btw fun fact inverse hyperbolic functions are actually "ar-" (artanh), not "arc-" like the regular functions since the former stands for "area" and the latter for "arcus"
Darn! You beat me to it. Every time he (it's happened more than once) does this, I make sure somebody tells him about this. It's just so irritating to me, for some reason (I know it's not rational), when people write arctanh instead of artanh. The stupidest part about all of this is that the Edge spellchecker likes arctanh but doesn't like artanh, meaning that even the people who made Microsoft's spellchecker got it wrong. Come on! I thought everyone knew about this. I mean, what arc would we be talking about. There are no arcs involved in the hyperbolic trig functions. The hyperbolic functions are defined using area under the curve and there's no arc anywhere to be found (with an arc being defined as a segment of a circle in my book.).
I originally solved it by expressing as an infinite series but by far the quickest way to evaluate it is using integration by parts. Let u=artanh(x^2) and v=(x-1)/x then the resulting integral is of a simple rational function
That's a clever usage of the integration constant in *v* - if we didn't choose *C = 1* in that case we would still be left with expressions tending to infinity at *x -> 1* ! -------------------------------------------------------- It's possible (but more technical) to do integration by parts without the clever integration constant. Notice the given integral is an improper integral to begin with as the *\artgh(..)* term diverges at *x -> 1* : *I = \lim_{b -> 1^-} \int_0^b x^{-2} \artgh{x^2} dx* Now we can do integration by parts with the more intuitive *v = -1/x* . The final limit *b -> 1^ -* will have two log-terms that both tend to infinity individually, but cancelling each other out.
In my attempt of the problem before watching the video I used the logarithmic form of arctanh(x^2), split it apart, and used the series representation of the logarithms. I then integrated, used partial fraction decomposition, and evaluated the series to get the answer.
You can also do this problem by setting I(t) = int(arctanh(x^2t)/x^2) from 0 to 1 The x^2 cancels out, and you should be able to integrate resulting definite integrals twice to get the desired result.
I think this is a lot easier if you split the log fraction in two and notice that it looks like a zeroth order integral, turn it into a double integral, and switch the order
I have absolutely no idea how you did the integration by parts at the end: the bottom part of the uv term has disappeared, the top part of the uv term equals 0 as far as I can tell, and the final integral term seems to diverge, so I've no idea how you get ln(2)
Since the integral is between 0 and 1, the series expansion of arctan(y) = y + (y^3)/3 + (y^5)/5-... is applicable. The integrand becomes: 1 + (x^4)/3 + (x^8)/5 .. +(x)^(4n)/(2n+1). The integral yields 1+1/(3.5)+1/(5*9)+1/(7*13)....+1/(2n+1)(4n+1)+.... which converges. Wolfram "sum from 0 to infinity 1/((4*n+1)*(2*n+1)" gives same result.
Substitution x=1/t. We get ∫ (from 1 to ∞) (1/2)*ln[(t^2+1)/(t^2-1)]dt= ∫ (from 1 to ∞) (1/2)*[ln(t^2+1)- ln(t^2-1)]dt . We integrate it in parts. To get rid of the divergences, one should calculate lim (a→1,b→∞){∫ (from a to b) ...dt} and , sometimes. sum up the problematic terms to the limit transition.
Feynman’s technique all the way. Or Leibniz Rule of Taking derivatives under the integral sign all the way. I love that really complicated integrals just have pi or e as an answer sometimes.
Indefinite integral can be easily calculated by parts And integration by parts approach is understandable also for those who begin with integral calculus F(x)=-1/x artanh(x^2) + Int(1/x * 2x/(1-x^4),x) F(x) = -1/x artanh(x^2) + 2Int(1/(1-x^4),x) F(x) = -1/x artanh(x^2) + Int(((1-x^2)+(1-x^2))/(1-x^4),x) F(x) = -1/x artanh(x^2) + Int(1/(1-x^2),x) + Int(1/(1+x^2),x) F(x) = -1/x artanh(x^2) + artanh(x) + arctan(x) + C limit(-1/x artanh(x^2) + artanh(x) + arctan(x) ,x = 1,left) - limit(-1/x artanh(x^2) + artanh(x) + arctan(x) ,x = 0,right)
When I saw the problem, what came to my mind is integration by parts. If we consider to differentiate arctanh(x^2) and integrate 1/x^2, we get -arctanh(x^2)/x + the integral of 2x/1-x^4 * 1/x which simplifies to 2/1-x^4 which is a simple partial fraction problem. Note: we can use d/dx arctanh(x) = 1/1-x^2 since the bounds are within -1 and 1.
Wait I tried my method and i seems to fail at the partial fraction part. We arrive at a point where we need to integrate 1/1-x from 0 to 1 which diverges. Now here is my question, if our integral at hand converges but this integration by parts method diverges, where is the point I have gone wrong?
arctanh has a great Maclaurin expansion: x + x^3/3 + x^5/5 ... so I did it like that. Ultimately got it down to the infinite sum from k=1 of 2/(4k-3) - 1/(2k-1), which would be easy if I was good at summation. Seems to be correct
It feels like we’re working towards instituting Integration Sundays. That’s fine by me. Also, I want more using Feynman’s technique. Thank you, professor.
