a classic integral without the standard tool -- antiderivative of ln x without integration by parts

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Michael Penn

Күн бұрын

Пікірлер: 70

@gravitas802 8 ай бұрын

Fubini didn't invent changing the order of integration so you could avoid integration by parts smh 😂

@robertl.crawford4369 8 ай бұрын

I'm saying.😅

@Bodyknock 8 ай бұрын

9:21 An interesting minor side note, at the end of the solution the expression a - a ln(a) is replaced by an arbitrary real number C. But recall that this derivation assumed in the beginning that a>0, so technically C can only take on values possible under that constraint. Which means you should probably prove that a - a ln(a) can in fact result in any real number for a>0. It pretty clearly can take on all negative numbers as a grows to infinity, so the main thing to prove is that the limit as a approaches 0+ of a - a ln(a) is positive infinity, which is equivalent to seeing if lim a ln(a) as a approaches 0+ is negative infinity. Lim a ln(a) = lim ln(a) / a ⁻¹ (By LHR) = lim -a ⁻¹ / a ⁻² = -lim 1/a⁻¹ = -lim a = 0 as a approaches 0+. But that means for a>0 the resulting constant is non-negative, so the derivation in the video doesn’t quite cover all the actual possible constants C could take in the indefinite integral.

@TheEternalVortex42 8 ай бұрын

There's no general requirement that the arbitrary constant must be able to take any value. For example, the integral of sin x clearly can only have bounded values for its constant.

@Bodyknock 8 ай бұрын

@@TheEternalVortex42 Exactly my point. We know that the indefinite integral of ln(x) can in fact have an arbitrary constant, but the derivation of the antiderivative in the video doesn't quite establish that.

@etienneparcollet727 8 ай бұрын

That limit is 0.

@Bodyknock 8 ай бұрын

@@etienneparcollet727 What limit is 0? The limit I was talking about in my comment isn't.

@etienneparcollet727 8 ай бұрын

@@Bodyknock 0 ln(0) is 0 famously. Makes all kind of things work like entropy. You made a mistake with your exponents it simplifies to a not 1/a.

@mtaur4113 8 ай бұрын

Since you are just looking for an antiderivative, you could simplify by using a=1 from the beginning and include +C at the end. The double integral is a simple triangle with only one region.

@mtaur4113 8 ай бұрын

This is actually done for arctan, where a=0 bases the triangle at the zero of arctan.

@nonentity168 7 ай бұрын

It was a bit confusing when I tried to know the general case for any value of a. It turns out to be definite integral, and we're finding the antiderivate so it doesn't matter what value we select for a.

@dusscode 27 күн бұрын

there is no guarantee that the final answer is off by a constant term

@mtaur4113 27 күн бұрын

@@dusscode integrating from a constant (a) to a variable is equal to the total change in an antiderivative. That total change +C is an arbitrary antiderivative. Using a particular convenient constant (such as 1) instead of (a) just leaves room for +C to do what it needs to.

@thephysicistcuber175 8 ай бұрын

I'm pretty sure this can be used to prove integration by parts.

@holyshit922 8 ай бұрын

but derivative of product is much simpler proof and better from teaching point of view

@thephysicistcuber175 8 ай бұрын

@@holyshit922 if the functions are not continuous I'm pretty sure this proof still holds.

@pierrot31511 8 ай бұрын

He can also derivate xln(x)

@merijnmartens6952 8 ай бұрын

@@thephysicistcuber175 The method from the video still assumes you know the derivative of the integrand. So the integrand has to be differentiable, hence continuous. Yes, you can use it to prove integration by parts, but the assumptions are the same as with the derivative of product method: The integrand has to be a product of two functions, one of which is integrable and the other differentiable.

@DarinBrownSJDCMath 8 ай бұрын

Hmmm... seems like this is integration by parts after all, no?

@goodplacetostop2973 8 ай бұрын

13:34

@YO-in2uw 8 ай бұрын

We can calculate the double integral as the big triangle (1 to x) minus the small triangle (1 to a). The latter is constant to x and bounded, so it can be treated as -C.

@matematicacommarcospaulo 8 ай бұрын

I liked this approach

@leif_p 8 ай бұрын

If we set I(x) = \int f(t) dt, then I'(x) = f(x) suggests the ansatz I(x) = x*f(x) + g(x). Taking the derivative, we get I'(x) = f(x) + x*f'(x) + g'(x) = f(x) => g'(x) = -x*f'(x). So if we know f'(x) and x*f'(x) is easier to integrate, we can find I(x). E.g. if f(x) = ln(x), then g'(x) = -x*1/x = -1. Then g(x) = -x, so: I(x) = \int ln(x) = x*ln(x) - x (+ C). This may also be equivalent to integration by parts.

