the failure of integration by parts!

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[thanderhop]

I don't see why someone would consider this to be an example of integration by parts failing. Both integrals are divergent, so it's not like we're saying you could start with one of those integrals, apply integration by parts, and get something wrong. An example of integration by parts failing would be if you had a convergent definite integral and applying integration by parts gave you infinity - infinity (one infinity from the boundary terms and the other infinity from the other integral), indicating that the integration by parts isn't capturing some delicate cancellation making the original integral convergent.

[corneliusgoh]

In French college math teaching, before integration the students must prove "continuity" in the interval, before attempting integration. That's rigorous math.

[wolfmanjacksaid]

I remember being so fascinated by integration by parts when finding the integral of (e^x)sin(x) and just rearranging for the answer.

[explsn]

In fact, the formal statement of ibp requires f, g to be differentiable and f', g' to be integrable. Given f, g are differentiable, f, g are continuous and thus integrable, and so fg' and f'g are also integrable. The differentiability assumption is obvious, but many people forget the integrability of derivatives is required as well.

[MercuriusCh]

Easier exploration of divergence:
cos^2(u) = (cos(2u) + 1)/2
cos(2u)/2u -- converge via Dirichlet Th.
But 1/2u -- divergent (integral is ln(u))

[mcqueen]

At 10:03, you need the function to also be bounded for what you said to be true. Otherwise take 1/x on [0, 1] which has countably many discontinuities (x=0) but isn’t integrable

[metalsonic]

Strictly saying, if common term of some series is less or equal than common term of divergent series, we can say nothing about convergence of the first series. Hovewer, if we can show inverse inequality (divergent <= some other), than this some other series diverges. Convergence of integral in this video can be easily checked by writing asymptotics of 1/x*(cos(1/x^4))^2 at 0 (which is just 1/x) and it's known that integral of 1/x diverges if bound of integration is 0

[MyOneFiftiethOfADollar]

Thinking of this as the reverse product rule of differentiation viz a viz u, v "parts" helps to spot circumstances where the integration by parts formulation can crap out or get iterative leading to a "looking back" algebraic solution.

[tomholroyd]

"up to a constant": I hear, "up to isomorphism"

[justinlink]

Pretty much _anything_ can be written as "infinity minus infinity" though, right?

[MarcoMate]

At 14:10 the inequality doesn't seem to be correct. For what we have just proved, the second function of the integrand, preceded by a minus sign, is less than or equal to that minus infinite sum, and for what Michael has just said the first part of the integrand has a finite positive integral on [0, 1], which can be bounded from above by 2 for example. So, a correct inequality would be to add 2 to that minus infinite sum. In reality, this error doesn't affect anything, because we will prove that that minus infinite sum is equal to minus infinity.
At 15:30 the first interval is actually not [-π, 0], but [2π - π/4, 2π].

[PRIYANSH_SUTHAR]

I think the - sign will be there. Because when you multiply original -sign with -¼, it became +. But then we reversed the bounds from ∞ →1 to 1 →∞. In other words it is (-)×(-)×(-). Otherwise the solution is neat. But then you brought back the -sign so it will not bother us 🙂

[luisaleman]

I'm sorry, what's the integral we "failed" solving with IBP? 🤷‍♂

[ojas]

If any two of the three definite integrals exist and are finite, so does the third and the Integration by parts Rule holds, enabling the third to be evaluated

[antonanton]

I guess a simple integral over interval (0, 1) of a constant function rewritten as X*(1/X) would be enough 😅

[FadkinsDiet]

11:14 what about the other branches of the 4th root?

[robshaw]

I was confused at first because I was overlooking the cos^2 always being positive, so the inequality is valid. If it was just plain cos, then there are intervals where the integral would be negative, and you don't know how they relate to the postively-valued intervals you are taking...

[SlipperyTeeth]

Well, if you were to use IBP here, say for f'g, you would get -infinity = (finite value)-(infinity).
Seems consistent to me.

[tomholroyd]

idk if I wanna watch this, failing at integration by parts is something I'm so familiar with it's triggering

[byronwatkins]

Infinity - infinity is a well-known indeterminate form; it can have any limit or no limit.