ฝัง
- เผยแพร่เมื่อ 11 ก.ย. 2024
- Here's another wonderful integral put to rest using Feynman's technique of differentiating under the integral sign. The integral itself looked absolutely gorgeous so no wonder the solution involves the Eular Masceroni constant and pi. The solution development also involves making use of the properties of the gamma function, including the awesome duplication formula
"I will prove this in a later video" - Top 10 famous last words.
There are actually quite a few videos on the gamma function that I want to do😂
Hopefully I'll get done with them soon
@@maths_505 I'm literally on the edge of my seat here sir! thanks for the amazing content G.
@@maths_505 Gamma function is one of those hero functions that gets buried and filed under "technicality/trick" waaaaay too often to justify.
Funny. I think I am developing an instinct. Although the title of your video puts the thumb on Feynman's technique with no mention of the Gamma function, just with a glance on the integrand I smelled there a Gamma function . Thanks for not disappointing me ! 😊
I deeply enjoy your solutions. Keep up with the good job!
"How could I ever forget the pi"? Well, this shows that even the best do forget a few things sometimes. Anyway, thanks for the great explanation and the awesome video.
OMG! These videos are addictive! Well done!
Thanks mate
I solved it like this:
Applying integration by parts we find that:
I= 2∫ x²exp{−x²}(log(x)−1)dx
Use the linearity of the integral and see that 2log(x)->log(x^2). Observe that
x²exp(−x²)log(x²)=∂ₜ(x²ᵗexp(−x²)) at t=1.
Define the function J(t) as the integral from 0 to infinity of x²ᵗexp(−x²)dx. Thus we find that
I = J’(1)-2J(1).
Setting u=x^2 for our definition of J(t), it is clear to see that it evaluates to
J(t)=[1/2]Γ(t+[1/2])
We can plug this into our equation for I to find that
I=1/2Gamma’(3/2)-Gamma(3/2).
Interestingly, we can generalize this easily to integrate e^(-x^n)ln(x) to find that
I_n=1/n Gamma’(1+1/n)-Gamma(1+1/n).
No Feynman needed! Just integration under the integral sign. Great problem!
There is a small mistake in the last picture. It should be " -(ln4+gamma)sqrt(pi) " at the 2nd line from bottom. The final result is correct and thank you very much for the video!
Bro, you are favourite math channel by far. Thank you for your work. I just keep finding awesome ideas. Also, I understood your approach in the beginning and solved it I think we don´t need to differentiate the recursive formula, because we can get gamma prime (1/2) from the duplication formula. We differentiate it and instead of t = 1 we just plug in t = 1/2. This way we don´t need the gamma prime (3/2) at all and even the gamma prime (2) for the solution.
P. S. patiently waiting for your book on Integration. Don't let us get old by then😂
Yeah that approach is definitely better. I think I should write a book by now😂
Hey maths 505! Another great video! I was surprised by the strategy to evaluate Γ’(1/2), you never disappoint. If you’re ever looking for more integrals, I’d like to suggest trying cosx/(x^4+1) from -inf to inf, since you already did cosx/(x^2+1). Or you could try a generalization for cosx/(x^2n+1), which gives a surprisingly compact result.
Noted
Glad you liked the strategy cuz I stumped myself in the first couple of tries before coming up with this😂
How did you do the 2n case? I tried integrating using the contour of a sector and got an integral I'm not sure how to do. Then I tried a semi-circular contour and got a summation I'm not sure how to simplify
@@two697 I actually also got the summation (the semicircular contour is definitely the way to go), it cant really be simplified to a greater extent, but it is at least a closed form with a sum of cosines
@Daniel Rosado I tried it and basically just got the residues multiplied by 2πi...
No, seriously - this case of the integral is always equal to its residues multiplied by 2πi, since if you're integrating over the semi-circular path, the integral over gamma (the large arc) always goes to zero. Which means that you're left with the statement that the line integral over the contour is simply equal to the actual integral from negative to positive infinity; i.e:
2πi ΣRes(f(z)) = ∫_(-inf)^(inf) cos(x)/(x^2n +1) dx.
Which (I believe) is independent of your choice of f(z), whether it be just the original function or its complex exponential counterpart. However, what I'll tell you is that these two options give you the same exact result, which is due to a neat little symmetry - and that I'll leave as an exercise to you :)
In this form, you really can't do anything but compute them manually each time to do these integrals, as the individual residues are pretty much unique, and I don't think Wolfram Alpha would like to do a bunch of indexed limits without telling you that it doesn't know what you want it to do.
