A crazy yet perfect integral

[@zunaidparker]

Amazing. Getting to a linear differential equation midway through the problem was pretty damn cool. And gotta love the dad joke at the end!

[@maths_505]

😂😂😂 Thanks mate

[@daniellindner826]

SICKKKK. integration really is an art.

[@Sugarman96]

After the u-sub, you get the product of two functions with known Fourier transforms, so the obvious approach is to use Plancherel theorem. The fourier transform of a Gaussian is a Gaussian, and the fourier transform of 1/(1+x^2) is e^(-|w|). Then you use the fact that the integrand is even to dumb the absolute value and be left with a Gaussian integral.

[@Priyam_B]

One the best videos I've experienced till now, maths is so cool and u make it even more amazing.

[@piplus2]

Amazing video! I just realised that after the first substitution, one can write I = exp(1) * integral exp(-(y^2 + 1)) / (y^2+1) dy which is closely related Owen's T function.

[@uggupuggu]

it came up in a problem i solved

[@quentinrenon9876]

Thank you very much! It's really amazing how everything comes together. I did it a somewhat different way and got the same result so that's cool!

[@TheMartinbowes]

It's the choice of a second order variable for the Feynman technique that has me intrigued. Doing it with a standard first order gets messy! Well played sir.

[@maths_505]

That's basically what I tried initially but it didn't work out

[@TheMartinbowes]

My final integral was of (e^-x)/✓x dx between 0 and t. A substitution of x = u^2 gets what you need ... But I'm not sure how to handle the bounds.

[@ericthegreat7805]

This made me realize that integration by parts is really a type of differential equation.

[@samning4141]

Wacky, wild, and fantastic.

[@MathOrient]

Outstanding end result :) Watched it all.

[@kappasphere]

Me watching Math 505 justify why 1-1=0 in the middle of his Feynman integration 👀

[@gaspell]

This is insane. Thank you very much Very impressive

[@holyshit922]

Firstly I observed that integrand is even on the interval symmetric around zero Second I substituted t = tan^2(x) Then i decided to play with Laplace transform I considered following function 1(t-1)/((sqrt(t-1)t) Laplace transform of this fuction gave me integral Int(exp(-sx)/(sqrt(x-1)x),x=1..infinity) Now I changed this single integral to double integral int(int(exp(-(s+y)x)/sqrt(x-1),x=1..infinity),y=0..infinity) int(exp(-(s+y)x)/sqrt(x-1),x=1..infinity) t = x - 1 dx = dt int(exp(-(s+y)(t+1))/sqrt(t),t=0..infinity) =exp(-(s+y))Int(exp(-(s+y)t)/sqrt(t),t=0..infinity) L(t^r) = Gamma(r+1)/s^(r+1) =exp(-(s+y))*Gamma(1/2)/(s+y)^(1/2) =exp(-(s+y))*Gamma(1/2)/sqrt(s+y) Gamma(1/2)Int(exp(-(s+y))/sqrt(s+y),y=0..infinity) s+y = u^2 dy = 2udu Gamma(1/2)Int(2uexp(-u^2)/u,u=sqrt(s)..infinity) 2Gamma(1/2)Int(exp(-u^2),u=sqrt(s)..infinity) This should give me Pi(1-erf(sqrt(s))) Now after using shifting property i had got Pi*exp(s)*(1-erf(sqrt(s))) Then after plugging s = 1 i had got the result PI*e*(1-erf(1)) Maybe Laplace transform is not necessary but double integral avoid this linear ode =2Int(exp(-tan(x)),x=0..Pi/2) u=tan(x) =2Int(exp(-u^2)/(1+u^2),u=0..infinity) =2Int(exp(-x^2)/(1+x^2),x=0..infinity) 1/(1+x^2) = Int(exp(-(1+x^2)y),y=0..infinity) 2Int(exp(-x^2)*Int(exp(-(1+x^2)y),y=0..infinity),x=0..infinity) 2Int(Int(exp(-x^2)*exp(-(1+x^2)y),y=0..infinity),x=0..infinity) 2Int(Int(exp(-x^2)*exp(-(1+x^2)y),x=0..infinity),y=0..infinity) 2Int(Int(exp(-y)*exp(-x^2(1+y)),x=0..infinity),y=0..infinity) 2Int(exp(-y)*exp(-x^2(1+y)),x=0..infinity) 2exp(-y)*Int(exp(-x^2(1+y)),x=0..infinity) u^2 = x^2(1+y) u=x*sqrt(1+y) sqrt(1+y)du=(1+y)dx dx = 1/sqrt(1+y) 2exp(-y)/sqrt(1+y)Int(exp(-u^2),u=0..infinity) Int(exp(-u^2),u=0..infinity) u^2=w 2udu=dw 2w^(1/2)du=dw du=1/2*w^(-1/2) 1/2*Int(w^(-1/2)exp(-w),w=0..infinity) 1/2*Gamma(1/2) Gamma(1/2)*Gamma(1-1/2) = Pi/sin(Pi/2) Gamma(1/2) = sqrt(Pi) Int(exp(-u^2),u=0..infinity)=1/2*sqrt(Pi) Int(sqrt(Pi)exp(-y)/sqrt(1+y),y=0..infinity) sqrt(Pi)Int(exp(-y)/sqrt(1+y),y=0..infinity) u = sqrt(1+y) u^2=1+y u^2 - 1 = y 1 - u^2 = -y 2udu = dy sqrt(Pi)Int(2uexp((1-u^2))/u,u=1..infinity) 2sqrt(Pi)exp(1)*Int(exp(-u^2),u=1..infinity) 2sqrt(Pi)exp(1)*(Int(exp(-u^2),u=0..infinity) - Int(exp(-u^2),u=0..1)) 2sqrt(Pi)exp(1)*(1/2*sqrt(Pi) - Int(exp(-u^2),u=0..1)) 2sqrt(Pi)exp(1)*(1/2*sqrt(Pi) - 1/2*sqrt(Pi)erf(1)) =Pi*exp(1)*(1-erf(1))

