an A5 Putnam Exam integral for calc 2 students
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- čas přidán 4. 01. 2018
- Integral of ln(x+1)/(x^2+1) from 0 to 1, this integral was on the Putnam exam back in 2005. We will use a trig substitution and some integral properties to take care of this problem!
This integral is also called the Serret's Integral. Subscribe to @blackpenredpen for more fun math videos.
Check out Mu Prime's video on the integral of ln(x^2+1)/(x+1) from 0 to 1, • It took me 3 hours to ...
Warning!!! For your mental safety do not try integration by parts
Luckily this is not in A level Mathematics. Integration is easy in A level, the hardest it probably gets is repeated integration by parts
By parts it is not hard. Try integrate by parts twice and you get a syllogism 0=0.
I used and got the answer
yup it doesn't give you an answer in terms of known functions, no matter which way you take it. You part of the final answer on the first by parts, which gives an interesting result for another integral, but you wouldn't know that without doing the way he did it in the video.
TheJampt can you please explain how it’s done by parts cause I get massively stuck when I have two of the same integral and can’t just add it to both sides or smthn
Historical note: this integral was first discussed by Bertrand in 1843 in the Journal de mathématiques pures et appliquées. Bertrand used the technique which is now often incorrectly attributed to Feynman by American sources. In the next issue of the same journal Serret solved the integral in a single line, which is why it is also sometimes referred to as Serret's integral.
Bertrand believed his technique was new, but in fact Euler had already discussed introducing an additional variable and differentiating under the integral sign in his 1775 paper Nova methodus quantitates integrales determinandi (E464 in the Eneström index). The title of his article indicates that Euler also believed the technique was new when he wrote his article. But Leibniz already used the technique in an appendix of a letter to Johann Bernoulli dated August 3rd 1697, so it is older than Euler and indeed it is therefore also known as the Leibniz integral rule. Clearly the technique dates back to the very beginnings of calculus as we know it, so there is no justification whatsoever to attribute this technique to Feynman. For the record, Feynman never claimed that he discovered the rule. He learned it from an old calculus text that his high school physics teacher had given him.
*References* (replace every + with .)
portail+mathdoc+fr/JMPA/PDF/JMPA_1843_1_8_A7_0.pdf
portail+mathdoc+fr/JMPA/PDF/JMPA_1844_1_9_A41_0.pdf
scholarlycommons+pacific+edu/cgi/viewcontent+cgi?article=1463&context=euler-works
archive+org/details/leibnizensmathe00leibgoog/page/n458/mode/1up?view=theater
hsm+stackexchange+com/questions/8132/why-is-differentiation-under-the-integral-sign-named-the-leibniz-rule
en+wikipedia+org/wiki/Leibniz_integral_rule
4:02 It can be solved really easily if at this moment you apply King's Property and get Integration as ln( tan(π/4 - x) +1 ) which is equal to ln ( (1 - tanx)/(1+tanx) +1 ) = ln( 2/(1 + tanx) ) so we get I = Integral of ln(2) from 0 to π/4 - I hence 2I = ln(2)π/4
Yeah bro I also think this method 👍👍👍👍
Yes property 4 yeah bro true
literally a standard integral atp pi/8 log 2
I'm not in college, I rarely have to use integrals, yet I make sure not to miss a single episode of this channel. Excellent brain gymnastics. And if I'm ever stuck in a deep dark forest and need to integrate some obscure function it will become helpful! :)
Thats so clever, I tried finding an antiderivative but is just way too hard and not feasible if you are not a machine. sometimes I forget definite integrals are just like calculating areas and you can clearly see that the area of the cos(x) function and the cos(x-pi/4) is the same on the interval [0,pi/4]. thanks for the video!
My pleasure!
It works in general too! When integrating f(x) from a to b we can also integrate f(a+b-x) from a to b to get the same answer. It's kind of like calculating the area from the other side :P
The antiderivative can be solved with the de Jonquiere's function, also called the polylogarithm:
Li_s(x) = sum_{n=1..inf} (x^n)/(n^s)
which is simple generalization of the logarithmic Mercator series:
-ln(1 - x) = sum_{n=1...inf} (x^n)/n.
In particular, one should note that int Li_s(x)/x dx = Li_(s+1)(x) + C.
