You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA== If you like the videos and would like to support the channel: www.patreon.com/Maths505
bro you explain just like a brother and also providing a best type of contain i know yur passionate about not only for math but also for beauty of math which you show in this lecture pls providing material such like this and that
This can also be done using contour integration, which is a more "brute force" approach that doesn't rely on being clever, and is just as fast if not faster. In the upper half plane there is a branch cut starting from +i1 and going upwards off to +i inf which we must go around. So, the contour is made up of: z from -inf to +inf, a quarter circle from +inf to + i inf, a line from +i inf to +i1, a line from +i1 to +i inf (on the other side of the branch cut), a quarter circle from +i inf to -inf. The -inf to +inf line integral is two times the desired integral, let's call this 2L. The "round trip" integral around the branch cut reduces to 2pi times integral of 1/(x^4+1) from 1 to infinity since integrating both sides of a log branch cut in opposite directions makes the log disappear. This can be evaluated by partial fraction decomposition into the denominators: (x-1/sqrt2-i/sqrt2), (x+1/sqrt2-i/sqrt2), (x+1/sqrt2+i/sqrt2), (x-1/sqrt2+i/sqrt2) by noticing that x^4+1 can be thought of as a difference of squares. These also happen to be the odd-numbered eighth-roots-of-unity. This is the only tedious step in the entire process. Let's call this value A. The quarter circle integrals are zero because we have z^4 in the denominator, which completes the closed contour. So, 2L+A=i2pi (sum of residues at 1/sqrt2+i/sqrt2 and -1/sqrt2+i/sqrt2). It is very easy to evaluate the residues, in fact, the partial fraction decomposition we did to evaluate A explicitly gives us all the residues.
Hi! Why does integrating both sides of a log branch cut in opposite directions make the log disappear? I follow all of your explanation other than that part :)
@@sss-ol3dl imagine there is no branch cut. if there is no branch cut, integrating forwards+backwards gives zero. the only reason the branch cut makes a difference is because the two sides of the branch cut have different values: logz vs logz+i2pi. since going the opposite direction gives a negative sign, you get logz-(logz+i2pi). To generalize, integrating a branch cut "round trip" means integrating the difference between branch cuts once. The log disappears because the difference between different branches (or riemann surfaces if you want to be fancy) is a constant i2pi
@@maths_505 hey man. big fan of your content. i was using contour integration, and i had a doubt about calculating the residues. you get log(1+i), and log(1-i) terms. when you expand them using log(z)=logIzI + iarg(z), how will the argument change here because of 2 branch cuts?
Ahh man believe I just heard it and I couldn't believe it 😂😂😂 I've switched up so many integrals and summation signs this past month I see em everywhere!!!😂😂😂
I loved your content and feynman technique is 1 of the best things seen in my life. Sir, can you tell me a book's name for learning different integration techniques plz.. Thanks to Math 505, Love from India.❤
Inside interesting integral by Paul Nahin Almost impossible integrals, sums and series by an author whose name I forgot 😂 You'll find it if you Google it
@@maths_505 how can "i forgot😂" be the name of author Just kidding😂. Thanks;i got the book on google I would read that integration book in reverse to learn differentiation.😂
I dont understand the final fraction we get with the partial fraction sums. it seems to me that when you replace back u with x^2 the fraction on the right disappears
Some results are used quite often here so I post write ups on Instagram instead of proving them again and again or making a video on something that simple.
@@maths_505Indeed! Just as we use various equivalent representations of integrals, whether another integral form, a function, a series representation or infinite sum/product etc as tools to substitute whenever necessary, why not use the results of known integrals as constants for substitution to save time? Only, one needs a methodology for cataloguing all these equivalent forms and known results to use it later as a crib. The information grows exponentially over time and I am afraid it would be difficult to remember even for geniuses who do maths for a living... I wonder how they manage it.
@@maths_505 suppose in the public key of elliptic curves.lets say i divide for example pubkey of 100 / 7 (has decimals) and pubkey of 100 / 10 (has no decimals), is there a way to tell the difference when you have the public keys only??
