TuringTest
  • TuringTest
Definite integral\[\int_0^1\frac x{x^3+1}dx\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • TuringTest
I've had this one on my head for a couple days now, so whoever can help me gets mad respect :)
experimentX
  • experimentX
you can do that using partial fraction. but this is not quite interesting.
TuringTest
  • TuringTest
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
Is PF really the only way I guess is my real question
TuringTest
  • TuringTest
it seems like there should be a trick
experimentX
  • experimentX
yeah ... very short and elegant method.
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
okay now I see they are the same, so either that was a dumb question, or there is a trick I want to know
experimentX
  • experimentX
there might be.
experimentX
  • experimentX
trig subs might work. but this seems more nasty.
dape
  • dape
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.
dape
  • dape
Muhahaha
amistre64
  • amistre64
... id have to brute away at it :/
dape
  • dape
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
dape
  • dape
I think it's a more elegant solution, despite it involving some pretty high level stuff.
dape
  • dape
That was fun, when I first saw that MacLaurin series I got some serious chills down my spine.
dape
  • dape
Oops, typo, the integral term on second line should be: \[ -\sum_{n=0}^\infty{\frac{(-1)^n}{3n+1}}\int_0^1{x^{3n+1}\,dx} \]
anonymous
  • anonymous
heh, was gonna say, you lost an x!
anonymous
  • anonymous
then i saw you just forgot it
experimentX
  • experimentX
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.
anonymous
  • anonymous
there's some extra stuff in there...
experimentX
  • experimentX
I'm not quite well with zeta functions ... looks impossible from Fourier series.
anonymous
  • anonymous
whatever, if it's actually supposed to be there, I can't figure out why..
dape
  • dape
Maybe I forgot something else, did it all on paper and copied it over.
anonymous
  • anonymous
seems it's just as experimentx did it, what's with the first series and the "(1-"
experimentX
  • experimentX
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.
anonymous
  • anonymous
If you've gone McLaurin - then why not go full complex contour integration. The poles are obvious -1, +third-root(-1), -third-root(-1)
dape
  • dape
Arent the poles \(-1,\sqrt[3]{-1},-(\sqrt[3]{-1})^2\)?
anonymous
  • anonymous
Isn't this thing just a trig substitution?
anonymous
  • anonymous
Never mind I see the x cubed. I can't see how to do this without partial fractions :/ .
TuringTest
  • TuringTest
nice @dape that was definitely more what I was looking for, now this integral is interesting again :)
anonymous
  • anonymous
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
TuringTest
  • TuringTest
Contour integral are something I certainly learn, but I would then like to see it that way too.
TuringTest
  • TuringTest
certainly need to*
anonymous
  • anonymous
|dw:1348327205764:dw|
TuringTest
  • TuringTest
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.
experimentX
  • experimentX
@Mikael how to you plan to evaluate that series via contour integrals? Don't you need need to change the series into integral?
anonymous
  • anonymous
The question - please READ the question of @TuringTest .

Looking for something else?

Not the answer you are looking for? Search for more explanations.