Evaluate $\int\frac1{1+x^n}dx$ for $n\in\mathbb R$

Question

I was wondering on how to evaluate the following indefinite integral for all $n\in\mathbb R$.

$$\int\frac1{1+x^n}dx$$

It seems to be peculiar in that we have

$$\begin{align} \int\frac1{1+x^{-1}}dx&=x-\ln(x+1)+c\\ \int\frac1{1+x^0}dx&=\frac12x+c\\ \int\frac1{1+x^{1/2}}dx&=2\sqrt x-2\ln(1+\sqrt x)+c\\ \int\frac1{1+x^1}dx&=\ln(x+1)+c\\ \int\frac1{1+x^2}dx&=\arctan(x)+c\\ \int\frac1{1+x^3}dx&=\frac13\ln(1+x)-\frac2{3\sqrt3}\arctan\left(\sqrt{\frac43}\left(x-\frac12\right)\right)+c \end{align}$$

Naturally, there appears to be some combination of $\ln$ and $\arctan$, though no simple formula arises to solve the general case.

It is, however, easy to see that

$$\int\frac1{1+x^{-n}}dx=\int1-\frac1{1+x^n}dx$$

So there is an easy enough connection between positive and negative $n$.

Also, it is easy enough to find the series expansion, taking advantage of the above connection we just made to circumvent problems concerning convergence.

$$\frac1{1+x^n}=1-x^n+x^{2n}-x^{3n}+\dots\forall\ |x|<1$$

$$\int\frac1{1+x^n}dx=c+x-\frac1{n+1}x^{n+1}+\frac1{2n+1}x^{2n+1}-\dots$$

$$=c+\sum_{k=0}^\infty\frac{(-1)^k}{kn+1}x^{kn+1}\ \forall\ |x|<1$$

Though this isn't very much along the lines of closed form.

For $n=\frac ab$, where $a$ and $b$ are whole numbers, we can use the substitution $x=u^b$ to get

$$\int\frac1{1+x^n}dx=\int\frac{bu^{b-1}}{1+u^a}du$$

~~though I'm unsure where that could lead.~~ This reduces the integral down to

$$\int\frac1{1+x^n}dx=b\int P(u)+\frac{u^{b-1-ak}}{1+u^a}du,\quad k\in\mathbb N$$

for some polynomial $P(u)$. Though I'm still clueless as to how this can be advanced.

How can I evaluate $\int\frac1{1+x^n}dx\ \forall\ n\in\mathbb R$ in closed form? Can someone prove there at least exists some closed form solution for all $n\in\mathbb Q$ if the above is not possible? If possible, use real numbers.

I'm curious about whether anything can be done with $\frac 1{1+x^{-n}}+\frac 1{1+x^n}=1$, this seems to be a relatively powerful identity that I hadn't considered before... — abiessu, Nov 05 '16 at 00:05
One way to connect all of these integrals is with the hypergeometric function. In particular, one may compute that $$ \int \frac{1}{1 + x^n},dx = x \cdot {}_2F_1(1,1/n;1 + 1/n; -x^n) + C $$ However, since the hypergeometric function is defined in terms of its power series, this is not a terribly satisfying epiphany. — Ben Grossmann, Nov 05 '16 at 00:06
I have a 150 page table of mathematical formulas in my bookshelf where 20 of those are devoted to indeterminate integrals. It lists this integral for $n = 1, 2, 3$ and $4$ separately, and that's it for your integrals (although the integrals for $1/(1-x^n)$ are also listed for the same $n$). I think that would suggest that no nice closed form was known at the time of printing, which was 2003. — Arthur, Nov 05 '16 at 00:10
I was looking at the same problem a while ago and posted a solution. See http://math.stackexchange.com/questions/1837784/help-verifying-equation-int-0-x-frac11tn-dt — 926reals, Nov 05 '16 at 01:08
@926reals That is quite nice. I too have spotted that solution, however, it is problematic that it doesn't really qualify as a closed form solution. — Simply Beautiful Art, Nov 05 '16 at 01:13
@Omnomnomnom See THIS ANSWER in which I developed a closed-form solution. ;-)) — Mark Viola, Nov 05 '16 at 02:09
@Simple art: It requires calculation of, cos(?), sin(?), Atan(??), and log (???), and knowing the value of PI. How do we do that. That was my quest, how to skip all of it. — 926reals, Nov 15 '16 at 02:01
@926reals IMO, it is easier than an infinite sum. Also thought you would like it. — Simply Beautiful Art, Nov 15 '16 at 14:23
@simple art, http://math.stackexchange.com/questions/2018468/solving-the-integral-int-fracx-alpha1-psi-x-mathrm-dx/2020799#2020799. — 926reals, Nov 19 '16 at 04:06
These two questions are also related (by not the same, since they do not request solution as combination of arctan and log): Solving this integral $\int\frac{1}{1+x^n} dx$? and How to find the indefinite integral $\int \frac{dx}{1+x^{n}}$? — Martin Sleziak, Jan 14 '17 at 09:16
Also related (I only stumbled upon this question now): https://math.stackexchange.com/q/2807302/96384 — Torsten Schoeneberg, Sep 30 '18 at 21:51
It is possible to simplify the hypergeometric function to an incomplete beta function, see Calculating integrals of the form $\int\frac{dx}{(1+x^{q})^\frac{p}{r}}$. One then gets $$\int \frac{dx}{1+x^n}=\frac{1}{n}B\left(\frac{x^n}{x^n+1};\frac{1}{n},1-\frac{1}{n}\right)+C.$$ — projectilemotion, Dec 14 '21 at 23:51
The answer is also expressible in terms of the Lerch Transcendent $\Phi(x,1,a)$ — Тyma Gaidash, Aug 04 '22 at 16:48

