According to Mathematica's implementation notes: "Many other definite integrals are done using Marichev-Adamchik Mellin transform methods. The results are often initially expressed in terms of Meijer G functions, which are converted into hypergeometric functions using Slater's theorem and then simplified." It also uses an extended version of the Risch algorithm for indefinite integrals. It only took 0.24 secs on my machine to get the answer so there wasn't massive recursion going on behind the scenes.
It's easier to compute the "tool integral" by using Feynman's trick. Here, you can simply take the derivative with respect to α. Then you need to do a partial fraction decomposition (which is easier than the one Michael is doing here) and then you see that the original integral is equal to pi/2 times the integral of 1/(α + √α), which is easily done using the substitution u = √α. The only remaining thing is to figure out the value of the original integral for α = 0 in order to show that the integration constant is zero. IIRC, Michael has done that integral before and shown that it is zero, but if not, that's also easy to show by using the substitution u = 1/x.
You can use substitution: t = x - 1/x . ln( x⁴ + 1 ) = ln( x² + 1/x² ) + 2lnx = ln (t² + 2) + 2lnx. Integral of lnx/(x²+1) is 0. So we will get Integral from 0 to inf of 2*ln(t²+2)/(t²+4). Substitution: t=2u, and we are done.
I always attempt the integral problems before watching the video, and I managed to do it starting with Feynman's technique with an extra parameter. I then used contour integration with complex analysis to evaluate that result, which had somewhat heavy algebra with complex numbers. I then integrated once more with respect to the extra parameter to get the final result. This one I used a technique from Blackpenredpen to integrate an integrand of the form (1-u^2)/(u^4+1). I also did this without the help of a previously known tool. This problem was a beast.
Some integrals that occur in quantum field theory are so complex. Can you post some integrals which occur in subjects like statistical physics, quantum field theory.
Great video as always! Thanks :) Were you just teasing us with that short a while ago, or can we actually expect a video on fractional derivatives at some point? Cheers!
I don't believe that usage of absolute values was legit for complex integrals... Say, we have log(abs(z)) there is no way in complex numbers to get 1/z from it. Or there is something in tool integral to be clarified.
What timestamp could this property have been applied? If you're referring to the very end when he combines logarithms using this property, we could have also combined them immediately as the presence of i doesn't change the log properties validity.
15:20 I think I've confused myself because I thought that the -pi to +pi convention was arbitrary, but it seems that the answer for this real integral changes if you take 0 to 2pi instead and therefore a different square root of -i. Is this resolved by the fact that our integral is along the positive real axis, meaning we want a branch cut out of the way i.e. along the negative real axis?
Suggestion: (though this may be too easy) arctan (sqrt(3)y / (2x+y)) + arctan (sqrt(3)x / (2y+x)) = ? Bonus points if you figure out the relationship between this equation and the tiling of equilateral triangles!
*Stop* your major cursing! It is ignorant and needless, especially in a mathematics forum. You should go back and edit it out. Manage to make civil posts.
@@robertveith6383 my language is perfectly fine and appropriate to the medium. sorry i have youtube notifications disabled so i didn't see your tears until i stumbled upon this video by chance. if you don't have anything pertinent to the mathematics to say, then you should probably go back into your hidey hole where people who aren't math geniuses like me hide until they work up the courage to poke out and cry about someone using big boy words like FUCK. incidentally, as i said in another comment, contour integration is a better way to do this problem. AND i figured out an even better way using an extension of integration by parts to three-factor integration that your dumb ass would never be able to grapple with. now FUCK off back to your hidey hole. math is the wild west. where only the toughest niggas ride and weak FUCKSs like you fall by the wayside. *fist bumps creator of video* don't worry about this kid. i took care of him for you.
Hi Michael, would you like to solve some Rational Geometry problems I had to solve in my tests from year 1 and 2 of high school? If you are interested, reply to my comment
Substitute u = 1/x and split the integral into integrals over the natural numbers (1st int from u=1 to u=2, 2nd int from u=2 to u=3, ...) For each integral from u=n to u=n+1: 7^[1/x] = 7^[u] = 7^(n+1), which can be pulled outside the integral The rest should be easy
According to Mathematica's implementation notes: "Many other definite integrals are done using Marichev-Adamchik Mellin transform methods. The results are often initially expressed in terms of Meijer G functions, which are converted into hypergeometric functions using Slater's theorem and then simplified." It also uses an extended version of the Risch algorithm for indefinite integrals. It only took 0.24 secs on my machine to get the answer so there wasn't massive recursion going on behind the scenes.
Not just A massive but a lot and so much of expand recursion instead of the integrand and integral formulation it sElf
This makes Computer Algebra sound fascinating
It's easier to compute the "tool integral" by using Feynman's trick. Here, you can simply take the derivative with respect to α. Then you need to do a partial fraction decomposition (which is easier than the one Michael is doing here) and then you see that the original integral is equal to pi/2 times the integral of 1/(α + √α), which is easily done using the substitution u = √α. The only remaining thing is to figure out the value of the original integral for α = 0 in order to show that the integration constant is zero. IIRC, Michael has done that integral before and shown that it is zero, but if not, that's also easy to show by using the substitution u = 1/x.
