contour integral with bounds [0,inf]

[mighty dew]

I was evaluating the integral

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[mighty dew]

And i got an answer that wasnt correct

[wind bronze]

Don't know complex analysis, but we can try the usual approach:
x = u^3
dx = 3u^2 du
x^(1/3) dx/(x^2 + 1) = 3u^3 du/(u^6 + 1)
Whike we can do partial feactions now, it's simpler if we take v = u^2, dv = 2udu first.
3u^3 du/(u^6 + 1) = (3/2)u^2 (2udu)/(u^6 + 1) = (3/2)vdv/(v^3 + 1)
And now you can do partial fractions.

[mighty dew]

i know we can do that, I just want to see how it would be evaluated using contour integration because the residue theorem doesnt give pi/sqrt(3)

[modern spade]

Well, in order to do such, we need to carefully handle the branch cut arising from the fractional power cube root(x).

[mighty dew]

yeah i know which is why we have 3 paths right?

[modern spade]

Firstly, consider the function in the complex plane:

[modern spade]

Yes.

[mighty dew]

i got gamma 1,2,3

one of them vanishes as it is the contour in the complex plane

but we have gamma 2 in the real axis so thats the original integral right?

so we are left with gamma 3

[modern spade]

So, the function z^1/3 has a branch point at z = 0. We will placa the branch cut along the negative real axis. We will use a keyhole contour that avoids the branch cut, consisting of:

• A line segment above the real axis from ɛ to R.
• A semicircle C_R of radius R centered around at the origin, going counterclockwise.
• A line segment below the real axis from R to ɛ.
• A small semicircle Cɛ of radius ɛ centered at the origin, going clockwise.

One moment, please.

The contour integral can be split into 4 parts.

[modern spade]

yes that works.

[mighty dew]

so we can use a keyhole contour as well instead of half a semi circle?

from 0 to inf?

which is where the fourth integral comes from

as we are going around the keyhole right?

[modern spade]

One moment, I'll present the residue calculation (1)

[mighty dew]

yeah i got that too

[modern spade]

,rotate

,rotate

,rotate

[modern spade]

yes

[mighty dew]

is that the answer?

[modern spade]

the fourth integral (around the small semicircle) comes from ensuring we avoid the singularity at z = 0 & close the contour properly

[mighty dew]

because the answer is what i posted

[modern spade]

those are the residue calculations.

not there yet 😃

[mighty dew]

i didnt see that but alright lets keep going

[modern spade]

that's just them summed up in the box.

[modern spade]

I thought i put it in this image lol

anyways

so we're evaluating the contour integral

do you know residue thereom? @mighty dew

[mighty dew]

yup

[modern spade]

cool, how are we applying this

[mighty dew]

i calculated the residues but then got stuck on the third gamma

but what do we do with the other paths?

[modern spade]

,rotate

,rotate

,rotate

[modern spade]

then you got real axis contribution

[mighty dew]

okay

[modern spade]

the Integrals along the line segments above and below the real axis are equal, so:

[mighty dew]

but how do we just conclude that

[modern spade]

so the Integrals along the large semicircle C_R (path gamma_2) and the small semicircle Cɛ(path gamma_4) both vanish as R --> infinity and ɛ-->0, respectively

the Integrals along the gamma_1 (just above the real axis) and gamma_3 (just below the real axis) combine to give the integral along the real axis from 0 to infinity, w/ a phase factor difference due to the branch cut

thus, you're left with the integral along the real axis, which is your original integral, plus the resides from the poles

[modern spade]

what do you think

point out where you don't understand the most, we can discuss each aspect over if you like.

[mighty dew]

i understood

[modern spade]

alright, if that concludes it and you have no other question, please use +close (& if you found my assistance satisfactory, please send a "thanks" 😀). good day to you.