There is a mistake at 8:14. It should be (pi/8)ln2+the integral from 0 to pi/4 of ln(tanx)dx, not minus that integral. (Notice on the previous board it is written correctly.) Fortunately the sign mistake is fixed at 12:30. That homework integral evaluates to -1/((2n+1)^2). However, Michael does not include this negative sign, so the two sign mistakes cancel, and the answer at the end is correct.
I didn’t notice that mistake at 8:14, so when I was doing the hw integral I was checking and checking if there was a minus sign I omitted, but fortunately I checked your comment in time before spending half an hour checking again.
I love when transcendental numbers/functions show up; there's always something so satisfying about breaking math, expanding your definitions, and repeating. Now you've got me going down the rabbit hole on Catalan's constant, thanks prof!! Exactly the kind of stuff I've been waiting for (whether in calc 2 integral videos or deeper abstract algebra applications) and it made my day. Can't wait to see some more! Have a great day:)
Pretty sure we don't know if G is transcendental. Regardless, you can make up so many definite integrals and the answer is more likely to be transcendental than not. It's only numbers like G and pi that are actually have useful applications outside of just being the value of an integral, which is why we label them.
omg I was stuck at evaluating integral of ln(tanx) from 0 to pi/4, never thought introducing infinite series expansion would help simplify the problem! that's awesome
log(1-x)/(1+x^2)= log((1-x)/(1+x))/(1+x^2)+log(1+x)/(1+x^2) first and second integrals can be evaluated using the change of variable y=(1-x)/(1+x) for the first one, one obtains \int_0^1 \frac{ln x}{1+x^2}=-G (Catalan constant) for the second one one gets \int_0^1 \frac{2/(1+x)}{1+x^2}dx=-log(2) times the integral, Thus its value is log(2)Pi/8
Yeah I noticed that too. Its almost like we split that ln(1-x) in the numerator to get (π/8)•ln(2) - int(0->1){ln(x)/(x²+1)}. Could we maybe get an idea from here about how to formulate ln(1-x), and further, even ln(a-b) while we're at it? Really interesting!
- "Can you guess the trick ?" - Watch video - Actualy, there's no trick, it needs 4 change of variable and 1 integration by parts, the exercise is incredibly difficult.
Found an alternative way to evaluate this with contour integration. First, do a substitution x = 1/y to transform this into an integral from 1 to infinity and split it up into two integrals using logarithm rules. Then, the first one can be done with contour integration and the second one is Catalan's constant.
Another nice way to do it involves a complementary integral. Set I= integral in the problem and J= int(0,1) ln(1+x)/(1+x^2) dx. Then solve I-J using t=(1-x)/(1+x) gives a famous integral representation for G which can be proved using a geometric series expansion. Additionally J=πln2/8 (Serret’s integral which can be proven in a number of ways including using the same t substitution). So then the value of I can be extracted.
expand the logarithm and the fraction 1/(1+x^2) as power series, since the integration interval allows it; then use Cauchy "convolution" formula to multiply both series; what ya think?
Before I even watched the video, I sketched a few points of the curve, and saw that a triangle defined by the points (0,0), (1,0), and (1,-1) looks like it contains a large proportion of the area between the curve and the x-axis. This triangle has an easily calculated area of 0.5. I then hoped that the area below that and between the curve and the vertical asymptote at x=1 converged, and estimated that additional area to be 0.15. Thus the value of the integral would be approximately -0.65, which turned out to be only off by 1% from the exact value (-0.6437673329...)
It says to guess, so I haven't watched the video yet, but I'm guessing it's an arctanh(x) kinda thing because of the sum of two squares in the denominator. Now I'll watch the video to see if I'm right! Edit: I was not right. I was thinking of 1/(1-x^2). I was also thinking of the differential equation of drag where you get t=tanh[v(t)]+c for the integral and solve for v(t) at the end. So I'm continuing my incredible streak of saying something dumb in every comment I've ever made on this channel.