Kind of true but if i recall corectly the one kinda similar with this one was from 1988 so things might have changed. Also kinda wish they would just make integrals harder and less combinatorics or discrete math. Gamma function and geometric series would go brrr
I hit a roadblock when using complex analysis. I used the same contour as the one in this vid except for the semi circle at the origin. ruclips.net/video/-6qs-XNvCMw/видео.htmlsi=yfsSullyuOoHH9_p There is a branch point at i, and a branch cut from i to i(inf).there is a pole in the first quadrant (1/√2)+i(1/√2). And one in the second quadrant -(1/√2)+i(1/√2). When we restrict the argument of log z about i, Do we evaluate the residue with the pole in the first quadrant as e^(i3π/4),{which lies outside our restricted domain} or as e^(i9π/4) {which lies within our restricted domain of argz}£ [π/2,5π/2]? Please help bro
Too long,but not difficult..anyway (pi/4rad2)(ln2+2ln(rad2+1))-1/8(pi^2/rad2)..mah..it's Easy wrong... I checked on internet.. Is correct... 0,4915483..
Hey can I ask you where can I learn this things?I mean in my country we are not taught a lot of things you use..🫠Can you tell me about some resources...
You can follow me on Instagram for write ups that come in handy for my videos:
instagram.com/maths.505?igshid=MzRlODBiNWFlZA==
If you like the videos and would like to support the channel:
www.patreon.com/Maths505
Loved it! So many steps, so many interesting integrals on the way and a beautiful result! I am elated. Thanks!
bro you explain just like a brother and also providing a best type of contain
i know yur passionate about not only for math but also for beauty of math which you show in this lecture
pls providing material such like this and that
That's the plan bro
Thanks
This can also be done using contour integration, which is a more "brute force" approach that doesn't rely on being clever, and is just as fast if not faster. In the upper half plane there is a branch cut starting from +i1 and going upwards off to +i inf which we must go around. So, the contour is made up of: z from -inf to +inf, a quarter circle from +inf to + i inf, a line from +i inf to +i1, a line from +i1 to +i inf (on the other side of the branch cut), a quarter circle from +i inf to -inf.
The -inf to +inf line integral is two times the desired integral, let's call this 2L.
The "round trip" integral around the branch cut reduces to 2pi times integral of 1/(x^4+1) from 1 to infinity since integrating both sides of a log branch cut in opposite directions makes the log disappear. This can be evaluated by partial fraction decomposition into the denominators: (x-1/sqrt2-i/sqrt2), (x+1/sqrt2-i/sqrt2), (x+1/sqrt2+i/sqrt2), (x-1/sqrt2+i/sqrt2) by noticing that x^4+1 can be thought of as a difference of squares. These also happen to be the odd-numbered eighth-roots-of-unity. This is the only tedious step in the entire process. Let's call this value A.
The quarter circle integrals are zero because we have z^4 in the denominator, which completes the closed contour.
So, 2L+A=i2pi (sum of residues at 1/sqrt2+i/sqrt2 and -1/sqrt2+i/sqrt2). It is very easy to evaluate the residues, in fact, the partial fraction decomposition we did to evaluate A explicitly gives us all the residues.
Wonderful explanation
Hi! Why does integrating both sides of a log branch cut in opposite directions make the log disappear? I follow all of your explanation other than that part :)
Hard to explain in a comment. Check out qncubed3's video on decreasing the power of a logarithm. I think he explained it there.
@@sss-ol3dl imagine there is no branch cut. if there is no branch cut, integrating forwards+backwards gives zero. the only reason the branch cut makes a difference is because the two sides of the branch cut have different values: logz vs logz+i2pi. since going the opposite direction gives a negative sign, you get logz-(logz+i2pi).
To generalize, integrating a branch cut "round trip" means integrating the difference between branch cuts once. The log disappears because the difference between different branches (or riemann surfaces if you want to be fancy) is a constant i2pi
@@maths_505 hey man. big fan of your content. i was using contour integration, and i had a doubt about calculating the residues. you get log(1+i), and log(1-i) terms. when you expand them using log(z)=logIzI + iarg(z), how will the argument change here because of 2 branch cuts?
