Now that was one seriesly awesome integration!!! And the appearance of pi at the end there was both unexpected and aesthetically pleasing. The reduction formula: www.instagram....
It can be calculated by parts and simple substitution Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=limit(-ln(x)*ln(x^2+1)/x,x=infinity)-limit(-ln(x)*ln(x^2+1)/x,x=0) + Int(1/x*(1/x*ln(x^2+1)+ln(x)*2x/(x^2+1)),x=0..infinity) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=Int(ln(x^2+1)/x^2,x=0..infinity)+2Int(ln(x)/(x^2+1),x=0..infinity) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=limit(-1/x*ln(x^2+1),x=infinity)-limit(1/x*ln(x^2+1),x=0)+Int(1/x*2x/(x^2+1),x=0..infinity)+2Int(ln(x)/(x^2+1),x=0..infinity) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)+Int(ln(1/u)/((1/u)^2+1)*(-1/u^2),u=infinity..0) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)+Int(ln(1/u)/(1/u^2+1)*(1/u^2),u=0..infinity) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)+Int(ln(1/u)/(u^2+1),u=0..infinity) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)-Int(ln(u)/(u^2+1),u=0..infinity) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2*(Pi/2) Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=Pi
@@meteor3033 This is Maple version but Mathematica should understand it also The solution above which i proposed doesnt use any advanced integral calculus techniques and can be understood also by beginners in integral calculus class
Find me a better duo than Maths 505 and invoking geometric series then switching up summation with integration...😂 But seriously almost any challenging integral is a combo of geometric series, gamma function and eta/zeta functions at the end.
Some of the reasons to like this channel, is not only Maths 505 solutions, but everyone else throwing in with another technique, or just really thoughtful questions. Good stuff.
Hi, friendly neighborhood contour integral enjooyer here :) Careful analysis of a semicircular contour in the upper half plane that goes around the vertical branch cut stemming from log(x^2+1) shows that twice the desired integral is equivalent to the real part of: integral i2pi log(x)/x^2 from i to i inf, which can be easily evaluated using integration by parts.
There's a much easier way to solve this by using feynman's technique...first set the integrand to be lnxln( tx^2+1)/x^2 then by differentiating we get lnx/tx^2+1 this can be computed easily by complex analysis or any real method then we'll get that I'(t)= -pi(lnt)/4sqrt(t) then using I (0)=0 and I (1)= I (our integral) and the gamma function or feynman's technique again and we'll get that our integral is indeed pi..
Majestic proof, but I have a question: my teachers always taught me to deal with these improper integrals first proving that they are convergent or divergent. Only then one could even possibly compute the exact value of an integral. I, instead, observe very often in this kind of video, that the first part is completely skipped, as if the integral is obviously convergent. Is it really that obvious?
Another beatiful integral. I solved it using the Feynman trick which was much easier. I have a @suggestion: add +1 to denominator. It won't be as cool as this integral but it's nice
It seems to me one should be more careful with uniform convergence. The geometric series does not uniformly convergent on the open interval (0,1), and one needs to work to switch sum and integral. The alternating series for log(2) that shows up does not convergence uniformly as well, indicating potential problems.
I was evaluating an integral from the Romanian mathematical magazine and got this integral en route to the final answer. Decided it was worth a video on its own
@@maths_505 I always chuckle when he leaves all the outtakes in. He starts saying something, doesn't like it, stops, looks annoyed, starts over, but leaves all that in. You just say "terribly sorry" lmao. What university are you at, if you don't mind sharing?
I'm so addicted with the reverse feynmann technique so as to defined an integral I(s, t) = int of (x^2+1)^t/x^s and said that our integral I is equal to the mixed partial fraction at (2,0). send help.
can u try the que from the JEE Adv 2016 Paper 2? it was a mixed question of limits and integral and considered the toughest que till date. I bet you will enjoy the que, and if possible do make a video on it.
The core difficulty was to split up the integration at x=1 (or adding another dimension a la Feynman). The rest is standard techniques. This split is indeed obvious when drawing the graph - unfortunately in this video it comes out of nothing and you act as if was totally natural to do so or as if the split was arbitrary and not important. That’s a bluff I think.
It can be calculated by parts and simple substitution
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=limit(-ln(x)*ln(x^2+1)/x,x=infinity)-limit(-ln(x)*ln(x^2+1)/x,x=0) + Int(1/x*(1/x*ln(x^2+1)+ln(x)*2x/(x^2+1)),x=0..infinity)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=Int(ln(x^2+1)/x^2,x=0..infinity)+2Int(ln(x)/(x^2+1),x=0..infinity)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=limit(-1/x*ln(x^2+1),x=infinity)-limit(1/x*ln(x^2+1),x=0)+Int(1/x*2x/(x^2+1),x=0..infinity)+2Int(ln(x)/(x^2+1),x=0..infinity)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)+Int(ln(1/u)/((1/u)^2+1)*(-1/u^2),u=infinity..0)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)+Int(ln(1/u)/(1/u^2+1)*(1/u^2),u=0..infinity)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)+Int(ln(1/u)/(u^2+1),u=0..infinity)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)+Int(ln(x)/(x^2+1),x=0..infinity)-Int(ln(u)/(u^2+1),u=0..infinity)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2Int(1/(x^2+1),x=0..infinity)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=2*(Pi/2)
Int(1/x^2*ln(x)*ln(x^2+1),x=0..infinity)=Pi
I seriously gotta start using IBP at the start of my solution developments😂
Good Job!
