## TuringTest Group Title Definite integral$\int_0^1\frac x{x^3+1}dx$ one year ago one year ago

1. TuringTest Group Title

I've had this one on my head for a couple days now, so whoever can help me gets mad respect :)

2. experimentX Group Title

you can do that using partial fraction. but this is not quite interesting.

3. TuringTest Group Title

the solution from PF I got seemed drastically different than wolf's elegant answer, though I could have just been unable to manipulate the algebra.

4. TuringTest Group Title

Is PF really the only way I guess is my real question

5. TuringTest Group Title

it seems like there should be a trick

6. experimentX Group Title

yeah ... very short and elegant method.

7. TuringTest Group Title

I know, that is why I want something besides hacking away with PF which gives http://www.wolframalpha.com/input/?i=integral+x%2F%281%2Bx%5E3%29

8. TuringTest Group Title

okay now I see they are the same, so either that was a dumb question, or there is a trick I want to know

9. experimentX Group Title

there might be.

10. experimentX Group Title

trig subs might work. but this seems more nasty.

11. dape Group Title

Yeah, PF would definitely work. Probably won't be able to do much, but I'll try to poke around and see if I can find a nicer method.

12. dape Group Title

Muhahaha

13. amistre64 Group Title

... id have to brute away at it :/

14. dape Group Title

It turns out after researching the integrand for a bit that it has an $$extremely$$ nice McLaurin series. I don't have the energy to type out the entire derivation, but you can check it's: $\frac{x}{x^3+1}=x-x^4+x^7-x^{10}+...=x\sum_{n=0}^\infty(-1)^nx^{3n}$ Throwing in this in the integral and doing a partial integration we get: $\int_0^1{\frac{x}{x^3+1}\, dx}=\int_0^1{x\sum_{n=0}^\infty(-1)^nx^{3n}\,dx} \\ =\sum_{n=0}^\infty \frac{(-1)^n}{3n+1}-\sum_{n=0}^\infty(-1)^n\int_0^1{x^{3n}\,dx} \\ =\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}\left(1-\frac{1}{3n+2}\right) \\ =\sum_{n=0}^\infty\frac{(-1)^n}{3n+2}$ Now that sum is a bit harder to calculate, it involves zeta-functions in combination with some subtitutions, but at least it comes out the same as wolf's answer :p

15. dape Group Title

I think it's a more elegant solution, despite it involving some pretty high level stuff.

16. dape Group Title

That was fun, when I first saw that MacLaurin series I got some serious chills down my spine.

17. dape Group Title

Oops, typo, the integral term on second line should be: $-\sum_{n=0}^\infty{\frac{(-1)^n}{3n+1}}\int_0^1{x^{3n+1}\,dx}$

18. Algebraic! Group Title

heh, was gonna say, you lost an x!

19. Algebraic! Group Title

then i saw you just forgot it

20. experimentX Group Title

this seems pretty straight forward $\int_0^1{x\sum_{n=0}^\infty(-1)^nx^{3n}\,dx} = \sum_{n=0}^\infty(-1)^n \int_0^1x^{3n+1}\,dx = \sum_{n=0}^\infty {(-1)^n \over 3n+2}$ I get chills when i see these.

21. Algebraic! Group Title

there's some extra stuff in there...

22. experimentX Group Title

I'm not quite well with zeta functions ... looks impossible from Fourier series.

23. Algebraic! Group Title

whatever, if it's actually supposed to be there, I can't figure out why..

24. dape Group Title

Maybe I forgot something else, did it all on paper and copied it over.

25. Algebraic! Group Title

seems it's just as experimentx did it, what's with the first series and the "(1-"

26. experimentX Group Title

still ... i never though of this technique. I haven't been using this technique lately. I'll try to think of this way too in the future. gotta sleep ... nearly 3.30 am here.

27. Mikael Group Title

If you've gone McLaurin - then why not go full complex contour integration. The poles are obvious -1, +third-root(-1), -third-root(-1)

28. dape Group Title

Arent the poles $$-1,\sqrt[3]{-1},-(\sqrt[3]{-1})^2$$?

29. Dido525 Group Title

Isn't this thing just a trig substitution?

30. Dido525 Group Title

Never mind I see the x cubed. I can't see how to do this without partial fractions :/ .

31. TuringTest Group Title

nice @dape that was definitely more what I was looking for, now this integral is interesting again :)

32. Mikael Group Title

Yes @dape these are the poles. Yet I am not certain there is an effective choice of contour so that we can find the needed contribution

33. TuringTest Group Title

Contour integral are something I certainly learn, but I would then like to see it that way too.

34. TuringTest Group Title

certainly need to*

35. Mikael Group Title

|dw:1348327205764:dw|

36. TuringTest Group Title

Having never taken complex analysis I really don't want to pretend that I understand much about integrating in the complex plane, though of course I understand the concept of writing complex numbers in them and roots of unity... basically.

37. experimentX Group Title

@Mikael how to you plan to evaluate that series via contour integrals? Don't you need need to change the series into integral?

38. Mikael Group Title