One should become familiar with that there are a lot more different mathematical functions in the toolchest than just sin, cos, tan, exp, and log, especially now with computers and math libraries to calculate them readily at hand. I would suggest BlackPenRedPen illustrate this. sin/cos/tan/exp/log works fine for Intro Calc 1 and 2 but when you get on up to stuff like Complex Analysis or out into the real world, you'll want to have a lot more in your bag with you.
Learn integration then bruh...
@@blackpenredpen m6 0l3àß743¡
It's m6¡
Flam's parametric method is very elegant.
BPRP trig substitution was clever and inspired.
I just solved it by brute force, but got some interesting results worth sharing...
First I replaced the whole integrand by its series expansion
Log[x+1]/(1+x^2)=
Sum[a_n x^n,{n,0,inf}]
Then the integral is just
Sum[a_n/(n+1) x^(n+1),{n,0,inf}]
Common sense indicates that the series should be centered in 0 or 1, so half of the terms disappear, but the fastest convergent solution was with the series centered in 1/2, as the coefficients get divided by powers of 2.
For the series centered in 0 (McLaurin) the convergence is extremely slow, as the coefficients converge to Log(2)/2 and pi/4 alternating signs. Curiously, the solution is the product of these 2 convergence points.
Of course, all these trials only gave a numerical approximation.
Then I tried to replace just the log for its series expansion, and integrate each term by doing long division. The integrals separate in 2 groups: log(2) and ArcTan(1), each one with a finite number of additional terms making a triangle. The funny thing is that both triangles complement each other, giving the series expansion for log(2)/2 and ArcTan(1), cancelling each other and giving the analytical solution!
This integral and the other Log integral by symmetry are amazing. You never fail to blow my mind BPRP!
sebmata thank you!!!!!
The whole solution is nothing short of miraculous, but once you wrote down the integral of log(cos(theta-pi/4)), the rest was completely trivial and waste of time. I mean, the limits are from zero to pi/4. You need no further explanation to cancel the log cos beasts.
Here is one more fantastic solution
czcams.com/video/u3HtCHjjSJs/video.html
This was an absolutely gorgeous integral. I got up to splitting it into the 3 separate integrals but I didn't think of working it as area at once. I wanted to get an indefinite integral and them substitute. Thanks for opening my mind to this way of thinking
Great video! I love these interesting integrals. Keep uploading my friend!
I honestly did not see that coming, that was incredible
*This Vietnamese brother knows his University Maths but @**03:35** when you wrote the limits of integration from 0-1 I was wondering "What the hell is he doing?" but when u corrected urself @**04:15** & I said "Ah yea, he's back on track, 0-pi/4 is correct" ! Good stuff.*
All your works are clear and easy to undestand. Good on you.
i mean... i understand it but i would never ever have came up with that on my own
That basically sums up all maths haha
You solved what took my friends and me a day in 19 mins
This integral is not so easy .
I solved under 5 minutes lol.... We don't have so much time in exam
A day lol this took me only 30 seconds,just substitute x=tanx,and use the king's rule
@@abhinavtushar305 there's no king rule.... It's just a property........ Bansal sir used to call it that..... It's not an internationally used term.....
@@pi17 😅😅
Thank you so much! I am a big fan of yours. That's it!! ^^
Had two awesome ways to solve it, thanks to u and to flammable math
Thank u for your patience with each little step!!!!!
You are awesome, and that is a cool mic! Thank you for the practice!
Never ceases to amaze me. Awesome...
Thank you!!!!!
You are a real teacher, I like it. Thanks a lot, it helped me preparing for my final G17 exam.
Awesome integral!
I solved this exact integral just the day before yesterday! Noticing the tan^2(x)+1=sec^2(x) for tangents is half the work. Keep up the good work!
THAT'S GENIUS !!!! In the final part i had a mindblow.
very nice, soon you'll have 100K subscribers. Great job!
There is actually a very lovely way of doing differentiating under the integral sign for this integral as well! if you let I(a) = integral of (ln(ax+1)/(x^2+1)) dx you can also get the answer!
this I would love to see!
This is how I originally solved the integral - Feynman's technique! I was curious to see how it could be solved using only calculus 2 techniques.