[@mathscribbles]

Bro you took us on an adventure

[@taterpun6211]

A similar path would be to do int(e^-x^2/(x^2+1))dx = eint((e^-(x^2+1)/(x^2+1))dx and use Feynman technique

[@dukenukem9770]

Beautiful!

[@txikitofandango]

The version of the joke I heard was: if you have an apple, then you have one apple, which means that x = 1x.

[@anonymous_0416]

when the integral is \int_{0}^{π/2}e^{-tan(x)}dx then how to evaluate ?

[@qsfrankfurt9513]

Bro which software are you using for writing out this mathematics?

[@robertsandy3794]

What sort of trick could you use if you couldn't bring the differential into the integral as a partial differential?

[@manstuckinabox3679]

Idk if this can be done with contour integration, I tried it and only got e*pi... but great video! I'm can feel the power of The Feynmann resonating! soon I'll be able to destroy any integral by just starring at it! instead of writing t for the parameter, it will be come t for trivial.

[@maths_505]

Inshallah

[@davidblauyoutube]

I also tried contour integration and got e*pi as well. Something fishy is going on here.

[@manstuckinabox3679]

نشكر الله على الأخ فاينمان Maths 505@

[@ARBB1]

What's the justification for transforming "u" back into "x" without any work?

[@co.c9412]

i think it's because it's a dummy variable

[@ガアラ-h3h]

Tbh looking at it I thought it could be done with gamma function but I kinda got stuck

[@Krishnajha20101]

Nice content. I just had a thought: this solution may have been a bit simplified if you just make f(t) to be exp(-tx^2) instead of exp(-t^2 x^2).

[@devd_rx]

Praise the Feynman

[@Calcprof]

This seem unnecessarily round about. I think you can go directly from the original integral (after the 1st tan substitution) to Erf (or Erfc)

[@MrWael1970]

Wacky integral, but perfect solution. Creativity is your trend. Thanks

[@tolberthobson2610]

Beautiful Result :)

[@조현욱-e5j]

not being unrepectful, but isn't the result beautiful and clean only because we have defined the dirty integral as erf?

[@ashishraje5712]

Great masterpiece

[@MrPabheredia]

When you multiply by e om both sides, why the denominator 2 dissapears?

[@hanibahout2217]

because the function is even he calculated half of it on the interval from zero to infinity and at the end he multiplied by 2 to get the good result.

[@anjanbiswas302]

Wonderful !!!

[@mathsrelated]

Is tan squared x even or odd function?? Similarly sin squared x? Even or odd?

[@zachbills8112]

Even

[@RanBlakePiano]

Creative !

[@cristoduke8953]

I think he forgot to multiply the integral by 2 in 1:57

[@orionspur]

2 returns at 2:52.

[@鲍辉仔]

I have a solution without solving the differential equation

[@aadarshpawar2632]

❤❤❤

[@orionspur]

P...erfc...t

[@whilewecan]

Impressive.

[@mrQueue78]

Okay cool

[@SatyanarayanaMudunuri]

pl write clearly. for example. your 2t and ∂t look the same. Also the change of pen in between is very distracting.