That’s a very nice solution
You can use substitution: t = x - 1/x . ln( x⁴ + 1 ) = ln( x² + 1/x² ) + 2lnx = ln (t² + 2) + 2lnx. Integral of lnx/(x²+1) is 0. So we will get Integral from 0 to inf of 2*ln(t²+2)/(t²+4). Substitution: t=2u, and we are done.
In the end you are using derived formula from the video. You were able to change in second degree of t. Well kudos to substitution.
@@anantyadav4739 c
18:20
LOL!
I always attempt the integral problems before watching the video, and I managed to do it starting with Feynman's technique with an extra parameter. I then used contour integration with complex analysis to evaluate that result, which had somewhat heavy algebra with complex numbers. I then integrated once more with respect to the extra parameter to get the final result. This one I used a technique from Blackpenredpen to integrate an integrand of the form (1-u^2)/(u^4+1). I also did this without the help of a previously known tool. This problem was a beast.
how old are you ? I'm 18 and I'm feeling kinda dumb that I'm having so much difficulty XD
@@4ptil195 I'm 36. There's no way I could have evaluated this when I was 18. It takes a lot of learning and practice.
Some integrals that occur in quantum field theory are so complex. Can you post some integrals which occur in subjects like statistical physics, quantum field theory.
Great video as always! Thanks :)
Were you just teasing us with that short a while ago, or can we actually expect a video on fractional derivatives at some point? Cheers!
Small mistake on 15.37
That is sqrt2/sqrt2 and not sqrt2/2
Wolfram Alpha:
integrate {(ln(x^4+1))/(x^2+1) between x=0,x=inf}
= 1/2 π log(6 + 4 sqrt(2))≈3.85771
I would love to suggest the sum. Of ln(1/n)/[1-n²] from n=2 to ♾️ using ln(1/x)=½ln[(1+y²]...🥺
I don't believe that usage of absolute values was legit for complex integrals... Say, we have log(abs(z)) there is no way in complex numbers to get 1/z from it.
Or there is something in tool integral to be clarified.
I like the shirt!
10:20
why is ln(y-a) evaluated 0 ln(a)? shouldnt it be ln(0-a)=ln(-a)?
I'm gonna hazard a guess and say that since the antiderivative contains an absolute value, you can use the fact that |-a| = |a|
@@trustnoone81 oh, youre right! i didnt pay enough attention to the absolute value, thanks!
The number of missing close parentheses and dx's make for a real cliffhanger! Even in the thumbnail...
Hi Michael, why can't we use the property of logarithm of lnA + lnB = lnAB right away? Is it because of the presence of i?
What timestamp could this property have been applied? If you're referring to the very end when he combines logarithms using this property, we could have also combined them immediately as the presence of i doesn't change the log properties validity.
We could, and he says as much, but he prefers to figure out what they are first
Good thnx🍺🍺🍻.
You removed the absolute value from the lan which changes the answer...
15:20 I think I've confused myself because I thought that the -pi to +pi convention was arbitrary, but it seems that the answer for this real integral changes if you take 0 to 2pi instead and therefore a different square root of -i. Is this resolved by the fact that our integral is along the positive real axis, meaning we want a branch cut out of the way i.e. along the negative real axis?
Suggestion: (though this may be too easy)
arctan (sqrt(3)y / (2x+y)) + arctan (sqrt(3)x / (2y+x)) = ?
Bonus points if you figure out the relationship between this equation and the tiling of equilateral triangles!
also why not do this with contour integration..
where the fuck does the 1/(y-a) come from at 4:14?
how does partial fraction decomposition give us this weird af coefficient??
*Stop* your major cursing! It is ignorant and needless, especially in a mathematics forum.
You should go back and edit it out. Manage to make civil posts.
@@robertveith6383 my language is perfectly fine and appropriate to the medium.
sorry i have youtube notifications disabled so i didn't see your tears until i stumbled upon this video by chance.
if you don't have anything pertinent to the mathematics to say, then you should probably go back into your hidey hole where people who aren't math geniuses like me hide until they work up the courage to poke out and cry about someone using big boy words like FUCK.
incidentally, as i said in another comment, contour integration is a better way to do this problem. AND i figured out an even better way using an extension of integration by parts to three-factor integration that your dumb ass would never be able to grapple with.
now FUCK off back to your hidey hole.
math is the wild west. where only the toughest niggas ride and weak FUCKSs like you fall by the wayside.
*fist bumps creator of video* don't worry about this kid. i took care of him for you.
Hi Michael, would you like to solve some Rational Geometry problems I had to solve in my tests from year 1 and 2 of high school? If you are interested, reply to my comment
Someone plz solve the integral
Int 0 to 1 (dx/ (7 ^[1/x])) where [ ] is greater integer function.
Asked in JEE Main 2022 27 June shift 2
Substitute u = 1/x and split the integral into integrals over the natural numbers
(1st int from u=1 to u=2, 2nd int from u=2 to u=3, ...)
For each integral from u=n to u=n+1:
7^[1/x] = 7^[u] = 7^(n+1), which can be pulled outside the integral
The rest should be easy
@@reeeeeplease1178 this is what I was going to suggest as well. It should be related to geometric series.
@@briemann4124 the series ends up being of the form of the Taylor series for ln(1-x)
@@calcul8er205 yep! Which is related to the geometric series.
I bet that the full name of the “Integral Suggester” has the initials M. P. 😁
this was the worst pres. ever - very confusing !
Please write in sentences.
2nd View 🤩😇