I love papa flammy!
Me too
😂 who doesn't? Except for his foul language sometimes...
1:30 it’s an unbreakable habit haha
Ahh man believe I just heard it and I couldn't believe it 😂😂😂
I've switched up so many integrals and summation signs this past month I see em everywhere!!!😂😂😂
@@maths_505 every time you close your eyelids you can see the switch-up being performed
@The_Shrike 😂😂😂
Good Job. It is stunning solution steps.
I loved your content and feynman technique is 1 of the best things seen in my life.
Sir, can you tell me a book's name for learning different integration techniques plz..
Thanks to Math 505, Love from India.❤
Inside interesting integral by Paul Nahin
Almost impossible integrals, sums and series by an author whose name I forgot 😂
You'll find it if you Google it
@@maths_505 how can "i forgot😂" be the name of author
Just kidding😂. Thanks;i got the book on google
I would read that integration book in reverse to learn differentiation.😂
ln(6+4root2) = 2ln(2+root2)
I dont understand the final fraction we get with the partial fraction sums. it seems to me that when you replace back u with x^2 the fraction on the right disappears
Very cool solution
Maths 505 quoting an Instagram post like a boss!
Some results are used quite often here so I post write ups on Instagram instead of proving them again and again or making a video on something that simple.
@@maths_505Indeed! Just as we use various equivalent representations of integrals, whether another integral form, a function, a series representation or infinite sum/product etc as tools to substitute whenever necessary, why not use the results of known integrals as constants for substitution to save time? Only, one needs a methodology for cataloguing all these equivalent forms and known results to use it later as a crib. The information grows exponentially over time and I am afraid it would be difficult to remember even for geniuses who do maths for a living... I wonder how they manage it.
the substitution u = x^2 => du = 2xdx => dx = du/2x => dx = du/2*sqrt(u) and so the integral doesn't seem to be that easy, or I missed something ?
ln((6-4*sqrt(2))/2) = ln(3-2*sqrt(2))
Insane
Can i ask you some qurstions regarding to elliptic curves please
Sure
@@maths_505 suppose in the public key of elliptic curves.lets say i divide for example pubkey of 100 / 7 (has decimals) and pubkey of 100 / 10 (has no decimals), is there a way to tell the difference when you have the public keys only??
@@exodus8213 I'm afraid I'm not familiar with elliptic curve cryptography
@@maths_505 its about some linear algebra
@@exodus8213 then I'm afraid I cant seem to grasp your question
Bro that looks like that putnam integral on steroids
Excellent description 😂
Though Putnam integrals aren't exactly that hard
Kind of true but if i recall corectly the one kinda similar with this one was from 1988 so things might have changed. Also kinda wish they would just make integrals harder and less combinatorics or discrete math. Gamma function and geometric series would go brrr
😂😂😂
I hit a roadblock when using complex analysis.
I used the same contour as the one in this vid except for the semi circle at the origin.
ruclips.net/video/-6qs-XNvCMw/видео.htmlsi=yfsSullyuOoHH9_p
There is a branch point at i, and a branch cut from i to i(inf).there is a pole in the first quadrant (1/√2)+i(1/√2). And one in the second quadrant -(1/√2)+i(1/√2).
When we restrict the argument of log z about i, Do we evaluate the residue with the pole in the first quadrant as e^(i3π/4),{which lies outside our restricted domain} or as e^(i9π/4) {which lies within our restricted domain of argz}£ [π/2,5π/2]?
Please help bro
Too long,but not difficult..anyway (pi/4rad2)(ln2+2ln(rad2+1))-1/8(pi^2/rad2)..mah..it's Easy wrong... I checked on internet.. Is correct... 0,4915483..
Hey can I ask you where can I learn this things?I mean in my country we are not taught a lot of things you use..🫠Can you tell me about some resources...
MIT lectures are the best