A nice integral from 1886
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- Опубликовано: 6 авг 2023
- Time travel seems to have applications for integral calculus. The integral here is pretty cool for an exam but it's a variation of it that got us one of those nice results we're used to seeing here on the channel.
![Skilla Baby - Misfit (Feat. Polo G) [Official Video]](http://i.ytimg.com/vi/IjsylSnNMyw/mqdefault.jpg)
In case anyone needs a proof for the series expansion of ln(2sinx):
ruclips.net/video/mqPTvELJPM0/видео.html
wouldn't that simply be one of the anti-derivatives of cot + ln(2)?
I doubt if this is really an examination question (without any hint) in Cambridge. I guess it will take a Math professor a day or half to figure out the possible solution. 😅
If you're allergic to trig like me, you can substitute u = 2-x to get an equivalent integral summed from -2 to 2, then substitue v=-u to get another equivalent integral, and then sum those 2 to see that twice the required integral is the integral of an even function from -2 to 2.
Allergic to trig😂😂😂
Nice one mate. I like the approach you've outlined.
I tried the same, but isn't it odd functions that 0-out across symmetric regions of integration?
Oops, you're right. I really shouldn't integrate at 1 in the morning. From where I left off, it looks like you do indeed have to use a trig sub to finish it off.
To show that the integral I is 0, you can also just substitute u=4-x to first get 2I=Int(ln(x*(4-x))/sqrt(x*(4-x)),x=0..4) after combining 2I=I+I. Then substitute x=2+u which gives 2I=Int(ln(4-u^2)/sqrt(4-u^2),u=-2..2)=2*Int(ln(4-u^2)/sqrt(4-u^2),u=0..2). Finally substitute v=4-u^2, which results in 2I=Int(ln(v)/(sqrt(v*(4-v))),v=0..4)=I. Thus I=0.
Which integral
Ah dear mister mathematics five-hundred and five, I consider your query a mere trivia to the winds of my supreme integration techniques.
Now that's the type of comment I wanted to read on a video about 19th century integral calculus exams🤣
Alternatively we can define the integral to be a more general form of this case: int 0 to A of ln(x)/sqrt(Ax-x^2) dx which upon separating the denominator into sqrt(x) sqrt(A-x) the substitution u=sqrt(x) is quite logical, leaving us with an integral which simplifies to int 0 to sqrt(A) 4ln(u)/(sqrt(A-u^2)) du which is begging for a trig sub u=sqrt(A)sin(t). Upon simplification we have int 0 to pi/2 4ln(sqrt(A)sin(t)) dt which using log rules can be separated into Euler's log trig integral, which is -pi/2 ln(2), and an easy integral which evaluates to pi/4 ln(A). Multiplying everything by the factor of 4 gives us pi(ln(A/4)) as the final result. Substituting A=4 gives us the desired result I=0.
That was so awesome, thanks for sharing!
Nice tricks!! I solved it with the beta function,no need for tricks, but alot of work and knowledge.😃💯
How other variables constant when we integrate with respect specific variables?
Good stuff
I remember seeing this in a book a while back, was able to do it via a substitution bonanza. Was shocked to see that the original integral is 0, but the added representation for G is even better…
Yeah I was kinda disappointed when I got a zero and then checked the solution given by nahin. So I tried playing around till I found a cool result.
@@maths_505 Keep up the good work! Definite integrals are probably my favorite computation to do, well except for some of the crazier contour integrals
We can get rid of radical using Euler substitution
sqrt(4x-x^2)=xt
t=sqrt(4x-x^2)/x
t=sqrt(4/x-1)
sqrt(4x-x^2)=xt
4x-x^2=x^2t^2
4x=x^2+x^2t^2
4x=x^2(1+t^2)
4/(1+t^2)=x^2/x
x = 4/(1+t^2)
xt = 4t/(1+t^2)
dx = -8t/(1+t^2)^2
Int(ln(4/(1+t^2))*(1+t^2)/(4t)*(-8t)/(1+t^2)^2,t=infinity..0)
2Int((2ln(2)-ln(1+t^2))/(1+t^2),t=0..infinity)
2ln(2)Int(1/(1+t^2),t=0..infinity)-2Int(ln(1+t^2)/(1+t^2),t=0..infinity)
Int(ln(1+t^2)/(1+t^2),t=0..infinity)
t = tan(u)
dt=(1+tan^2(u))du
Int(ln(1+tan^2(u)),u=0..Pi/2)
Int(ln(1/cos^2(u)),u=0..Pi/2)
-2Int(ln(cos(u)),u=0..Pi/2)
Noice
Sir I didn't get the step when you put 2sinu in ln how can you please explain me
Since the original integral evaluates to zero and the function which is integrated has a zero at x = 1, it follows that the integrals over that function from 0 to 1 and from 1 to 4 cancel each other. So I thought it would be interesting to calculate one of these integrals on its own; obviously, I chose the simpler one from 0 to 1. Progressing by the same methods, finally I have to evaluate the following infinite sum:
-2 sum_{k \ge 1} sin(k pi/3) / k²
Examining the possible values of sin(k pi/3), this can be written as the following infinite sum:
-sqrt(3) sum_(n ge 0} (1/(6n+1)² + 1/(6n+2)² - 1/(6n+4)² - 1/(6n+5)²)
And now I'm stuck. :D (I very vaguely remember that I've seen Michael Penn evaluate a similar sum some months ago, but I don't remember either when nor how he did do it.)
