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

1. Jun 19, 2010

### Gregg

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

2. Jun 20, 2010

### arildno

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. Jun 20, 2010

### arildno

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. Jun 20, 2010

### 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. Jun 20, 2010

### arildno

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. Jun 20, 2010

### Gregg

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. Jun 20, 2010

### arildno

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

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

8. Jun 20, 2010

### arildno

9. Jun 20, 2010

### Gregg

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. Jun 21, 2010

### ross_tang

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 http://www.voofie.com/content/82/how-to-find-the-integral-of-1xna/" [Broken] for step by step details.

Last edited by a moderator: May 4, 2017
11. Jun 21, 2010

### arildno

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.

Last edited by a moderator: May 4, 2017
12. Jun 21, 2010

### Gregg

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. Jun 21, 2010

### arildno

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

Is an infinite series simpler??

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

14. Jun 21, 2010

### Gregg

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

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

Last edited by a moderator: May 4, 2017
15. Jun 21, 2010

### Gregg

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. Jun 21, 2010

### ross_tang

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. Jun 21, 2010

### Gregg

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. Jun 21, 2010

### ross_tang

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. Jun 21, 2010

### Gregg

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

I understand that

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

20. Jun 22, 2010

### ross_tang

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.