My idea was this. After substitution z=1/x we have integral argtgh(1/z^2) on(1; Infinity) We can use Lebesgue integral then. We have integral sqrt(1/tgh y) on(0; Infinity) After substitution sqrt( tgh y)=r we have integral 2/(1-r^4) dr on (0;1) Then we use partial......,we have the same result of origin integral.
The calculation of f(1) went completely off-road :-P. Your anti-derivative of 1 should be y-1 instead of just y, otherwise the final integral does not converge (you missed the boundary at y=1 which does not exist). Probably substituting y=1/u and using the series expansion is the easiest way though.
Just a minor comment but it is not "Feynman's integration technique" as the method you used is not formally or conventionally attributed to him as and nor is it it a mathematical discovery of his, even if he did employ a method for solving particular indefinite integrals
"Feynman's integration technique" is surely A name for it; I knew what he planned to do just from the name, likely you as well. English is not prescriptive you know :)
@@bsmith6276 Only in US it is called Feynman's integration technique because even in other English spoken countries like Australia is called after its inventor With L'Hopital rule is different thing No one name it after Bernoulli
Yeah i have no idea what any of this means, and I have absolutely no idea how maths is something that people agree on, or how it is the ‘languages of languages’ as Joshua Bach says
btw fun fact inverse hyperbolic functions are actually "ar-" (artanh), not "arc-" like the regular functions since the former stands for "area" and the latter for "arcus"
Or argth with arg like argument.
Darn! You beat me to it. Every time he (it's happened more than once) does this, I make sure somebody tells him about this. It's just so irritating to me, for some reason (I know it's not rational), when people write arctanh instead of artanh. The stupidest part about all of this is that the Edge spellchecker likes arctanh but doesn't like artanh, meaning that even the people who made Microsoft's spellchecker got it wrong. Come on! I thought everyone knew about this. I mean, what arc would we be talking about. There are no arcs involved in the hyperbolic trig functions. The hyperbolic functions are defined using area under the curve and there's no arc anywhere to be found (with an arc being defined as a segment of a circle in my book.).
artanh reminds me of Ar Di (our Lady Diana) don't ask
I originally solved it by expressing as an infinite series but by far the quickest way to evaluate it is using integration by parts. Let u=artanh(x^2) and v=(x-1)/x then the resulting integral is of a simple rational function
That's a clever usage of the integration constant in *v* - if we didn't choose *C = 1* in that case we would still be left with expressions tending to infinity at *x -> 1* !
--------------------------------------------------------
It's possible (but more technical) to do integration by parts without the clever integration constant. Notice the given integral is an improper integral to begin with as the *\artgh(..)* term diverges at *x -> 1* :
*I = \lim_{b -> 1^-} \int_0^b x^{-2} \artgh{x^2} dx*
Now we can do integration by parts with the more intuitive *v = -1/x* . The final limit *b -> 1^ -* will have two log-terms that both tend to infinity individually, but cancelling each other out.
In my attempt of the problem before watching the video I used the logarithmic form of arctanh(x^2), split it apart, and used the series representation of the logarithms. I then integrated, used partial fraction decomposition, and evaluated the series to get the answer.
16:10
You can also do this problem by setting
I(t) = int(arctanh(x^2t)/x^2) from 0 to 1
The x^2 cancels out, and you should be able to integrate resulting definite integrals twice to get the desired result.
I think this is a lot easier if you split the log fraction in two and notice that it looks like a zeroth order integral, turn it into a double integral, and switch the order
14:47 you missed a y in the denominator
8.20 u missed dv after writing
du=-1/v^2
I have absolutely no idea how you did the integration by parts at the end: the bottom part of the uv term has disappeared, the top part of the uv term equals 0 as far as I can tell, and the final integral term seems to diverge, so I've no idea how you get ln(2)
Prof you're buffed
Michael knows how to calculate quite easy integral in complicated way
Since the integral is between 0 and 1, the series expansion of arctan(y) = y + (y^3)/3 + (y^5)/5-... is applicable.
The integrand becomes: 1 + (x^4)/3 + (x^8)/5 .. +(x)^(4n)/(2n+1).
The integral yields 1+1/(3.5)+1/(5*9)+1/(7*13)....+1/(2n+1)(4n+1)+.... which converges.
Wolfram "sum from 0 to infinity 1/((4*n+1)*(2*n+1)" gives same result.
@15:38 The integral needs to have a -1/2 in front, not 1/2.
Substitution x=1/t. We get
∫ (from 1 to ∞) (1/2)*ln[(t^2+1)/(t^2-1)]dt= ∫ (from 1 to ∞) (1/2)*[ln(t^2+1)- ln(t^2-1)]dt .
We integrate it in parts.
To get rid of the divergences, one should calculate lim (a→1,b→∞){∫ (from a to b) ...dt} and , sometimes. sum up the problematic terms to the limit transition.