Yo this is so hard to read
@@meteor3033 This is Maple version but Mathematica should understand it also
The solution above which i proposed doesnt use any advanced integral calculus techniques and can be understood also by beginners in integral calculus class
@@holyshit922 Yeah I figured it must be for something like that, but Im on a phone and it displays as practically illegible 😂
Find me a better duo than Maths 505 and invoking geometric series then switching up summation with integration...😂 But seriously almost any challenging integral is a combo of geometric series, gamma function and eta/zeta functions at the end.
Some of the reasons to like this channel, is not only Maths 505 solutions, but everyone else throwing in with another technique, or just really thoughtful questions. Good stuff.
Hi, friendly neighborhood contour integral enjooyer here :) Careful analysis of a semicircular contour in the upper half plane that goes around the vertical branch cut stemming from log(x^2+1) shows that twice the desired integral is equivalent to the real part of: integral i2pi log(x)/x^2 from i to i inf, which can be easily evaluated using integration by parts.
Noice
There's a much easier way to solve this by using feynman's technique...first set the integrand to be lnxln( tx^2+1)/x^2 then by differentiating we get lnx/tx^2+1 this can be computed easily by complex analysis or any real method then we'll get that I'(t)= -pi(lnt)/4sqrt(t) then using I (0)=0 and I (1)= I (our integral) and the gamma function or feynman's technique again and we'll get that our integral is indeed pi..
feynmann technique is so smooth
Majestic proof, but I have a question: my teachers always taught me to deal with these improper integrals first proving that they are convergent or divergent. Only then one could even possibly compute the exact value of an integral. I, instead, observe very often in this kind of video, that the first part is completely skipped, as if the integral is obviously convergent. Is it really that obvious?
Indeed it is.....you never really need to prove convergence as you can do so using software like wolfram alpha
Nice integral,and the result is even nicer😃
Another beatiful integral. I solved it using the Feynman trick which was much easier. I have a @suggestion: add +1 to denominator. It won't be as cool as this integral but it's nice
It seems to me one should be more careful with uniform convergence. The geometric series does not uniformly convergent on the open interval (0,1), and one needs to work to switch sum and integral. The alternating series for log(2) that shows up does not convergence uniformly as well, indicating potential problems.
Life is beautiful😍
If you use leibniz rule and then put ax=u you end up getting I'(a)=-pi*log(a) which then is straight forward.
Yes this works very well
Should combine the two integrals on step to cancel the entire ln(1+x^2) term
Good Job. Thanks.
Where are you getting these from? :D I remember a book by Boris P. Demidovich that had a lot of integrals that we used in our undergrad calc course.
I was evaluating an integral from the Romanian mathematical magazine and got this integral en route to the final answer. Decided it was worth a video on its own
@@maths_505 This integral sure is a beast. You're like Michael Penn, on speed.
Nah bruh penn is clear. That guy makes like 20 different substitutions and the integral turns elementary 😂
@@maths_505 I always chuckle when he leaves all the outtakes in. He starts saying something, doesn't like it, stops, looks annoyed, starts over, but leaves all that in. You just say "terribly sorry" lmao.
What university are you at, if you don't mind sharing?
@@Kapomafioso I just graduated with a major in applied math. University of Punjab in Lahore (Pakistan)
I'm so addicted with the reverse feynmann technique so as to defined an integral I(s, t) = int of (x^2+1)^t/x^s and said that our integral I is equal to the mixed partial fraction at (2,0). send help.
Why is this integral "logarithmic"? Because the function starts with log? What if the log appeared in the middle of the expression?
Can be done with feynman's trick
1:20 Can you use it though? the upper limit of the integral is 1.
upper limit meaning x approaches 1 from the left so it's perfectly valid.
@@maths_505 yeah, but what about in 1? That's what I meant before.
It diverges so we can't use the geometric series.
8:15 there's a minus sign missing
If you ever taste my mom’s apple pie then you would know that there’s nothing wrong with pi!
can u try the que from the JEE Adv 2016 Paper 2? it was a mixed question of limits and integral and considered the toughest que till date. I bet you will enjoy the que, and if possible do make a video on it.
The core difficulty was to split up the integration at x=1 (or adding another dimension a la Feynman). The rest is standard techniques. This split is indeed obvious when drawing the graph - unfortunately in this video it comes out of nothing and you act as if was totally natural to do so or as if the split was arbitrary and not important. That’s a bluff I think.
I got pi^2/(2sqrt(2)) + pi 😭😭😭
So you're only from actual value of 3 by 3 (taking sqrt(2) to be 1.5😂😂😂
Learn how to write ln(x) please
3/4 days and i solve It...
Why 3/4???
微積サークルから来た
Daaaamn. 🔥🔥🔥