4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Try completing the square and then trig substituting. $\endgroup$ – Kemono Chen Nov 12 '18 at 2:43
9
$\begingroup$

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 $b=1$.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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 $\vert b\vert \gt \vert a\vert$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.