@@leif1075 it's not TOO dumb, but I knew better. The issue is that in terms of differential equations exponential solutions go with real numbers, and sinusoidal solutions go with imaginary numbers. I.E. it's confusing e^(ix) with e^x. Of COURSE the sum of two squares is the sinusoidal one because it has complex factors. It was silly.
@@jeremyredd4232 that's not dumb at all though if you think about it..and indidnt see anything complex about this solution didn't he use tangent of x and not hyperbolic tangent anyway..so why is there anything complex involved?
Indeed. I just skimmed the video knowing how the solution would go and got a bit concerned as to why it's taking so long until I saw Catalan's constant pop out at the end.
10:21 I don't understand how the dominated convergence theorem is used. do we go, like, every other partial sum and take our function ( ln(x)/(1+x^2) ) as a bound? I'm confused because I do not see easily any integrable function (the hard part is that the integral is finite). Also, this sequence would be monotonic, and this I can see working. Another option would be ln(x)/(1-x^2), and I do not see immediately why it is integrable (the Lebesgue meaning of integrable, has finite integral). It has a constant sign I guess, but I don't know if it helps. I'm confusion, help.
I don't understand also, but you can use the Monotone Convergence Theorem instead on the partial sum from 0 to 2m-1. (it works because x is between 0 and 1)
I couldn't find a way to do it with Feynman's technique. Either everything cancels out or I get a differential equation with a series involving the lower incomplete gamma function.
Arctan doesn’t have the same nice properties as tangent; tangent^2 has the trig identity with sec^2, which is the same as its derivative. This allows the cancellation of the two in this integral, thus simplifying. Unfortunately, arctan doesn’t have those same properties, so it’s less useful for this integral. Great question!
Pretty much same question came in fiitjee aits (jee test series) only difference was there was ln(1+x) and they substituted x=1-u/1+u in the solution, i think similiar approach can be used here to get a simpler solution
@@pablosarrosanchez460 I don't know exactly, but that's a Mobius transformation. It maps quadratic functions to quotients of quadratic functions, so maybe when you have an integrand of the form f(linear or quadratic)/quadratic, you can construct a Mobius transformation that gives you the original integral + something (hopefully) simpler? It's also useful in complex analysis to transform contour integrals (maybe for a related reason it works here, idk).
The integral of ln(trig(x))dx appears frequently on this channel; this video included. Interestingly, Catalan's constant is equal to one of these integrals. The integral from 0 to pi/4 of ln(cotx)dx is Catalan's constant. Furthermore it is the negation of the ln(tanx) integral we saw in this video.
"How many u-substitutions do you need to solve the integral?"
"Yes."
Thank you Michael ! Evaluate integrals are my favorite series !
There is a mistake at 8:14. It should be (pi/8)ln2+the integral from 0 to pi/4 of ln(tanx)dx, not minus that integral. (Notice on the previous board it is written correctly.) Fortunately the sign mistake is fixed at 12:30. That homework integral evaluates to -1/((2n+1)^2). However, Michael does not include this negative sign, so the two sign mistakes cancel, and the answer at the end is correct.
I didn’t notice that mistake at 8:14, so when I was doing the hw integral I was checking and checking if there was a minus sign I omitted, but fortunately I checked your comment in time before spending half an hour checking again.
I love when transcendental numbers/functions show up; there's always something so satisfying about breaking math, expanding your definitions, and repeating. Now you've got me going down the rabbit hole on Catalan's constant, thanks prof!! Exactly the kind of stuff I've been waiting for (whether in calc 2 integral videos or deeper abstract algebra applications) and it made my day. Can't wait to see some more! Have a great day:)
Pretty sure we don't know if G is transcendental. Regardless, you can make up so many definite integrals and the answer is more likely to be transcendental than not. It's only numbers like G and pi that are actually have useful applications outside of just being the value of an integral, which is why we label them.
