a classic integral without the standard tool -- antiderivative of ln x without integration by parts
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Fubini didn't invent changing the order of integration so you could avoid integration by parts smh 😂
I'm saying.😅
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.
This is actually done for arctan, where a=0 bases the triangle at the zero of arctan.
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.
I'm pretty sure this can be used to prove integration by parts.
but derivative of product is much simpler proof
and better from teaching point of view
@@holyshit922 if the functions are not continuous I'm pretty sure this proof still holds.
He can also derivate xln(x)
@@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.
Hmmm... seems like this is integration by parts after all, no?
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.
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.
@@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.
That limit is 0.
@@etienneparcollet727 What limit is 0? The limit I was talking about in my comment isn't.
@@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.
I liked this approach
13:34
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.
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.
Cool Method ❤
Is It possible use that thecnic in the sec integral ?
Since you could have chosen any value for a > 0, why not just choose a = 1 and get rid of the rectangular area?
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.
It seems to converge toward ~1,994955. Fun stuff!
10:57 shouldn't it be the other triangle since u always starts at 0?
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?
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?
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.
hmm why would the second part integral has u as lower bound and not a
🔥🔥
What about guess and check? My kids told me that GaC was forbidden in college, which left me unable to help them.
Why is the lower bound 1 for the integral representation of ln x?
Because ln1=0
Because it's the definition of ln x.
ln 1 = 0 is a consequence of that definition.
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
I would have used the formula for the integral of an inverse function, which gives the results immediately (in all cases).
Just take u = ln x and integrate by part the gamma like function.
Good lesson 😊
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
in the first bit you have to assume that a>1!
or do a case where 0
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
Whoa, 6 AM Mike?
interesting
This problem reminds me of a mane stripped lion or one whose powers have been stripped and revoked. Nonetheless, the problem is still solvable.
so what's standard tool?
A complete reading of the title of the video answers your question.
@@xizar0rg no answer at all
@@user-yu1ic2ip4h 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".
Correct me if I'm wrong, but you actually did an integration by parts...
Maybe we can use the identity G(x)=x*f-1(x)-F(f-1(x)) where F is an integral of f
Isn’t that just integration by parts?
@@stephenbeck7222No, it's a formula used to find the integral of an inverse function
@@axeitorits derived from integration by parts :). try it yourself.
I think all integration by parts can be interpreted like this.
no that's not a good place to stop
looks awfully like integration by parts tho
This is just integration by parts with extra steps
Slenderman is approaching 💀💀💀
So stupid and complicated method , to deliver some one good knowldge ..use easy approch. Its harsh method
Yes,but he wants more views , and this gives him alot of viewers.
I thought you would do some manipulation in expansion