A crazy yet perfect integral

Here's another awesome integral with a beautiful solution development using Feynman's trick and the error function.

[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!

[daniellindner]

SICKKKK. integration really is an art.

[Sugarman]

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.

[worldnotworld]

This sort of mathematics aesthetically resembles nothing more than Baroque art. Ornaments, colors, figurations all constantly re-appearing with surprising but always perfect appropriateness: halos, angels, Virgin Mary's, castles in the distance, travelling kings, ornate trills, rampant complex polyphony, naked cherubs fluttering everywhere (the square root of pi is a good example of a naked cherub), the incarnate Logos lying in a manger (the Second of Three Roots of Unity), peasants out walking and dancing, flowers carved in stone... it's all there!

[quentinrenon]

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!

[piplus]

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.

[MathOrient]

Outstanding end result :) Watched it all.

[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.

[ericthegreat]

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

[gaspell]

This is insane. Thank you very much
Very impressive

[kappasphere]

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

[holyshit]

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)


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))

[taterpun]

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

[anonymous_]

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

[ARBB]

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

[Krishnajha]

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).

[ガアラ-hh]

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

[robertsandy]

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

[txikitofandango]

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