A tricky integral from MIT

This one's from the semi finals of the 2024 MIT integration finals. Full solution development leading to a beautiful result.

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Advanced MathWear:

[Mathematician]

Hey I solved it in a different way. It's an interesting one. See patiently.

Let I(a) = intgrl e^(-ax^2) /(x^2 + 1/2)^2 dx from 0 to equ 1

See we can split this function into

2 { e^(-ax^2) /(x^2 + 1/2) - x^2 e^(-ax^2)/ (x^2 +1/2)^2 }

So

I(a) /2 = intgrl e^(-ax^2) /(x^2 + 1/2) dx from 0 to infinity - intgrl x^2 e^(-ax^2)/ (x^2 +1/2)^2 dx from 0 to infinity

From equ 1,
I'(a) = intgrl - x^2 e^(-ax^2)/ (x^2 +1/2)^2 dx from 0 to infinity

So, we can say

I(a) /2 = intgrl e^(-ax^2) /(x^2 + 1/2) dx from 0 to infinity [***now let this be S] + I'(a) .... equ 2

Now let's do integration by parts for I'(a), take xe^(-ax^2) to be differentiated and - x /(x^2 + 1/2)^2 to be integrated

After some calculations,

You will find I'(a) = intgrl ae^(-ax^2) dx from 0 to infinity - (a+1)*S /2

I'(a) = sqrt(api) /2 - (a+1)*S /2

So, we get S= sqrt(api) /(a+1) - 2I'(a)/(a+1)

Put this in equ 2, simplyfying we get


I(a) /2 = sqrt(api) /(a+1) + (a-1)I'(a)/(a+1)

put a=1

I(1)/2 =sqrt(pi) /2 +0

I(1)=sqrt(pi)

[CM_France]

hi,

"ok, cool" : 2:45, 4:42, 5:48, 6:21, 7:40, 8:30,

"terribly sorry about that" : 2:57 .

[lakshay]

2:35 Here we have (1+2x²)² in the denominator, we can replace it with the derivative of the geometric series and then put x²->x and we have the gamma function directly

[MrWael]

Thank you for your featured effort. It is a nice approach.

[Babyshark-coks]

How do they possibly expect you to find the idea AND evaluate these integrals in like 4 minutes?!

[holyshit]

1=2*1/2
1/2 = 1/2 + x^2 - x^2
-
Integration by parts
then there are multiple ways such as double integral or reflection formula for Gamma

[polpotify]

Eid Mubarak and thanks for such an interesting integral

[julianwang]

Alternatively, to evaluate the original integral K, consider J = integral_0^inf{dx e^(-x^2)/(x^2+1/2)}=K/2 + integral{dx x^2 e^(-x^2)/(x^2+1/2)^2} = K/2 - x e^(-x^2)/2/(x^2+1/2)|_0^inf + J/2 - integral{x^2 e^(-x^2)/(x^2+1/2)}. The second term is just zero, and the last term equals -integral{e^(-x^2)} + J/2 = sqrt(pi)/2 + J/2. Putting everything together, we have J = J + K/2 - sqrt(pi)/2. So, obviously K = sqrt(pi).

[divyaanshu]

What if we try substituting x=(1/sqrt2)tant, then taylor expansion of exp to form a beta function to eventually lead to gamma func, applying reflection formula to get infinite summation of terms

[adoanmian]

Which calculus books are you using to solve these crazy integral problems?

[xleph]

At 6:41 I don't think you explained very well what happened to the scalar 2 outside the integral. When you brought in the 2 (to make t -> 2t and 2t -> 1 + 2t - 1) it was obvious what you were doing. It's not really bothersome, but more verboseness would have been nice.

[momchi]

The solving is fine and all, it's doable, but how did they expect someone to just spontaneously think of the conversion idea? You need to just have that integral of t*e^-ktdt from 0 to infinity is 1/k^2 in the back of your head just in case? Or do they really just want to stop anyone who doesn't have Ramanujan like intuition on how to solve equations?

[edmundwoolliams]

Could you use Feynman trick (before doing anything else)? It looks like it's asking to be differentiated twice to cancel that denominator!

[TheAzwxecrv]

Wonderful! U see, my friend, I'm a physicist, but I want to plan a course MATH / PHYS 505 Advanced Methods of Integration or simply Advanced Integrals, and collect all your videos to present in the course!

[silas]

does anyone have an explanation for the simplification done to the denominator at 2:25 ?

[aleksandervadla]

Could you use planckerals identity and fourier transform on this?

[giuseppemalaguti]

Con feyman ho l'integrale .I(0)=π/√2..adesso faccio gli altri calcoli.mi risulta l'equazione ah goodnight..ho rifatto i calcoli e sono arrivato alla equazione si risolve se il termine forzante è una radice e non un polinomio in a?

[archiebrew]

i found a very quick way,

Let I = integral of (e^-(x^2))/(x^2+1/2) from 0 to infinity

Integrate by parts with u = (e^-(x^2))/(x^2+1/2) and dv = 1

The answer pops out immediately.

[hutus]

Can someone solve this problem by using Cauchy integral ?

[holyshit]

You complicated a lot this integral
That is result of avoiding integration by parts with all costs