Question

Definite integral\[\int_0^1\frac x{x^3+1}dx\]

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

it seems like there should be a trick

Answer 6

yeah ... very short and elegant method.

Answer 7

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

Answer 8

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

Answer 9

there might be.

Answer 10

trig subs might work. but this seems more nasty.

Answer 11

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.

Answer 12

Muhahaha

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

heh, was gonna say, you lost an x!

Answer 19

then i saw you just forgot it

Answer 20

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.

Answer 21

there's some extra stuff in there...

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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

Answer 29

Isn't this thing just a trig substitution?

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

certainly need to*