/sci/ - Science & Math

>>10834286
0, by symmetry

>>10834286
[math] xD [/math]

>>10834303
Nay

>>10834303
uh no

>>10834303
/thread

>>10834286
Did you try partial sums?

>>10834286
u^3 = x
3u^2 du = dx
now we are integrating 3u / (1 + u^3)
hmm, a contour integral would work here i expect. just go along the positive real axis, then along a third circle, then back along e^(i*2pi/3). apply the residue theorem. send radius to infinity. done.

>>10834741
Nope, there is a generalization for the integral though

>>10834804
>nope
what do you mean nope? what could possibly be wrong about what i said?
i don't want to compute this, but i will if needed.

>>10834286
no, you can't, don't even try

>>10834821
There is an answer though :) that is if you know the restrictions

>>10834812
Maybe what you said isn't wrong but an exact value is expected

>>10834855
fine, i'll do it properly
first note that the integral is improper of both types, so in fact the bounds are m and M and we'll send each back to where it needs to be. now make the substitution u^3 = x. then 3u^2 = dx/du, so the integral becomes [eqn] \lim_{m, M} \int_{m^{1/3}}^{M^{1/3}} \frac{3u^2}{(1 + u^3)u}du = 3 \int_0^\infty \frac{u}{1 + u^3} du [/eqn]
It is likely that one may solve this through more elementary means, but that sounds annoying, so I'm going to use residue calculus. Merely consider the contour C given by moving R along the real axis, then along the circle of radius R ccw until reaching Re^(i*2pi/3), then back to the origin along that ray. Within this contour for R > 1 we have one simple pole, namely e^(i*pi/3), and the residue of z/(1 + z^3) here is just e^(i*pi/3) / (3e^(i*2pi/3)) = (1/3)e^(-i*pi/3). So we get that
[eqn] \int_C \frac{z}{1 + z^3} dz = 2\pi i \frac{1}{3}e^{-i \pi/3} = \frac{2}{3} \pi e^{i \pi / 6} [/eqn]
Now, let's break down the contour integral. On the circular path we have the integral
[eqn] \int_0^{2\pi/3} \frac{Re^{i\theta}}{1 + R^3e^{i3\theta}} iRe^{i\theta}d\theta \leq \frac{2\pi}{3}\left[\frac{R^2}{1 - R^3}\right] \to 0 [/eqn]
as R goes to infinity.
Then, the other parts of the path are moving from 0 to R, in which we now take R to infinity, as well as moving from Re^(i*2pi/3) to 0, which is the integral
[eqn] \int_R^0 \frac{e^{i2\pi/3}z}{1 + z^3}e^{i2\pi/3}dz[/eqn]
which is just the other integral times -e^(i*4pi/3). So now we have (1 - e^(i*4pi/3)) * integral = 2pi/3 * e^(i*pi/6), and of course we also have to multiply by 3. Note 1 - e^(i * 4pi/3) = 1 - ( - 1/2 - sqrt(3)/2 i) = 3/2 + sqrt(3)/2 i = sqrt(3)e^(i*pi/6). so dividing, we find the final value of [math] \frac{2\pi}{\sqrt{3}} [/math].

>>10834895
High fucking effort anon

>>10834912
it took quite a while to do, yes, and now i'm going to sleep later than i'd hoped. but i do think this captures a very general and very useful method rather thoroughly.

>>10834895
OP here Yes this is correct. Well done anon.

>>10834895
thank you