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I found out that the integral has the following closed form. My question is can the hypergeometric function be simplified?

$$\int \frac{1}{(1+x+x^2)^n}dx=2^{2n-1}3^{-n}(2x+1){}_2F_1\left(\frac12,n;\frac32;-\frac13(2x+1)^2\right)$$

banana
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Gabriel Demirdag
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  • I think the n and 2 should be switched or am i wrong? – Gabriel Demirdag Oct 10 '23 at 20:41
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    Via a substitution we may as well ask the question about $\int \frac{du}{(1 + u^2)^n} = u {}_2 F_1\left(\frac12,n;\frac32;-u^2\right) + C$. (Cf. https://math.stackexchange.com/questions/3697659/prove-formula-for-int-fracdx1x2n, which gives a reduction formula but not an explicit formula.) – Travis Willse Oct 10 '23 at 22:01
  • (Also, via the trig substitution $u = \tan \theta$, the question is equivalent to integrating $\int \cos^{2 n - 2} \theta ,d\theta$.) – Travis Willse Oct 11 '23 at 00:18
  • https://math.stackexchange.com/questions/2170676/integrate-int-cosn-x-dx – Travis Willse Oct 11 '23 at 01:19

2 Answers2

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let's denote the integral by $I_n(x)$ $$ I_n(x)=\int \frac{1}{(x^2+x+1)^n} dx=\int \frac{1}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \right)^n} dy $$ now put , $x=\frac{\sqrt{3}}{2} \tan y-\frac{1}{2}$ $$ I_n(x)=\int \frac{1}{\left(\frac{3}{4} \tan^2y+\frac{3}{4} \right)^n} \frac{\sqrt{3}}{2}\sec^2y dy $$ So $$ I_n(x)=\frac{2^{2n-1}}{3^{n-\frac{1}{2}}}\int (\cos y)^{2n-2} dy $$ for $n$ is a natural number we have here the formula $$ \int (\cos y)^{2n} dy=4^{-n}\sum_{k=1}^{n} \binom{2n}{k+n} \frac{\sin(2k y)}{k}+\binom{2n}{n} 4^{-n} y+c$$ So $$ I_n(x)=\frac{2^{2n-1}}{3^{n-\frac{1}{2}}} \left(4^{1-n}\sum_{k=1}^{n-1} \binom{2n-2}{k+n-1} \frac{\sin(2k y)}{k}+\binom{2n-2}{n-1} 4^{1-n} y+c \right) $$ finally $$ I_n(x)=\frac{2}{3^{n-\frac{1}{2}}} \sum_{k=1}^{n-1} \binom{2n-2}{k+n-1} \frac{\sin\left(2k\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right)}{k}+\frac{2}{3^{n-\frac{1}{2}}} \binom{2n-2}{n-1} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}} \right)+c=A_n(x)+c $$ to get the sine function in polynomial form see the formula there which give us $$ \sin\left(2k\tan^{-1}x\right)=\frac{1}{(x^2+1)^{k}}\sum_{p=1}^{k} \binom{2k}{2p-1} (-1)^{p-1} x^{2p-1}$$ So $$ \sin\left(2k\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) \right)= \frac{3^{k+\frac{1}{2}} 4^{-k}}{(x^2+x+1)^{k}}\sum_{p=1}^{k} \binom{2k}{2p-1} (-3)^{p-1} (2x+1)^{2p-1} $$ then to find the hypergeometric function in terms of $A_n(x)$ we need to evaluate $c$ $$ 2^{2n-1}3^{-n}(2x+1){}_2F_1\left(\frac12,n;\frac32;-\frac13(2x+1)^2\right)=A_n(x)+c$$ by putting $x=-\frac{1}{2}$ give us $c=0$ , therefore $$ {}_2F_1\left(\frac12,n;\frac32;-\frac13(2x+1)^2\right) =\frac{2^{1-2n} 3^n}{2x+1} A_n(x) $$ for example

let $n=1$ we have $A_1(x)=\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}} \right)$ So $$ {}_2F_1\left(\frac12,1;\frac32;-\frac13(2x+1)^2\right)=\frac{\sqrt3}{2x+1} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}} \right) $$

Faoler
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  • What happens if we integrate it from 0 to +inf? It has a form P* π * sqrt(3)-Q – Gabriel Demirdag Oct 14 '23 at 17:15
  • using the series I used for $I_n$ and putting $y=\frac{\pi}{2} , y=\frac{\pi}{6}$ then $$ p=\frac{2}{3^{n+1}} \binom{2n-2}{n-1} \quad , \quad Q=\frac{2}{3^n} \sum_{k=1}^{n-1} \binom{2n-2}{k+n-1} \frac{\sin \left(\frac{\pi k}{3} \right)\sqrt{3}}{k}$$ – Faoler Oct 15 '23 at 10:18
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Since$$x^2+x+1=\frac 1 4 \left((2 x+1)^2+3\right)$$ let $$2x+1=\sqrt 3\, \sinh(t)$$ $$\int \frac{1}{(1+x+x^2)^n}\,dx=2^{2 n-1} 3^{\frac{1}{2}-n}\int \text{sech}^{2 n-1}(t)\,dt$$

for which exist a reduction formula but no general expression.

Claude Leibovici
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