Happy new year BlackpenRedpen. That was really nice integral, it was awesome!!! Greetings from Greece!
If I hadn't seen this video and this we're on a test I'd probably drop that class lmao. Amazing work, crazy how powerful substitution is. Hope I can apply this knowledge one day!
Man this was good thanks . I have been trying to solve this integral for hours .
I'm in school vacation here in my country(Brazil), and I should have been resting, but I just can't stop watching these integral videos. Nice job BPRP 😁
whoa whoa whoa... i thought this was black pen red pen, not black pen red pen blue pen!!
hehehehe
lol
I dont know how I was recommended this video considering I have probably 70% of my subscribed channels never being recommended to me..... This is a good video for students learning calculus.
A graph would have been icing on the cake. Thank you.
Beautiful and elegant. More putnam problems please.
Beautiful !
Very convoluted explanation. Using property of definite integral, the problem can be solved in less than a minute
Brilliant job
OMG congratulations , you are really amazing
who else thinks this is a little too easy for Putnam?
yeah man, I already did this in school... So way tooooo eady for Putnam... But, if true, I can maybe then get rid of my fear of Putnam
I think, it's become a famous question after it was asked in Putnam. But at the time, it might have been a very new question.
But yes, it is indeed very easy for Putnam.
Yes A easy one
Yes, easy because it's the "standard" question when you're introducing students to Leibniz rule. Quite sure back then it would've stumped most of us (it's Putnam after all).
@@chandrabhan7212 jþt
Very Niceeeeee i think i would like to see more of these Math competition problems , as this had a very satisfying way of solving it .
Thanks!!!!
ؤ
The first substitution was really clever, I like how you could then solve the rest of the integral using the knowledge from previous integrals ;)
Amazing!! Thank you from México
Happy to see some really hard integrals returning, they had been gone for a while... please do more putnam!
Thanks! I will try.
Nicely done!
Fantastic explanation !
Very nice - I thought there was no hope I would understand as you started the second substitution but I made it through!
That was so satisfying.
Excellent!
Ehrenmann , sweeet stuff dude, sugar
That was so... Elegant
Ok that was actually amazing
Amazing!
Aha what a legend! How do you actually think so cleverly? When I watch this it makes sense but its so clever man!!! Keep it up I marvel at your work!
It's known as the Serret’s integral for some people who are still wondering.
Oh nice!! I didn't know about this.
Sir just awesome
It is a beautiful integral
Outstanding..... 😍😍
Interesting integral and interesting solution.
that was beautiful :')
Great video!
your videos are addictive.
Well... this IS a potential AP Calculus Question (I take AB).
I’m glad I found you because I can keep my Calculus strong during Spring Break!
Now I will just see more videos to know how to do harder integrals.
Impressive!! Subscribed!
Radio TV thanks!!!!!
at the beginning of the year I was taking calc because I had to. Now I'm spending my free time watching videos of people solving calc
Very motivating integral. Thanks a lot.
Miroslav Dimitrov my pleasure!!
it was on my exam omfg many thxs
Great. It is really helpful. Wonder you could also do another version based on complex analysis.
Amazing.
Ohh my god this question came in my today's exam.The exact same question THANK YOU SIR!
What kind of exam?
Good one!
Great video
very nice!
wow this is art
A video to make people feel good about Putnam exam. This is like ABC of what's asked in their
If math problems had imposter syndrome
Awesome!
Whoever was the first person to figure that out should get a star for their forehead
He makes calculus so easy, if only he was actually my home tutor
our* home tutor
Very good.
What an ingenious tricks! Firstly the change of variable x = tanθ , secondlysinθ + cos θ = √ (2) cos( θ - π/4), thirdly making use of cosine is even, i.e. cos (- Ø ) = cos (Ø ), plus other minor tricks. Impressive! This integral is one hardest out of integrals solved on terms of elementary functions I ever seen.
Can you please do another Putnam integral, the integral of dx/(1+(tan x)^sqrt2) from 0 to pi/2 ? I'm really amazed at how you come up with these clever solutions, really mind blowing!
