integrate $\int \frac{1}{e^{x}+e^{ax}+e^{a^{2}x}} \, dx$

Asked May 22 '15 at 11:16

Active May 22 '15 at 15:55

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I've been trying to integrate $$ \int \frac{1}{e^{x}+e^{\omega x}+e^{\omega^{2}x}} \, dx $$ where $\omega=e^{2i\pi/3}$

but to no avail. I've tried substituting in $u=e^{(1+\omega)x}$ but ended up with an integral which looked like: $$ \int \frac{1}{u^{\frac{2+\omega}{1+\omega}}+u^{\frac{1+2\omega}{1+\omega}}+1}\,du $$ multipled by some constant including $\omega$. Does this integral have a closed form in terms of the hypergeometric function or otherwise?

EDIT

Let the integral be denoted by I then $$ I=-\omega\int\frac{1}{u^{1-\omega}+u^{1-\omega^2}+1}\,dx=-\omega\int\frac{1}{u^{1-\omega}(1+u^{1+\omega})+1}\,dx $$ using the notation from earlier.

edited May 22 '15 at 15:55

asked May 22 '15 at 11:16

Calimops

$$\frac{1}{e^x +2e^{-x/2}\cos\frac{x\sqrt{3}}{2}}$$ does not look to have a nice primitive, but maybe the integral over $\mathbb{R}^+$ is more good-looking. – Jack D'Aurizio May 22 '15 at 11:28
I just want to clarify, is $\omega$ itself supposed to be an exponential, or is it only the exponent? This is reminiscent of discrete fourier transform transfer functions. – Gappy Hilmore May 22 '15 at 11:35
$\omega$ is a constant (cube root of 1) that is multiplied by the $x$ in the exponential – Calimops May 22 '15 at 11:40
what a monster... – tired May 22 '15 at 14:42
Expanding the inverse of the denominator as a series or using binomial theorem implies we need na double sum, like this. – Тyma Gaidash Jan 04 '24 at 01:30

integrate $\int \frac{1}{e^{x}+e^{ax}+e^{a^{2}x}} \, dx$

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