Feynman’s technique all the way. Or Leibniz Rule of Taking derivatives under the integral sign all the way.
I love that really complicated integrals just have pi or e as an answer sometimes.
Indefinite integral can be easily calculated by parts
And integration by parts approach is understandable also for those who begin with integral calculus
F(x)=-1/x artanh(x^2) + Int(1/x * 2x/(1-x^4),x)
F(x) = -1/x artanh(x^2) + 2Int(1/(1-x^4),x)
F(x) = -1/x artanh(x^2) + Int(((1-x^2)+(1-x^2))/(1-x^4),x)
F(x) = -1/x artanh(x^2) + Int(1/(1-x^2),x) + Int(1/(1+x^2),x)
F(x) = -1/x artanh(x^2) + artanh(x) + arctan(x) + C
limit(-1/x artanh(x^2) + artanh(x) + arctan(x) ,x = 1,left) - limit(-1/x artanh(x^2) + artanh(x) + arctan(x) ,x = 0,right)
When I saw the problem, what came to my mind is integration by parts. If we consider to differentiate arctanh(x^2) and integrate 1/x^2, we get -arctanh(x^2)/x + the integral of 2x/1-x^4 * 1/x which simplifies to 2/1-x^4 which is a simple partial fraction problem. Note: we can use d/dx arctanh(x) = 1/1-x^2 since the bounds are within -1 and 1.
Υep, less uselessly convuluted than my idea
Wait I tried my method and i seems to fail at the partial fraction part. We arrive at a point where we need to integrate 1/1-x from 0 to 1 which diverges. Now here is my question, if our integral at hand converges but this integration by parts method diverges, where is the point I have gone wrong?
@@SartajKhan-jg3nz -arctanh(x²)/x also diverges, but in the other direction; you need to take a limit.
@@talinuva yes but using L'hopital i think we get that it approaches o as x tends to 0
@@SartajKhan-jg3nzTake limits of the whole expression. The divergent parts cancel.
arctanh has a great Maclaurin expansion: x + x^3/3 + x^5/5 ... so I did it like that. Ultimately got it down to the infinite sum from k=1 of 2/(4k-3) - 1/(2k-1), which would be easy if I was good at summation. Seems to be correct
You could use the digamma function
It feels like we’re working towards instituting Integration Sundays. That’s fine by me.
Also, I want more using Feynman’s technique.
Thank you, professor.
My wife thinks I am crazy because I watch Prof. Penn's channel because it is fun (and educational). Lifetime learning.
@@MathHammer heheh, same here.
You could have used 1/x^2+a^2 integral = 1/a arctan(x/a)
A very elegant video if I do say so myself
Integrating by parts is much easier, it will give 1/(1-x^4) under the 2nd integral, which is not hard to do...
Does anybody other than the integral suggestor get their suggestions for problems turned into videos?
My idea was this.
After substitution z=1/x we have integral argtgh(1/z^2) on(1; Infinity)
We can use Lebesgue integral then.
We have integral sqrt(1/tgh y) on(0; Infinity)
After substitution sqrt( tgh y)=r we have integral 2/(1-r^4) dr on (0;1)
Then we use partial......,we have the same result of origin integral.
There is also a mistake at 15.35
It should be 1/(y-1)-1/(y+1) and not 1/(y+1)-1/(y-1)
Eso altera el resultado de la integral. Su valor es pi/4.
"nice"
The calculation of f(1) went completely off-road :-P. Your anti-derivative of 1 should be y-1 instead of just y, otherwise the final integral does not converge (you missed the boundary at y=1 which does not exist). Probably substituting y=1/u and using the series expansion is the easiest way though.
asnwer= 1 isit mlddle isit hmm
This requires that Z isn’t equal to 1 so that we can divide by Z-1
So how can we evaluate f(Z) at Z=1 later in the video
Differentian Under the Integral Sign is more accurate btw, and is accredited to one of the co-discoverers of Calculus... I think you know the one :)
Just a minor comment but it is not "Feynman's integration technique" as the method you used is not formally or conventionally attributed to him as and nor is it it a mathematical discovery of his, even if he did employ a method for solving particular indefinite integrals
"Feynman's integration technique" is surely A name for it; I knew what he planned to do just from the name, likely you as well. English is not prescriptive you know :)
"Feynman's integration technique" is like "L'Hopital's Rule". Neither one was created by their namesake but the names still persist.
@@bsmith6276 historically speaking, the latter has been established for centuries now and is now a convention
@@bsmith6276 Only in US it is called Feynman's integration technique because even in other English spoken countries like Australia is called after its inventor
With L'Hopital rule is different thing No one name it after Bernoulli
Yeah i have no idea what any of this means, and I have absolutely no idea how maths is something that people agree on, or how it is the ‘languages of languages’ as Joshua Bach says
I think you have a red sticker on shirt lol
It's actually from the laser site of the terrorist who was forcing him to do this
@@jackw7714 at first I thought it was a laser site 😂
i think its part of the shirt design