Wolfram alpha
Apply King's identity for definite integration
Use that the integral remains same by putting x as (pi/2 - x) and add both of them. The numerator and denominator cancel out....the answer is pi/4.☺☺
Simple:
Do the substitution x=Tan§
And then after obtaining the expression just use King rule and just add both , and we have achieved our goal
For those who don't know the king's rule, it states that If you have an integral with bounds a and b f(x)dx its the same as integral bounds a and b f(a+b-x)dx.
Nice evaluation..
I used Leibniz rule introducing ln(tx+1) and it worked marvelously
Math is amazing😍😍
Awesome. That trig identity for sin theta + cos theta rings a very faint bell. Well done.
I wrote my intermediate second year(='class 12' as known in many other countries) board examination mathematics paper-II today.
I saw this video yesterday.
In the exam we were asked to compute the integral of ln(1+tanx) w.r.t. x from 0 to pi/4 ( which is exactly the same question!!).
But, I solved it differently by using this property:
integral of f(x) dx from a to b = integral of f(a+b-x) dx ("King's property" as it is popularly called)
Yes this ques is in Rd Sharma also
I did this with Feynman's technique, I(b) = int [0,1] (ln(bx+1)/(x^2+1))dx, but it only worked because the endpoints were the same as the parameters you need to take the integral to 0 and to equal the original integral.
I was just about to give up until I saw this comment and finally noticed the thing about the endpoints
Integrating by parts, it's very easy to prove that the integral between 0 and 1 of
ArcTan[x]/(1+x)
it's also pi/8*Log[2]
Sergio Korochinsky it's very easy
two integrals for the price of one...
how would you prove that
@@kne-si5zj ...just integrate by parts.
dv = 1/(1+x) then v = Ln[1+x]
u = ArcTan[x] then du = 1/(1+x^2)
u v = ArcTan[x] Ln[1+x]
(between 0 and 1) = pi/4*Log[2]
and the integral of v du is solved in the video... pi/8*Log[2].
This is great! I’ve only take AP Calculus AB before and wanted to see how much I’d get (obviously barely anything), I’m taking calc 2 freshman year in university and wanted to make myself less scared
Excellent work as always! I wonder if the fact that the two cos (...) integrals cancel each other out could not be recognized earlier, e.g. by some symmetry property.And by the way: could you make videos on the Fourier series in 2018? Thx and BR
Ooo !what elegant is bprp and also his equation !
wow!!! Thanks :)
thank you. Really huge insight when dealing with bounded integrals. would it have worked had the integral had no boundaries?
can you please do int (sqrt (1+sqrtsecx))dx?
Hi bprp, wonderful number game with angle functions...
Thank you 👍
You always have a tendency to talk fast. You clearly love doing math, and I love that, keep up the good work.
Very nice
There is another easy way to solve this. After the theta substitution once we get the integral as integral from 0 to pi/4 ln(1+tanx) dx (using x for theta), we can use a property of definite integral that integral 0 to a f(x)dx is the same as integral 0 to a f(a-x)dx. So the 0 to pi/4 ln(1+tanx) (let us call it capital I), is the same as 0 to pi/4 ln(1+tan(pi/4 -x)). If you use the trigonometric formula for tan(pi/4 - x) and simplify the integral becomes integral 0 to pi/4 ln(2/(1+tanx)) which is integral 0 to pi/4 ln2 - 0 to pi/4 ln(1+tanx). This means
I = 0 to pi/4 ln2 - I, means 2I = Integral of 0 to pi/4 ln2
means I = pi/8 times ln2.
Another approach is to expand ln(1 + tan(theta)) as a power series and integrate term by term. You end up having to combine two infinite triangular arrays but there is a neat trick and the answer drops out easily once you spot it. You need to use the identities 1-1/2+1/3-1/4+......= ln(2) and 1-1/3+1/5-1/7...= pi/4.
Man I solved it in 1min it boosted my confidence man. Now I would focus on physics portion for my exam.
Ok, because it's the new year and you did a great job (as always) extra for you I unsubscribe, so that I can subscribe again
HAHAHAHAHAHAHA!! Thank you. I wish all my subscribers can do the same for me!
= DDDDDD
AndD whats the point of subtracting then adding back the same quantity since the original quantity will remain unaltered? well thats exactly what ur doing here AndD, by unsub. then subscriging again!
@@ivornworrell Obvious. It's like that integral you have to add 1 and then subtract 1 and you get a nicer little integral...