0

I need to know whether this monster is integrable.

My fail attempts included: substituting $x^n$ by $z$ and also substituting for some smaller power of $n$, somehow using the fact that this is integrable for when n=1,2,3. I searched a lot online, but could not managed to get an answer either.

Any help is appreciated.

Barreto
  • 866
  • 1
    For this to be possible we need to find $$\int \frac{dx}{1+x^n}$$ which I highly doubt is possible. WolframAlpha gives you a closed form using hyper geometric functions if that’s what you’re interested in, however. – phi-rate Feb 24 '22 at 00:15
  • 1
    If $n$ is an integer, then the integrand is a rational function, which are always integrable by partial fraction decomposition. – David H Feb 24 '22 at 00:15
  • Thanks a lot for the comments! @DavidH yes, $n$ is assumed to be an integer, thank you. Do you know however some general solution for it? Or we can just state a closed form solution exists? – Barreto Feb 24 '22 at 00:18
  • 1
    This answer might be useful. – Robert Lee Feb 24 '22 at 00:24
  • 1
    You can always split the integral, into $\int \frac{adx}{b+x^n} +\int\frac{dx}{\frac{b}{x^n}+1}$. Suppose $b$ is a value such that an $n$-th root can exist, you can do the substitutions $x=\frac{u}{b^{1/n}}$ for the first integral and $ x = \frac{b^{1/n}}{u}$ for the second integral. This should turn it into $\frac{a}{b^{1+1/n}}\int\frac{du}{1+u^n} - \int \frac{du}{u^2(1+u^2)}$. Where you can use pfd like DavidH suggested and hypergeometric functions like intellect4 suggested. – Doge Chan Feb 24 '22 at 00:28
  • 1
    @Sarah A general solution for arbitrary $n$ requires we do two things: 1) find a general formula for the roots of the denominator, and 2) a general formula for the p.f.d. coefficients. The first part is easy since they’re just roots of unity. The second part is much hairier. See: https://en.wikipedia.org/wiki/Partial_fraction_decomposition#Residue_method – David H Feb 24 '22 at 00:40
  • 1
    The result for the hypergeometric function is easy to transform into an lerch transcendent as seen in a linked answer which usually simplifies down. – Тyma Gaidash Feb 24 '22 at 01:36

1 Answers1

2

$$I=\int \frac{a + x^n}{b + x^n}dx=\int \frac{b+(a-b) + x^n}{b + x^n}dx=\int dx+(a-b)\int \frac{ dx }{b + x^n}$$ Let $x=y \,b^{\frac{1}{n}}$ $$I=x + (a-b)\, b^{\frac{1}{n}-1}\int \frac {dy}{1+y^n}$$ For the remaining integral, you could use partial fraction decompositions with the roots of unity (not the most pleasant) or use the gaussian hypergeometric function $$\int \frac {dy}{1+y^n}=y \, _2F_1\left(1,\frac{1}{n};\frac{n+1}{n};-y^n\right)$$ where $n$ can be any number

Claude Leibovici
  • 260,315