-infty}frac{1}{x^2+2ax+b^2}dx=dfrac{d}{db}frac{pi}{sqrt{b^2-a^2}}$$
$$int^{infty}_{-infty}frac{-2b}{(x^2+2ax+b^2)^2}dx=frac{-2bpi}{2sqrt{(b^2-a^2)^3}}$$
or
$$int^{infty}_{-infty}frac{1}{(x^2+2ax+b^2)^2}dx=frac{pi}{2sqrt{(b^2-a^2)^3}}$$
and one another derivative gives following result
$$int^{infty}_{-infty}frac{1}{(x^2+2ax+b^2)^3}dx=frac{3pi}{8sqrt{(b^2-a^2)^5}}$$
now you have the answer with $a=dfrac12$ and $b=1$.

For real $x$, we have $x^2 + x + 1 = (x+frac12)^2 + frac34 ge frac34$.

This implies for any $t in (0,frac34)$, following expansion in $t$ converges:

$$frac{1}{x^2+x+1 - t} = sum_{k=0}^infty frac{t^k}{(x^2 + x + 1)^{k+1}}$$

Since everything on RHS is non-negative, by Tonelli, we can integrate them term by term:

$$int_{-infty}^infty frac{dx}{x^2+x+1 - t} = sum_{k=0}^infty t^k int_{-infty}^infty frac{dx}{(x^2+x+1)^{k+1}}
$$
Set $a = frac12$ and $b = sqrt{1-t}$. Notice $b > |a|$, the formula you have
tell us the integral on LHS is

$$frac{pi}{sqrt{b^2 - a^2}} = frac{pi}{sqrt{frac34 - t}}
= frac{2pi}{sqrt{3}sqrt{1 - frac{4t}{3}}}$$

Recall $displaystyle;frac{1}{sqrt{1-4s}}$ is the generating function for the central binomial coefficients:

$$frac{1}{sqrt{1-4s}} = sum_{k=0}^infty binom{2k}{k} s^k$$

This leads to

$$sum_{k=0}^infty t^k int_0^infty frac{dx}{(x^2+x+1)^{k+1}} =
frac{2pi}{sqrt{3}}sum_{k=0}^infty binom{2k}{k}frac{t^k}{3^k}
$$
By comparing coefficients of $t^k$ on both sides, we obtain

$$int_{-infty}^infty frac{dx}{(x^2+x+1)^{k+1}}
= frac{2pi}{3^ksqrt{3}}binom{2k}{k}quadtext{ for } k in mathbb{N}
$$

In particular, for $k = 2$, this give us

$$int_{-infty}^infty frac{dx}{(x^2+x+1)^3} = frac{2pi}{3^2sqrt{3}}binom{4}{2} = frac{4pi}{3sqrt{3}}$$

For any $A>0$ we have $int_{-infty}^{+infty}frac{dz}{z^2+A} = frac{pi}{sqrt{A}}$, hence by applying $frac{d^2}{dA^2}$ to both sides we get
$$ int_{-infty}^{+infty}frac{dz}{(z^2+A)^3}=frac{3pi}{8A^2sqrt{A}}.tag{1} $$
On the other hand
$$ int_{-infty}^{+infty}frac{dx}{(x^2+x+1)^3}stackrel{xmapsto z-frac{1}{2}}{=}int_{-infty}^{+infty}frac{dz}{left(z^2+tfrac{3}{4}right)^3}=frac{4pi}{3sqrt{3}}.tag{2}$$

$$I(r)=int_{-infty}^{infty} frac {dx}{left(left(x+frac 12right)^2+frac 34right)^r}=int_{-frac 12}^{infty} frac {dx}{left(left(x+frac 12right)^2+frac 34right)^r}+int_{-infty}^{-frac 12} frac {dx}{left(left(x+frac 12right)^2+frac 34right)^r}$$

Substitute $x+frac 12=u$

$$I(r)=int_0^{infty} frac {du}{left(u^2+frac 34right)^r}+int_{-infty}^0 frac {du}{left(u^2+frac 34right)^r}$$

Since the integrand is even we get $$I(r)=2int_0^{infty} frac {du}{left(u^2+frac34right)^r}=frac {2cdot 4^r}{3^r} int_0^{infty} frac {du}{left(frac {4u^2}{3}+ 1right)^r}$$

On substituting $frac {4u^2}{3}=t$ we get $$I(r)=frac {2cdot 4^{r-1}}{3^{r-frac12}} int_0^{infty} frac {t^{-frac 12} dt}{(1+t)^r}$$

Now using the definition of Beta function and it's relation with Gamma function we get $$I(r)=frac {2cdot 4^{r-1}}{3^{r-frac12}} Bleft(frac 12,r-frac 12right) =frac {2cdot 4^{r-1}}{3^{r-frac12}} frac {Gammaleft(frac 12right)Gammaleft(r-frac 12right)}{Gamma(r)}=frac {2cdot 4^{r-1}}{3^{r-frac12}} frac {sqrt {pi}cdot Gammaleft(r-frac 12right)}{Gamma(r)}$$

In general it can be proved that $$I(a,b,r)=int_{-infty}^{infty} frac {dx}{(x^2+2ax+b^2)^r} = frac {sqrt {pi}cdotGammaleft(r-frac 12right)}{Gamma(r)} left(frac {1}{b^2-a^2}right)^{r-frac 12}$$ if $vert bvert gt vert avert$