# Y' = 1/(x^N+1)

1. ### Gregg

462
How would I solve $$y'(x) = \frac{1}{x^N+1}$$ ?

2. ### arildno

12,015
Re: y'=1/(x^N+1)

Note that the polynomials $$x^{n}+1$$ can be factorized in terms of the nth-roots of (-1), i.e, as:
$$x_{j,n}=e^{i\pi\frac{1+2{j}}{n}}, i=\sqrt{(-1)}, j=0,...n-1$$

You can then factorize your polynomial denominator, use fractional decomposition, and perform termwise integration.
Be particularly aware of the pitfalls involved in complex logarithms.

Last edited: Jun 20, 2010
3. ### arildno

12,015
Re: y'=1/(x^N+1)

Let us take the case with n=4.
Then, the roots are:
$$x_{0,4}=e^{i\frac{\pi}{4}}=\frac{1}{\sqrt{2}}(1+i)$$
$$x_{1,4}=e^{i\frac{3\pi}{4}}=\frac{1}{\sqrt{2}}(-1+i)$$
$$x_{2,4}=e^{i\frac{5\pi}{4}}=\frac{1}{\sqrt{2}}(-1-i)$$
$$x_{3,4}=e^{i\frac{7\pi}{4}}=\frac{1}{\sqrt{2}}(1-i)$$

We may factorize x^4+1 into 4 complex-valued first-order polynomials:
$$\frac{1}{x^{4}+1}=\frac{1}{(x-\frac{1}{\sqrt{2}}(1+i))(x-\frac{1}{\sqrt{2}}(1-i))(x-\frac{1}{\sqrt{2}}(-1+i))(x-\frac{1}{\sqrt{2}}(-1-i))}$$

If you wish to work with real valued polynomials, multiply together the complex conjugates, and get:
$$\frac{1}{x^{4}+1}=\frac{1}{(x^{2}-\sqrt{2}x+1)(x^{2}+\sqrt{2}x+1)}$$

A partial fractions decompositions would then proceed as follows:
$$\frac{1}{x^{4}+1}=\frac{Ax+B}{(x^{2}-\sqrt{2}x+1)}+\frac{Cx+D}{(x^{2}+\sqrt{2}x+1)}$$

You may then determine what A,B, C and D must be by multplying the whole equation with x^4+1, getting, by rearrangement:
$$0=(A+C)x^{3}+(\sqrt{2}{A}-\sqrt{2}C+B+D)x^{2}+(A+C+B-D)x+(B+D-1)$$
The coefficients to each power of x must be 0, yielding:

$$B=D=1/2, C=-A=\frac{1}{2\sqrt{2}}$$

In general, your anti-derivatives will be sums of logaritms and arctan-functions.

Last edited: Jun 20, 2010
4. ### jostpuur

Re: y'=1/(x^N+1)

I see that when $N$ is chosen and fixed, the task can be completed with finite amount of effort. But is there more general results? Is there theory about these coefficients that you encounter in fractional decomposition?

5. ### arildno

12,015
Re: y'=1/(x^N+1)

Probably.

I don't know of it, though I'm sure a generalized result, in terms, perhaps, of a complex finite sum as a function of N has been made by somebody

6. ### Gregg

462
Re: y'=1/(x^N+1)

I could see the cases N=2,3 etc. but Mathematica gives

$$\int \frac{dx}{x^N+1} = x\left({}_2 F_1 \left(\frac{1}{N},1,1+\frac{1}{N},-x^N\right)\right)$$

$$\text{Hypergeometric2F1}(a,b,c,z) = \, _2F_1(a,b;c;z)$$

7. ### arildno

12,015
Re: y'=1/(x^N+1)

And that is the closed-form solution, with a truly nasty function called Hypergeometric2F1 in its expression..

12,015
9. ### Gregg

462
Re: y'=1/(x^N+1)

$$\, _2F_1(a,b;c;z) = 1+\frac{a b}{1!c}z + \frac{a(a+1)b(b+1)}{2!c(c+1)}z^2 +\cdots$$

So I think it looks something like this

$$\, _2F_1(a,b;c;z) =\displaystyle 1+\sum _{k=0}^{\infty } \frac{\prod _{j=0}^{(k)} \frac{(a+j)(b+j)}{(c+j)}}{(k+1)!}z^k$$

$$\int \frac{1}{x^N+1} \, dx = x\left(\displaystyle 1+\sum _{k=0}^{\infty } \frac{\prod _{j=0}^k \frac{(\frac{1}{N}+j)(1+j)}{(1+\frac{1}{N}+j)}}{(k+1)!}(-x^N)^{(k+1)}\right)$$

Last edited: Jun 20, 2010
10. ### ross_tang

65
Re: y'=1/(x^N+1)

Actually, this integral can be computer if N is integer, without using hypergeometric function.

$$\int \frac{dx}{x^N+ a}=\sum _{k=0}^{N-1} b_k\ln \left(x-(-a)^{\frac{1}{N}}e^{\frac{2k \pi }{N}i}\right)$$

and

$$b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}$$

Please refer to this question in Voofie for step by step details.

11. ### arildno

12,015
Re: y'=1/(x^N+1)

Thank you, ross tang.
I mentioned, in post 2, that by using factorization by means of first order complex polynomials, you'd get a sum of complex logarithms as your answer.