Hm, the sum of sin(k x)/k^2 does not seem to have a closed-form solution (if you don't admit poly logarithm function family), not sure if specifically for x = pi/3 it does...
@@richardheiville937 Thanks, I just found out for myself that the result is -2 times the Clausen function of pi/3. And apparently one can't express that in closed form, using other known constants like pi or Catalan's constant. :(
@@richardheiville937 actually, that's a good point, Michael Penn did a video on Clausen2 function recently. It can be expressed via an integral or a sum, that's it.
You could have expanded the intergrand ln(1-sinx) as a power series in sinx and used tricks to evaluate that. I think that would have been easier.
Indeed
10:40 most important explanation of the video, for those wondering.
Finally someone adds a timestamp
Can you make a video where you demonstrate when and when not use interversion of limits sum and integrals among each other , that will really be helpful
I definitely agree with this one. 👍💯💯
If the structure converges you can switch up limits of operators involved. If it doesn't, the switch up is invalid.
@@maths_505 Some counterexamples would be really helpful for new viewers, though.
Aight bro
Thank you for your faithful effort. At 14:49 the index of summation shall be in terms of n but not k.
Oh yes indeed
Thanks for pointing that out doctor
As a mathematician I feel these advanced calculus questions are not taught in Indian Education syllabus till now It should be taught in MSc or PhD in foreign universities😎😎
Not it shouldn't
I can't understand ur comment🤔🤔
Nice david goggins reference with the logs lmao
They really don't me son😂
The integral that took Paul J. Nahin 5 hours to evaluate
Only reason I did it faster was cuz I was listening to eminem while penning the solution development 😂
@@maths_505 took me 20 or so minutes as well. It was 4-5 am in the morning I just couldn't sleep that night I was solving integral after integral that night.
Integral bros😎
How you just instinctively knew that a u-sub of x-2=2sin(u) would make everything cancel out makes me feel wholly inadequate. What made you think of a trig-identity there?
Bro why would you feel that way??
C'mon don't be hard on yourself I've done about 300 videos and I've been solving integrals for years just for the fun of it. Just keep grinding, you'll develop an intuitive sense with time.
Think of it as hitting the gym.... initially you're gonna mess up most exercises with bad form and be clueless about volume but later you develop an intuitive sense of body building, get better at technique (watching lots of Jeff Nipard and mike isratel 😂) and you figure out how much volume and rest is optimal for you.
It's not obvious to me why you choose a trig function for the substitution.
14:48 There's a tiny glitch in the solution. Your sum should begin with k>=0 not k>=1. And to be more correct, n>=0. Thanks
From Bangladesh 🇧🇩
Areehhh... Tumi eikhane!
how u solve integral(lncosudu)
You mean from zero to pi/2?
That's a pretty common result. I made a video on it a long while back. I think there's a post on my Instagram that evaluates it too
Hi good work can i solve them using resdue cauchy theorem❤
It's hard to do so because of the limits.
@@maths_505 thanks
Andrew Tate has been preaching at Cambridge since 1886.
Nd goggins carried the logs that made the buildings
are you really a Ronaldo fan or was that a nice joke that fit the situtation?
SUIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
x=2+2sinA....I=0, boh a me risulta 0.. Magari rifaccio i calcoli domani e troverò l'errore
Right on for the first integral
@@maths_505 are you sure?
Ok...😂😂
I(0to2)=-2G=-1,82193...
your 'u' and 'n' are indistinguishable. very annoying.
Muy mala explicación y mala letra.