I looked up Catalan's constant and was surprised to learn that it's not even known if it's irrational or not
i had a proof of it's rationality but i didn't have enough paper to write it on
@@BenDover69831 Why do I have this feeling of dejavu
@@BenDover69831 My dog ate my proof.
its surprising if it was irrational, its a sum of rational numbers
@@Walczyk So is e, but that's irrational
omg I was stuck at evaluating integral of ln(tanx) from 0 to pi/4, never thought introducing infinite series expansion would help simplify the problem! that's awesome
log(1-x)/(1+x^2)= log((1-x)/(1+x))/(1+x^2)+log(1+x)/(1+x^2) first and second integrals can be evaluated using the change of variable y=(1-x)/(1+x) for the first one, one obtains \int_0^1 \frac{ln x}{1+x^2}=-G (Catalan constant) for the second one one gets \int_0^1 \frac{2/(1+x)}{1+x^2}dx=-log(2) times the integral, Thus its value is log(2)Pi/8
Nice y=(1-x)/(1+x) is a very helpful substitution.
@@pauldifolco5736 For sure but it's the subtitution linked to u=\tan(pi/4-x) in trigonometry.
Loved this!
Thank you, professor.
This also means that (π/8)ln2 = integral from 0 to 1 of {ln(x(1-x))/(x^2+1)}dx.
Yeah I noticed that too. Its almost like we split that ln(1-x) in the numerator to get (π/8)•ln(2) - int(0->1){ln(x)/(x²+1)}. Could we maybe get an idea from here about how to formulate ln(1-x), and further, even ln(a-b) while we're at it? Really interesting!
- "Can you guess the trick ?"
- Watch video
- Actualy, there's no trick, it needs 4 change of variable and 1 integration by parts, the exercise is incredibly difficult.
Why not just use integration by parts in the last step since x2n can just be derivstive of x to the 2n plus 1 you don't need the y substitution..
Found an alternative way to evaluate this with contour integration. First, do a substitution x = 1/y to transform this into an integral from 1 to infinity and split it up into two integrals using logarithm rules. Then, the first one can be done with contour integration and the second one is Catalan's constant.
Another nice way to do it involves a complementary integral. Set I= integral in the problem and J= int(0,1) ln(1+x)/(1+x^2) dx. Then solve I-J using t=(1-x)/(1+x) gives a famous integral representation for G which can be proved using a geometric series expansion. Additionally J=πln2/8 (Serret’s integral which can be proven in a number of ways including using the same t substitution). So then the value of I can be extracted.
13:19
expand the logarithm and the fraction 1/(1+x^2) as power series, since the integration interval allows it; then use Cauchy "convolution" formula to multiply both series; what ya think?
Before I even watched the video, I sketched a few points of the curve, and saw that a triangle defined by the points (0,0), (1,0), and (1,-1) looks like it contains a large proportion of the area between the curve and the x-axis. This triangle has an easily calculated area of 0.5. I then hoped that the area below that and between the curve and the vertical asymptote at x=1 converged, and estimated that additional area to be 0.15. Thus the value of the integral would be approximately -0.65, which turned out to be only off by 1% from the exact value (-0.6437673329...)
It says to guess, so I haven't watched the video yet, but I'm guessing it's an arctanh(x) kinda thing because of the sum of two squares in the denominator. Now I'll watch the video to see if I'm right!
Edit:
I was not right. I was thinking of 1/(1-x^2). I was also thinking of the differential equation of drag where you get t=tanh[v(t)]+c for the integral and solve for v(t) at the end. So I'm continuing my incredible streak of saying something dumb in every comment I've ever made on this channel.
How is that dumb it's tan x you were close tanh x Is almost tanx
@@leif1075 it's not TOO dumb, but I knew better. The issue is that in terms of differential equations exponential solutions go with real numbers, and sinusoidal solutions go with imaginary numbers. I.E. it's confusing e^(ix) with e^x. Of COURSE the sum of two squares is the sinusoidal one because it has complex factors. It was silly.
@@jeremyredd4232 that's not dumb at all though if you think about it..and indidnt see anything complex about this solution didn't he use tangent of x and not hyperbolic tangent anyway..so why is there anything complex involved?
