generalizing a Calculus 2 integral

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[@ItzMezzo]

you forgot +C

[@Monster-nn.007]

😂😂😂😂

[@tigrexgaming21]

At 5:04 there shouldn't be a x - omega^2n factor, since its a duplicate of the x - 1 factor.

[@skylardeslypere9909]

Luckily it doesn't matter since they get divided out, but I was screaming this at my screen as well

[@cv990a4]

@@skylardeslypere9909 Correct!

[@gp-ht7ug]

I need to watch this video at least ten other times

[@Alan-zf2tt]

Ah - that makes at least two of us

[@CM63_France]

Hi,
It is a real feat to have dealt with this in the general case! Bravo!

[@gp-ht7ug]

An example would help understand

[@goodplacetostop2973]

18:15

[@pubkeybreaker]

Express x^n+1 in product form in terms of its roots and then decompose in terms of partial fractions.

[@bsmith6276]

Very nice. I remember doing something like this years ago when trying to evaluate the indefinite integral of (tan x)^(1/n). [n=2 is the classic sqrt(tan x)]. But this was a bit smoother than what I did to get the fractional composition.

[@stephenhamer8192]

Has anyone checked Penn's final result by differentiating back? Nah, just kidding

[@Patapom3]

Amazing!

[@roastedchicken9096]

Sir nice video.
Pls make more of calc 2 integral

[@Keithfert490]

You specified that the primitive nth root of unity is the number that
1. Yields 1 when raised to the nth power; and
2. Does not yield 1 when raised to any lower power.
In 2, the lower power can be any real number, right? If you take (what I consider to be) the more obvious interpretation that the power is an integer (since n is an integer), then the number specified by 1. and 2. is not unique. For example, if n is prime, all nth roots of unity meet these conditions.

[@Alan-zf2tt]

Okay - Warning: Michael's on fire
One of these days I may just be able to fill in all of the blanks but I think I will need a large scratchpad and lots of time

[@M.Z.M.N.]

Or you could just use the β function

[@richardheiville937]

Here, it's an indefinite integral, not integral on [0,infinity[ so your method doesn't apply here.

[@randomguy2836-g3i]

@@richardheiville937 He meant incomplete Beta function

[@IoT_]

2:36 a typo, should've been 2iw

[@davinheagertans4275]

Why are these u-tube math guys so frantic? Red bull?

[@jamesfortune243]

Is it reasonable to conclude that because of the pi in the calculation of residues, integrals of functions with poles often contain pi?

[@aadfg0]

Formula has a mistake. int_0^inf dx/(x^3+1) converges, but according to your formula it diverges to infinity due to the ln(x^2-2cos(pi*k/n)+1) term.

[@deinauge7894]

In the formula (with n=3) there are 3 terms of this form. 2 of them with a factor of 1/2 (= cos(pi/3)=cos(5pi/3)), the other one with a factor of -1 (=cos(pi)).
Thus the divergence cancels out.

[@aadfg0]

​@@deinauge7894 Got it. Now I watched the whole video and didn't spot any global errors. As a bonus, finding I = int_0^inf dx/(x^n+1) with contour integration and comparing leads to a new proof of the closed form formula for some messy trig sum.

[@simoncrouch3549]

A good place, surely, to introduce students to hypergeometric functions?

[@TomFarrell-p9z]

Fails for n = 0

[@yashvardhanbiyani7617]

At n=1, the antiderivative should be ln(x+1), however that doesn't seem to workout from the final formula derived on plugging n=1. For me it gives, ln(x^2+3). Am i missing something?

[@mrgold4678]

Well, he missed an “x” in final formula in log term: log(x^2 - 2 cos(pik/n)x + 1). But still, plugging n = 1 gives 2 ln (1 + x) and I’m not sure where it goes wrong…

[@deinauge7894]

​@@mrgold4678there's always a 2 or a minus sign in the way of a perfect calculation 😅

[@zucazx]

At 12:55 he could think that you are summing a complex to its conjugate, since the conjugate of w^m is w^(-m) and the conjugate of a product is the conjugate of the products. If he did it, x would need to be real to you declare that w^(-m)log(x-w^(-m)) is the conjugate of w^m log(x-w^m).

[@bridgeon7502]

I'm not the target audience but I guess I'm early?

[@Alan-zf2tt]

I think that makes at least 3 of us

[@EtienneSturm1]

nice but brutal LOL

[@LucasSilva30]

I'm not happy with the final formula 😢

[@Dionisi0]

Jesuschrist is always the answer
remember that
💀💀

[@fxrce6929]

u forgot the + C video ruined fell off