Integrate 1/(x^2n + 1) with De Moivre's Theorem

[FedEx]

Homework Statement

$$\int\frac{1}{x^{2n} + 1}$$

Homework Equations

De moivre's theorem

The Attempt at a Solution

I am sorry moderators. I had posted this problem earlier on PF but i got no replies so i had to do it again.

I have tried almost every possible substitution. But in vain. So just one way left. Factorize the denominator using de moivre's theorem. I don't know why i am using de moivre but i do think that we may get some ease doing this integral by this method

 

[rock.freak667]

Did the question state specifically to use De Moivre's theorem?
What substitutions did you happen to use?

If you used De Moivre's theorem where did you reach when you put $x=cos\theta + isin\theta$ ?

 

[Dick]

Are you doing the definite integral from -infinity to +infinity? In that case, sure, write down the poles using deMoivre and use the residue theorem.

 

[FedEx]

Are you doing the definite integral from -infinity to +infinity? In that case, sure, write down the poles using deMoivre and use the residue theorem.

No. Its an indefinite integral. And we haven't been taught with the residue theorem.

 

[FedEx]

Did the question state specifically to use De Moivre's theorem?
What substitutions did you happen to use?

If you used De Moivre's theorem where did you reach when you put $x=cos\theta + isin\theta$ ?

$$x = (-1)^{\frac{1}{2n}}$$

$$\int\prod^{2n-1}_{k=1}\frac{1}{(x - cos\frac{2k\pi + \pi}{2n} + isin\frac{2k\pi + \pi}{2n})}$$

 

[MathematicalPhysicist]

Well, you say De-Moivere's?
If you were to define this integral by I(n).
Then I(0)=x (upto a constant), do this by induction and by parts.
That will do the trick.
I(1)=arctg(x), I(2)=(do it by residue theorem for 1/(1+x^4)) for I(3)=S(1/(x^6+1)), the same.

You can do this with De-Moivere, but I need to think it over.

 

[MathematicalPhysicist]

I think you need to do the substituition x=cosh(t) or sinh(t) and use the identity cosh^2(t)-sinh^2(t)=1.

 

[Hurkyl]

I confess that I would solve this problem with power series. But that's probably a bad idea if you haven't learned methods for manipulating series efficiently, and I'm not sure if I got the answer in the simplest form either.

 

[FedEx]

Well, you say De-Moivere's?
If you were to define this integral by I(n).
Then I(0)=x (upto a constant), do this by induction and by parts.
That will do the trick.
I(1)=arctg(x), I(2)=(do it by residue theorem for 1/(1+x^4)) for I(3)=S(1/(x^6+1)), the same.

You can do this with De-Moivere, but I need to think it over.

We have been not taught with hyper trigonometric functions nor we have been taught to the residue theorem. How ever i would love to learn that. Any good resource which you could recommend me with? And would it be able for you to elaborate the above quoted post?

 

[FedEx]

I confess that I would solve this problem with power series. But that's probably a bad idea if you haven't learned methods for manipulating series efficiently, and I'm not sure if I got the answer in the simplest form either.

To hell with the answer.I would love to solve the sum even if the answer come in complex numbers. But you could possibly tell me what do you mean by power series.