@@leif1075 1+x^2=(1+ix)(1-ix). 1-x^2=(1+x)(1-x). Tan(x) goes with the complex one and tanh(x) goes with the real one.
@@leif1075 it's also worth noting that tan(x)=i tanh(ix).
Yes is the answer to the question posed.
This is almost the exact same as that one Putnam integral
Indeed. I just skimmed the video knowing how the solution would go and got a bit concerned as to why it's taking so long until I saw Catalan's constant pop out at the end.
I think that it wouldn't hurt to say about transit from 1/(x^2+1) to geometric series that it converges when x
Furthermore the integral is taken from 0 to 1, so there will not be a convergence issue.
Wouldnt it be easier to just expand the natrul log directly? OH Wait there is a convergence issue we got around by doing it this way
SUBSTITUTION IS KING OF INTEGRAL KINGDOM
What’s the website where you get those kind of integral ?
10:21 I don't understand how the dominated convergence theorem is used. do we go, like, every other partial sum and take our function ( ln(x)/(1+x^2) ) as a bound? I'm confused because I do not see easily any integrable function (the hard part is that the integral is finite). Also, this sequence would be monotonic, and this I can see working.
Another option would be ln(x)/(1-x^2), and I do not see immediately why it is integrable (the Lebesgue meaning of integrable, has finite integral). It has a constant sign I guess, but I don't know if it helps.
I'm confusion, help.
I don't understand also, but you can use the Monotone Convergence Theorem instead on the partial sum from 0 to 2m-1. (it works because x is between 0 and 1)
You can integrate by parts ln(x)/(1+x^2) and then integrate by power series
I couldn't find a way to do it with Feynman's technique. Either everything cancels out or I get a differential equation with a series involving the lower incomplete gamma function.
that last bit sounds like fun lol
Can you send me your work, genuinely interested.
Is it improper integral at x=1????
I think it is 1st or 2nd
This is calculus 1 or 2 or 3?
The dot above made me think the upper limit was i.
Approx -0.644
Got the answer but I took x = -tany
But both are same.
This quesn was there in my textbook
Can we perform
asnwer=1 but mlddle isit
Could you use an arctan substitution rather than tangent? Just curious
Arctan doesn’t have the same nice properties as tangent; tangent^2 has the trig identity with sec^2, which is the same as its derivative. This allows the cancellation of the two in this integral, thus simplifying. Unfortunately, arctan doesn’t have those same properties, so it’s less useful for this integral. Great question!
Pretty much same question came in fiitjee aits (jee test series) only difference was there was ln(1+x) and they substituted x=1-u/1+u in the solution, i think similiar approach can be used here to get a simpler solution
Every where is indian
hoy does that substitution work? in which cases is it useful?
@@pablosarrosanchez460 ye a teacher at my school said he spent an hour and a half looking at that integral to spot this substitution
This doesn't have an elementary answer no matter how you solve it
@@pablosarrosanchez460 I don't know exactly, but that's a Mobius transformation. It maps quadratic functions to quotients of quadratic functions, so maybe when you have an integrand of the form f(linear or quadratic)/quadratic, you can construct a Mobius transformation that gives you the original integral + something (hopefully) simpler? It's also useful in complex analysis to transform contour integrals (maybe for a related reason it works here, idk).
Next time please put more space between the lines-> 0:01 you are writing it like the integral rums from 0 to i.
Show! Maicão Caneta!
Good
You gotta French up the pronunciation of Catalan.
No, I can not guess that
kinda easy intergral ngl, trig substitution is immediatly obvs.
Question appeared in jee test series
No it didnt
With 1+x
i solved it in two minutes
Good for you
Really? Why don't you post your two minute solution, then?
And you should not have any likes for your comment.
@@forcelifeforce im joking man dont be angry
The integral of ln(trig(x))dx appears frequently on this channel; this video included. Interestingly, Catalan's constant is equal to one of these integrals. The integral from 0 to pi/4 of ln(cotx)dx is Catalan's constant. Furthermore it is the negation of the ln(tanx) integral we saw in this video.
Yes indeed, I also solved this integral by two methods