That one was from experience. :(
@@daddy_myers yep, the residues being equal to the whole integral is pretty much par for the course in terms of complex integrals. If the upper curve doesn’t evaluate to 0, there’s not much you can typically do to evaluate it. And of course the real part is equal to the whole function, because the imaginary part is the integral of sinx/x^2n+1, an odd function. And yes, you have to compute it manually. But you can have a very compact form of simply a sum from 1 to n of sines.
very nice solution
Awesome calculus, thanks a lot for this great video, Kamaal! I've wondered whether a closed-form solution can be obtained for a generalized version ∫_b ^inf exp(-x^2)*ln(x+c) dx, involving a positive lower bound [b>0] for the integral and a positive offset parameter [c>0] applied to the argument of the logarithm. With straigthforward modification of the technique applied in the first part of the video, it is easy to obtain ∫_b ^inf exp(-x^2)*ln(x) dx = (1/4) * ∂/∂t Γ_inc(t=1/2, b^2), Γ_inc(t,x) denoting the upper incomplete gamma function. However, it may not be possible to resolve the problem caused by adding the offset term [c]. Using the same technique as in the video would now require defining the integral I(t) = ∫_b^ inf exp(-x)*(x+c)^t dx, but then the terms exp(-x) and (x+c)^t are shifted as regards their arguments [x], so evaluating (x+c)^t at [t=-1/2] doesn't produce the desired term x^(-1/2) that would be necessary due to the change of variable in the very first step. Any ideas? I've tried to type in the problem to Wolfram Alpha's online integral calculator but it doesn't provide any closed-form solution, either.
I think the best we can do is the form in terms of the incomplete gamma function
Maybe a transformation of the variable and splitting up the resulting integral could yield more insight, but then we'd have a result in terms of an integral over a bounded interval which I'm not a big fan of.
By the way, there is another blind mathematician that did huge contributions - Lev Pontryagin.
A slightly different approach which could've reduced the Algebra significantly, and would've been more efficient to compute:
*Start by defining I(α) to be equal to:*
∫_0 ^inf e^(-t) * t^α * ln(t) dt
*Upon observation:*
I(α-1) = Γ'(α)
*Which is useful, since:*
I(-½) = Γ'(½)
*Using the duplication formula:*
Γ(α) Γ(α+½) = π½ * 2^(1-2α) Γ(2α)
*Differentiate with respect to α:*
Γ'(α) Γ(α+½) + Γ(α) Γ'(α+½) = -ln(4)π½ Γ(2α) + π½ * 4^(1-α) Γ'(2α)
*Plugging in α=½:*
Γ'(½) Γ(1) + Γ(1/2) Γ'(1) = - π½ * ln(4) * Γ(1) + 2π½ Γ'(1)
*Simplifying:*
Γ'(½) - γ * π½ = - π½ * ln(4) - 2γ * π½
*Solving for Γ'(½):*
Γ'(½) = -π½ (ln(4) + γ)
*Remembering that:*
I = ¼ * I(-½) = ¼ * Γ'(½)
*Then:*
I = - ¼ * π½ (ln(4) + γ)
*_QED_*
QED Suiiiiiiiiiiiiiiii
Euler-Mascarpone constant
We should make a comic on a superhero named feyn-man that stomps literally every villanous integral ever.
on a side note, thanks for making me feel better after getting my behind kicked by the easiest two freaking diff equations ever:
1- 4xy''(x)+y(x)=0.
as a hint, it says to try a sub of y = tu(t) where t = sqrt(x), this should turn it into a equation of a bassel function solution, although when I tried it, it's giving a t^3 for the coef of u''. idk why I can't
t find a way lol
2- (1−x^2)y ′ −2xy' + 2y = 0 ; y(1)=2, y′(0)=−4.
Which is literally freaking x, when using reduction of order, what I get is -1 + 1/2 ln(x+1/x-1) which is undefined for 1! idk man I think I'm doing something wrong, but for some reason felt like sharing my defeat.
It's cool bro
You'll get it eventually
my strategy would be use laplace for everything (lol), you can always get great simplification to a first order differential equation as long as y and it’s derivatives are multiplied only by x and not x^2
@@danielrosado3213 gigachad level strategy
For the second DE, you were right about the ln((1-x)/(1+x)). Using either an exact second order differential equation formula (which I used) or using reduction of order like you did, we find that a general solution is c1 + c1x/2*ln|(1-x)/(1+x)|) + c2x. In this case, c1 = y(0) and c2= y’(0). So you can plug in those values to find the particular solution.
Hey loved your video but I don’t understand how you went from “ e^-u to e^-x when u = x^2 at min 1:32 thanks for the vid btw I’m starting to like math thanks to your vids
In definite integrals, the variable is just a dummy variable so you can call it whatever you want
Its not the case for indefinite integrals which are antiderivatives.
Antiderivatives depend on variables whereas definite integrals represent structures so once you get a new structure after substitution like I got at the 1:32 mark, you can throw that substitution out of the window and name the variable whatever you want
And thank you so much for that compliment!
Means alot to hear stuff like this
@@maths_505 okay thanks so much I understand now. Thanks !
PS: just gained a subscriber
Without the derivation of the duplication formula, something is missing :)
What's the software that you're using for writing the notes? It's cool
I proved in a video that can be found in the proofs playlist
So you are saying and have proven that Euler was the “Bruce Lee” of mathematics.
That's actually an excellent approximation of Eular😂
I love everything about your video except how you write t! If you ever have to write a słowo in Polish, you're going to be in trouble 😆
😂😂😂
Got a better parameter?
@@maths_505 in all seriousness, learning the Russian alphabet has significantly helped my algebra. It's like the more types of letter pool I have the more types/contexts of mathematical object I can juggle. When I feel brave enough, I want to try the traditional chinese pallette... but tbh it scares the pants off me, its like the niagra falls of rabbit holes with a sign that says "indefinitely inextricable side-track" overhead.
@@tinkeringtim7999 learning Russian it is then😂