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!

  • $\begingroup$ Try completing the square and then trig substituting. $\endgroup$ – Kemono Chen Nov 12 '18 at 2:43

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

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

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

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

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

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

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


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 $$ \int_{-\infty}^{+\infty}\frac{dz}{(z^2+A)^3}=\frac{3\pi}{8A^2\sqrt{A}}.\tag{1} $$ On the other hand $$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+x+1)^3}\stackrel{x\mapsto z-\frac{1}{2}}{=}\int_{-\infty}^{+\infty}\frac{dz}{\left(z^2+\tfrac{3}{4}\right)^3}=\frac{4\pi}{3\sqrt{3}}.\tag{2}$$


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

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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