@dougrife8827 8 ай бұрын

There’s a simpler derivation: Notice that integrating ln(u) from 1 to x equals the area of the rectangle x*ln(x) minus the area under the inverse function e^u between zero and ln(x). That is, this definite integral equals x*ln(x) - (x -1) or, x*ln(x) - x + 1. In this case the constant of integration happens to be unity but that’s of no importance since the antiderivative is the same x*ln(x) - x. This method of integration is very similar to implicit differentiation. It takes advantage of the fact that the antiderivative of the inverse function is known or easy to evaluate just as implicit differentiation makes use of the derivative of the inverse function to find the derivative of the given function.

@Czeckie 8 ай бұрын

9:17 I find it curious that not every antiderivative is achieved like that (not every real C has a corresponding value a). I wonder if you can characterize for which functions every antiderivatives can be expressed as a definite integral. We need a surjective antiderivative. Eg. it's possible for x^2, but not possible for x or any bounded function. Can you check for surjectivity of the antiderivative without knowing the antiderivative?

@CielMC 8 ай бұрын

10:57 shouldn't it be the other triangle since u always starts at 0?

@ChaoticNeutral6 8 ай бұрын

Why is the lower bound 1 for the integral representation of ln x?

@Noam_.Menashe 8 ай бұрын

Because ln1=0

@ericbischoff9444 8 ай бұрын

Because it's the definition of ln x. ln 1 = 0 is a consequence of that definition.

@JCCyC 8 ай бұрын

I inevitably (as always happens when you release a Calculus video) started to play randomly with integrals in Desmos, and eventually came across what seems to be a fun fact: the integral from zero to infinity of 1/(x^x) seems to be 2. Worth a video? Edit: it MAY not be exactly 2.

@SylvainBerube 8 ай бұрын

It seems to converge toward ~1,994955. Fun stuff!

@vczh 8 ай бұрын

I don't quite understand why int(lnxdx) = int(intdt, a, x)? Is that because C is replaced with int(intdt, 0, a)? So this technique only works for lnx but not all functions?

@GrouchierThanThou 8 ай бұрын

Since you could have chosen any value for a > 0, why not just choose a = 1 and get rid of the rectangular area?

@VicTheMathMan 8 ай бұрын

Is It possible use that thecnic in the sec integral ?

@ayylmao2410 8 ай бұрын

hmm why would the second part integral has u as lower bound and not a

@txikitofandango 4 ай бұрын

The drawing looks very similar to geometric interpretations of integration by parts that I've seen

@RSLT 8 ай бұрын

Cool Method ❤

@victorpaesplinio2865 8 ай бұрын

Using the inverse function theorem you can get this result. But it kind of uses integration by parts to prove the general result, so I dont think it is fair in this case

@ingiford175 8 ай бұрын

You said you can do some work to get 0 included. Are you sure? from 0 to any point looks like it has an area of infinity

@scottmiller2591 8 ай бұрын

I would have used the formula for the integral of an inverse function, which gives the results immediately (in all cases).

@bassamxp 8 ай бұрын

Just take u = ln x and integrate by part the gamma like function.

@get2113 7 ай бұрын

What about guess and check? My kids told me that GaC was forbidden in college, which left me unable to help them.

@Oskar-zt9dc 8 ай бұрын

in the first bit you have to assume that a>1!

@Oskar-zt9dc 8 ай бұрын

or do a case where 0

@manuelosuna1562 8 ай бұрын

Given that lnx is continuous on (0,∞), any a>0 suffices for the antiderivative to equal the integral of lnt from a to x. In particular, choosing a>1 is what is done in the video but as you pointed out, it will be interesting to see if it is possible to reach the same result taking 0

@真正的粉丝-d5y 8 ай бұрын

so what's standard tool?

@xizar0rg 8 ай бұрын

A complete reading of the title of the video answers your question.

@真正的粉丝-d5y 8 ай бұрын

@@xizar0rg no answer at all

@xizar0rg 8 ай бұрын

@@真正的粉丝-d5y The structure of the title, "A Classic Integral without the standard tool -- antiderivative of ln x without integration by parts", is "without the standard tool " -- "without " If that context is insufficient, perform an internet search on the phrase "what is the standard tool for integrating the natural logarithm".