I mean, if the integral is divergent anyway, then whether or not you can use Feynman’s rule is kinda moot, isn’t it? It’s basically saying “Feynman’s rule can’t help us find the answer because the answer doesn’t exist.” Which doesn’t really have anything to do with the rule itself. You’d have to find a function with a limit that converges if you do it in one order, but diverges as a mixed limit, for the hypotheses of the rule to matter.
When changing x^2 into x^s, the integral has the closed form: 1/4*GAMMA(1/2*(s+1)/s)*2^(1/s)*(2*ln(2)+Psi(1/2*(s+1)/s)+Psi(1/2*(2*s-1)/s))*Pi^(1/2)/s^2/GAMMA(1/2*(2*s-1)/s) for s in R, s > 1 (and provides the analytic continuation to a wider domain). Psi(z) is the Digamma-function. Limit s -> 1 yields -gamma.
Thanks for teaching us all this methods, have been from HUGHE help for me to continue studying calculus by myself, I'm so grateful by all your effort, keep it up man :>
Watching the plot of this thing I can barely believe someone managed to computer its integral exactly… it looks like the crazy functions I had fun coming up with in high school just to see how they’d look like.
I did it a bit differently. I=integral (laplace(sin(x))*inverse_laplace(log(x)/sqrt(x)). The inverse laplace transform of log(x)/sqrt(x) can be solved by assuming it is equal to (a+b*log(x))/sqrt(x) (a bit similar to log(x)/x) then applying the laplace transform, we can find a and b. The integral after that becomes easier to solve.
You can pretty easily show the integral with the complex exponential is equal to the same one but with a negative exponent (gamma function integral form) multiplied by a constant factor using complex integration.
Great one. The integral-calculator site had it slightly differently, basically reverting the sign in the parentheses by prepending them by (i - 1)(i + 1) and then rewriting the denominator as 2^(9/2), and rewriting 2*ln(4) as 4*ln(2). I'm amazed that it found an analytical solution. Now I'm curious how that Maxima computer algebra system works.
Do one integral where feynman’s trick fails due to divergence
Integral from 0 to infinity (e^-x-1)/x
I mean, if the integral is divergent anyway, then whether or not you can use Feynman’s rule is kinda moot, isn’t it?
It’s basically saying “Feynman’s rule can’t help us find the answer because the answer doesn’t exist.” Which doesn’t really have anything to do with the rule itself.
You’d have to find a function with a limit that converges if you do it in one order, but diverges as a mixed limit, for the hypotheses of the rule to matter.
When changing x^2 into x^s, the integral has the closed form: 1/4*GAMMA(1/2*(s+1)/s)*2^(1/s)*(2*ln(2)+Psi(1/2*(s+1)/s)+Psi(1/2*(2*s-1)/s))*Pi^(1/2)/s^2/GAMMA(1/2*(2*s-1)/s) for s in R, s > 1 (and provides the analytic continuation to a wider domain). Psi(z) is the Digamma-function. Limit s -> 1 yields -gamma.
I love the laplace transform ❤
Thanks for teaching us all this methods, have been from HUGHE help for me to continue studying calculus by myself, I'm so grateful by all your effort, keep it up man :>
Watching the plot of this thing I can barely believe someone managed to computer its integral exactly… it looks like the crazy functions I had fun coming up with in high school just to see how they’d look like.
Well that's one way of generating cool integrals
Great work
I did it a bit differently. I=integral (laplace(sin(x))*inverse_laplace(log(x)/sqrt(x)). The inverse laplace transform of log(x)/sqrt(x) can be solved by assuming it is equal to (a+b*log(x))/sqrt(x) (a bit similar to log(x)/x) then applying the laplace transform, we can find a and b. The integral after that becomes easier to solve.
Thank you for your innovative video.
You can pretty easily show the integral with the complex exponential is equal to the same one but with a negative exponent (gamma function integral form) multiplied by a constant factor using complex integration.
how? can you explain more? i want to try without laplace way
@@192chickenking search "Cauchy integration thereon". It's an analytic function and use a quarter circle contour.
@@Noam_.Menashe thanks, i see . i thought there was another substitution instead of contour integral lol
I hope you get better soon, prof. Bro.
Ah such a satisfying Journey, really brings out the ln(x) in me.
At 7:30 ..... Cos is taken as an imaginary part?? ?? please explain sir...i didn't understand
Phase shift of pi/2 turns a trig ratio into its co ratio
@@maths_505 ohh...yes...got it...mind lag....😂😂
❤
Why is no one in the comments using the Ramanujan Master Theorem???
Nice solution, but i think the Laplace transform part was unncessary, you could have just spammed integration by parts
We watch for elegance, not for vigor
I do also prefer elegance, but using higher level stuff when you can do it equally easy with elementary tools is not that elegant
Still, thats my opinion, yours may differ
Great one. The integral-calculator site had it slightly differently, basically reverting the sign in the parentheses by prepending them by (i - 1)(i + 1) and then rewriting the denominator as 2^(9/2), and rewriting 2*ln(4) as 4*ln(2). I'm amazed that it found an analytical solution. Now I'm curious how that Maxima computer algebra system works.
Integral (-1)^[x]?
This what?
Sweat integral ❤
Sweat
Figured out the solution in few seconds. Still, cool stuff!
Guess I've left a bad impression on the homies😂
Nice one! I used Ramanujan master theorem btw
My way is similar to how you Integrate sinx*lnx/x from 0 to inf.