It is nice to see the closed form solution in this case as well.

12. ### Gregg

462
Re: y'=1/(x^N+1)

If N is an integer isn't this even simpler?

$$y=\int \frac{1}{x^N+1} dx$$

$$y=\int (1-x^N+x^{2N}-x^{3N}+O(x^{4N})) dx$$

$$y=x-\frac{x^{N+1}}{N+1}+\frac{x^{2N+1}}{2N+1}-\frac{x^{3N+1}}{3N+1}+O(x^{4N+1})$$

$$y=\displaystyle \sum_{k=0}^{\infty} \frac{x^{kN+1}}{kN+1}$$

13. ### arildno

12,015
Re: y'=1/(x^N+1)

Is an infinite series simpler??

Furthermore, the anti-derivative you make there implies that |x|<1

14. ### Gregg

462
Re: y'=1/(x^N+1)

How do you go from

$$\sum _{k=0}^{N-1} b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)=1$$

to

$$b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{j=0j\neq k}^{N-1} \left(e^{\frac{i}{N}(\theta +2n \pi )}-e^{\frac{i}{N}(\theta +2j \pi )}\right)}$$

15. ### Gregg

462
Re: y'=1/(x^N+1)

Ah you actually go from

$$\sum _{k=0}^{N-1} b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)=1$$

$$\Rightarrow b_n=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{j=0j\neq k}^{N-1} \left(e^{\frac{i}{N}(2n \pi )}-e^{\frac{i}{N}(2j \pi )}\right)}$$

I can see

$$\sum _{k=0}^{N-1} b_k=\frac{1}{(-a)^{\frac{1}{N}}\prod _{j=0j\neq k}^{N-1} \left(e^{\frac{i}{N}(2n \pi )}-e^{\frac{i}{N}(2j \pi )}\right)}$$

But I can't see how you get bn

Last edited: Jun 21, 2010
16. ### ross_tang

65
Re: y'=1/(x^N+1)

@Gregg,

Have you try putting $$x = \sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}$$?

If you do that, you can see for every other term in the sum, i.e. when $$k \neq n$$

$$b_k\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(\sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}-\sqrt[N]{A}e^{\frac{\theta +2j \pi }{N}i}\right) = 0$$

Since there is a factor of

$$\sqrt[N]{A}e^{\frac{\theta +2n \pi }{N}i}-\sqrt[N]{A}e^{\frac{\theta +2n\pi }{N}i}$$

when j = n.

Only when k = n, the term j = n is gone, since the sum excluded the factor.

17. ### Gregg

462
Re: y'=1/(x^N+1)

Can you write it as this?

$$\sum _{k=0}^{N-1} b_k=\frac{A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2k \pi )}}{\prod _{j=0}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)}$$

So every time that $$k \ne n$$ it is possible for $$j=n$$ and therefore a factor of zero in the product. With $$k=n$$ since $$j \ne k = n$$ there is no zero factor.

$$b_n=\frac{1}{\prod _{j=0k\neq j}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)}$$

Then i get

$$b_n=\frac{1}{ A^{\frac{1}{N}}e^{i \frac{\theta }{N}}\prod _{j=0k\neq j}^{N-1} \left(e^{\frac{(2n i \pi )}{N}}-e^{\frac{(2j i \pi )}{N}}\right)}$$

$$b_n=\frac{1}{(-a)^{\frac{1}{N}}\prod _{j=0k\neq j}^{N-1} \left(e^{\frac{(2n i \pi )}{N}}-e^{\frac{(2j i \pi )}{N}}\right)}$$

which is different so what have I missed?

Last edited: Jun 21, 2010
18. ### ross_tang

65
Re: y'=1/(x^N+1)

I am sorry. I think my notation confused you.

I wrote this as the equation:

$$\sum _{k=0}^{N-1} b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)=1$$

But in fact, i really means this:

$$\sum _{k=0}^{N-1} \left( b_k\prod _{j=0j\neq k}^{N-1} \left(A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2n \pi )}-A^{\frac{1}{N}}e^{\frac{i}{N}(\theta +2j \pi )}\right)\right)=1$$

19. ### Gregg

462
Re: y'=1/(x^N+1)

I understand that

where do you get $$(-a)^{{N-1}\over{N}}$$ from?

20. ### ross_tang

65
Re: y'=1/(x^N+1)

I am sorry. I have little bit of typo in the answer.
It should be like this:
$$\Rightarrow b_k=\frac{1}{(-a)^{\frac{N-1}{N}}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2k \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)}$$
You don't understand why there is a factor of $$(-a)^{\frac{N-1}{N}}$$. Actually it is just property of the product notation.

The step you missed is this one:
$$\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(\sqrt[N]{A}e^{\frac{\theta }{N}i}\left(e^{\frac{2 n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)\right)$$
$$\Rightarrow \left(\sqrt[N]{A}e^{\frac{\theta }{N}i}\right)^{N-1}\prod _{\underset{j\neq k}{j=0}}^{N-1} \left(e^{\frac{2 n \pi }{N}i}-e^{\frac{2j \pi }{N}i}\right)$$
When you take out a factor in the product sign, you are not just taking 1 factor out. Since in the product, there are N-1 factors, so you are take N-1 factors out instead. Hope you can understand.