# Finding a closed form for this integral if one exists.

Find a closed form for this integral $$\int \frac{dx}{(1+x)(1+x^a)}$$

This integral has the possibility of not having a closed form in which case can it be proven?

Feeble attempt so far: $$\int \frac{dx}{(1+x)(1+x^a)}=\frac{\log(x+1)}{1+x^a}-\int\frac{ax^{a-1}\log(x+1)}{(x^a+1)^3}dx$$It is starting to feel analytical.

WA is not happy with it. No elementary function representation found it says. WA

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@GitGud Mr WA says cant do it – Ali Caglayan Nov 17 '13 at 21:18
I do not think you can get a nice closed form? – Mhenni Benghorbal Nov 17 '13 at 23:14
Judging by the plethora of different expressions obtained for various values of a, both integer as well as fractional, I'd say that a general form is out of the question. – Lucian Jan 21 '14 at 21:35

It is clearly known that this integral should have closed form when $a$ is a rational number.

Let $a=\dfrac{p}{q}$ , where $p\in\mathbb{Z}$ , $q\in\mathbb{Z}^+$ and $\text{gcd}(p,q)=1$ ,

Then $\int\dfrac{dx}{(1+x)(1+x^a)}=\int\dfrac{dx}{(1+x)(1+x^\frac{p}{q})}$

Let $u=x^\frac{1}{q}$ ,

Then $x=u^q$

$dx=qu^{q-1}~du$

$\therefore\int\dfrac{dx}{(1+x)(1+x^\frac{p}{q})}=\int\dfrac{qu^{q-1}}{(1+u^q)(1+u^p)}du$ , which is an integral of rational function and it should have closed form.

When $a$ is an irrational number, it is afraid that you can only solve this integral by these approaches:

When $|x|<1$ and $a>0$ ,

Then $\int\dfrac{dx}{(1+x)(1+x^a)}$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{1+x^a}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{n+ak}~dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n+ak+1}}{n+ak+1}+C$

When $|x|>1$ and $a>0$ ,

Then $\int\dfrac{dx}{(1+x)(1+x^a)}$

$=\int\dfrac{dx}{x^{a+1}(1+x^{-1})(1+x^{-a})}$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-n}}{x^{a+1}(1+x^{-a})}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n-ak}}{x^{a+1}}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{-n-a(k+1)-1}~dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n-a(k+1)}}{-n-a(k+1)}+C$

$=-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}}{(n+a(k+1))x^{n+a(k+1)}}+C$

When $|x|<1$ and $a<0$ ,

Then $\int\dfrac{dx}{(1+x)(1+x^a)}$

$=\int\dfrac{dx}{x^a(1+x)(1+x^{-a})}$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{x^a(1+x^{-a})}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n-ak}}{x^a}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{n-a(k+1)}~dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n-a(k+1)+1}}{n-a(k+1)+1}+C$

When $|x|>1$ and $a<0$ ,

Then $\int\dfrac{dx}{(1+x)(1+x^a)}$

$=\int\dfrac{dx}{x(1+x^{-1})(1+x^a)}$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-n}}{x(1+x^a)}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n+ak}}{x}dx$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{-n+ak-1}~dx$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n+ak}}{-n+ak}+C$

$=-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}}{(n-ak)x^{n-ak}}+C$

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