Question

Integral...

Answer 1

\[\int\limits_{-\infty}^{\infty} \frac{d \xi}{1+\xi^6}\]

Answer 2

Is that\[\xi^{6}\]on the bottom?

Answer 3

Yes indeed.

Answer 4

1+xi^6 is the denominator.

Answer 5

residues?

Answer 6

Please show me, I can't get a good grasp on it to save my life. I was going to factor the bottom and use partial fractions with complex numbers :/

Answer 7

that might work as well

Answer 8

i always have to teach myself this again
you need the poles in the upper half plane, which is easy enough
then you compute the residues at each pole, then the integral is \(2\pi i \sum res\)

Answer 9

in the upper half plane the poles are \(i=e^{\frac{\pi}{2}i}\) , \(e^{\frac{\pi}{6}i}\) and \(e^{\frac{5\pi}{6}i}\) if i am not mistaken
each of them are simple poles

Answer 10

then you need to compute the residues, which is not too bad, because since they are simple poles you can do this by evaluating \(\frac{1}{6z^5}\) at each pole

Answer 11

@Zarkon stop me if i am putting my foot in my mouth

Answer 12

you are doing great :)

Answer 13

it has been a long while for sure

Answer 14

\[\frac{1}{6e^{\frac{5\pi i}{6}}}=\frac{1}{6}e^{-\frac{5\pi i}{6}}=\frac{1}{6}\left(-\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)\]

Answer 15

that is one residue, you should check it to make sure it is right

Answer 16

ok well we lost @malevolence19 so no need to go further
i will do it for homework

Answer 17

lol

Answer 18

i think that residue is right
then a raft of stuff cancels when you add if i am not mistaken

Answer 19

it is correct...it all woks out nicely

Answer 20

don't you end up with something like \(2\pi i\times -\frac{1}{6} i \) ?

Answer 21

no...but close

Answer 22

am i missing another \(\frac{1}{6}\) ?

Answer 23

yes

Answer 24

i suppose i could actually do it instead of guessing

Answer 25

\[\pm\frac{\sqrt{3}}{12}-\frac{1}{12}i\]

Answer 26

oh doh i forgot about \(\frac{1}{6i^5}\)

Answer 27

so you end up with \(-\frac{1}{12}i-\frac{1}{12}i-\frac{1}{6}i=-\frac{1}{3}i\) multiply by \(2\pi i\) and get \(\frac{2\pi}{3}\)

Answer 28

nice

Answer 29

not bad for 25 years later....

Answer 30

lol

Answer 31

really!

Answer 32

I taught complex variables a year or so ago...so I still have some of it in my head

Answer 33

i taught solving linear equations recently
can do that pretty well now...

Answer 34

gotcha

Answer 35

well apparently you still have some complex skills ;)

Answer 36

Couldn't this also be computed by partial fraction decomposition?

Answer 37

yes, if you allow factoring over complex numbers

Answer 38

I'm sorry, the site was down and my computer was acting up. I thought the residue for order k was given by:
\[\text{Res}_k=\lim_{z \rightarrow z_0}\frac{d^{k-1}}{dz^{k-1}} (z-z_0)^kf(z)\]
Then from there you just have:
\[R_i \text{ = ith residue}; \text{ solution = } 2 \pi i\sum_j R_j\]

Answer 39

Oh, since k is order 1 for all of them you get something really simple... nevermind. Thanks guys!

Answer 40

But with that being said, how do you do stuff like this with logarithms and things? My friend in complex was showing me integrals you couldn't do without complex analysis (or estimation).