score 47 · Accepted Answer · edited Apr 13 '17 at 12:21

47

In THIS ANSWER, I showed that that the indefinite integral of interest is given by

$$\bbox[5px,border:2px solid #C0A000]{\int\frac{1}{x^n+1}dx=-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C} $$

for $n\ge 1$, where $x_{kr}$ and $x_{ki}$ can be written, respectively, as

$$x_{kr}=\cos \left(\frac{(2k-1)\pi}{n}\right)$$

$$x_{ki}=\sin \left(\frac{(2k-1)\pi}{n}\right)$$

For $n<0$, we simply write

$$\int \frac{1}{1+x^{-|n|}}\,dx=x-\int \frac{1}{1+x^{|n|}}\,dx$$

and use the aforementioned result with $n$ replaced with $|n|$.

edited Apr 13 '17 at 12:21

Community

1

answered Nov 05 '16 at 02:05

Mark Viola

179,405

3

Wow, that is quite beautiful. Mind me asking if you can do anything for $n\in\mathbb Q$ or $\mathbb R$? – Simply Beautiful Art Nov 05 '16 at 13:03
@SimpleArt Thank you for the nice comment! Much appreciated. Now, unfortunately, I don't know how to generalize this result to the case for non-integer values of $n$ without appealing to the hypergeometric function. But that is unlikely the goal here. -Mark – Mark Viola Nov 05 '16 at 15:05
@marcocantarini Thanks Marco. Much appreciative. – Mark Viola Nov 15 '16 at 14:19
1

Would you be so kind as to look at $\displaystyle\int\frac{x^b}{1+x^a}dx$ for $a,b\in\mathbb N$? – Simply Beautiful Art Nov 19 '16 at 00:23
@simpleart First, thank you so much for awarding this the 100 point bounty! If you post the generalization, I'd be happy to have a look. -Mark – Mark Viola Nov 19 '16 at 00:26
You're welcome, you deserved it. And it's already in under the $n\in\mathbb Q$ section. – Simply Beautiful Art Nov 19 '16 at 01:58
2

Very nice! How in the world could I have missed this precious answer to this question? +1 – DonAntonio Jan 13 '17 at 21:25
@DonAntonio Thank you my friend! And don't dismay. There were several users that posted comments on the referenced answer who claimed that no closed form is possible. – Mark Viola Jan 13 '17 at 22:18
You may be interested in Evaluating $\int\frac{x^b}{1+x^a}~dx$ for $a,b\in\Bbb N$ – Simply Beautiful Art Aug 28 '17 at 22:52

score 11 · Answer 2 · answered Nov 05 '16 at 00:10

11

For positive integers $n$ you can write $$ \dfrac{1}{1+x^n} = \sum_{\omega} \dfrac{r(\omega)}{x - \omega}$$ where the sum is over the $n$'th roots of $-1$ and $r(\omega)$ is the residue of $1/(1+x^n)$ at $x = \omega$, so that your integral is $$ c + \sum_\omega r(\omega) \log(x - \omega)$$

You can also express the power series solution in terms of the Lerch Phi function: $$ c + \dfrac{x}{n} {\rm LerchPhi}(-x^n,1,1/n) $$

answered Nov 05 '16 at 00:10

Robert Israel

448,999

Ah, interesting idea! Basically equivalent to partial fraction decomposition. Though it suggests I should use complex logarithms to write my $\arctan$. If possible, keeping everything under real numbers is recommended. – Simply Beautiful Art Nov 05 '16 at 00:18
Robert, SEE THIS, in which I developed a closed-form solution. ;-)) – Mark Viola Nov 05 '16 at 02:12

score 1 · Answer 3 · answered Jul 18 '19 at 08:56

Slightly different approach to Robert Israel's. Using $$\frac{1}{1+x}=1-x+x^2-x^3+x^4-\cdots,\tag{1}$$ where $|x|<1$ you can substitute $x^n$ for $x$ to obtain $$\int\frac{1}{1+x^n}dx=\int 1-x^n+x^{2n}-x^{3n}+x^{4n}-\cdots dx$$

So we end up with

$$\int\sum_{k=0}^\infty (-1)^kx^{kn}dx=\sum_{k=0}^\infty(-1)^k\frac{x^{kn+1}}{kn+1}+c=\frac{x}{n} \Phi \left(-x^n,1,\frac{1}{n}\right)+c.$$ This actually works for $x,n\in\mathbb{R}$, but you'd have to prove that for $x$ since it does not follow from (1).

The series diverges when $|x|\ge 1$. – Mark Viola Dec 22 '22 at 19:11 — Mark Viola, Dec 22 '22 at 19:11

Evaluate $\int\frac1{1+x^n}dx$ for $n\in\mathbb R$

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