<s>[INST] Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. 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 $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tried:
$\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)}dx=\frac{2\pi}{\sqrt3}$. 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!
 [/INST] $$I(r)=\int_{-\infty}^{\infty} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}=\int_{-\frac 12}^{\infty} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}+\int_{-\infty}^{-\frac 12} \frac {dx}{\left(\left(x+\frac 12\right)^2+\frac 34\right)^r}$$
Substitute $x+\frac 12=u$
$$I(r)=\int_0^{\infty} \frac {du}{\left(u^2+\frac 34\right)^r}+\int_{-\infty}^0 \frac {du}{\left(u^2+\frac 34\right)^r}$$
Since the integrand is even we get $$I(r)=2\int_0^{\infty} \frac {du}{\left(u^2+\frac34\right)^r}=\frac {2\cdot 4^r}{3^r} \int_0^{\infty} \frac {du}{\left(\frac {4u^2}{3}+ 1\right)^r}$$
On substituting $\frac {4u^2}{3}=t$ we get $$I(r)=\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} \int_0^{\infty} \frac {t^{-\frac 12} dt}{(1+t)^r}$$
Now using the definition of Beta function and it's relation with Gamma function we get $$I(r)=\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} B\left(\frac 12,r-\frac 12\right)  =\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} \frac {\Gamma\left(\frac 12\right)\Gamma\left(r-\frac 12\right)}{\Gamma(r)}=\frac {2\cdot 4^{r-1}}{3^{r-\frac12}} \frac {\sqrt {\pi}\cdot \Gamma\left(r-\frac 12\right)}{\Gamma(r)}$$
In general it can be proved that $$I(a,b,r)=\int_{-\infty}^{\infty} \frac {dx}{(x^2+2ax+b^2)^r} = \frac {\sqrt {\pi}\cdot\Gamma\left(r-\frac 12\right)}{\Gamma(r)} \left(\frac {1}{b^2-a^2}\right)^{r-\frac 12}$$ if $\vert b\vert \gt \vert a\vert$
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