We can find I without using the Gamma function by writing I as int of (1/(sqrt i)) ( e^(-(sqrt i)x)^2) d((sqrt i) x) = sqrt pi /2 (sqrt i) by the Gauss formula
This is how I solved them, by starting from the Laplace transform, making a substitution of x=-iu, and then equating the real and imaginary parts. The only problem I had with this is that when making that substitution, the limit of the integral should go from infinity to (i)*infinity. But if we just treat i as a real number the limit stays as infinity and we end up with the right solution. Is there an explanation on why this works and why we're allowed to do it?
You don't have to make that substitution. The "s" parameter of the laplace transform is a complex number so you can let s=i. That'll get rid of pesky limit transforms.
I think something which should have been mentioned is that (i)^(-0.5) can represent two different complex numbers and we are looking for the one which has positive real part and negative imaginary part. So, for your approach to be fully rigorous I think it needs to be shown independently that the two Fresnel integrals are actually non-negative. Can you give a short proof of this? Other than that, nice work!
Your comment reminded me when my father gave me a book of logarithms and a slide rule in elementary school some 60 years ago. Also, we lived near the Chemical Rubber Company (CRC) who gave all the kids a copy of “the CRC” which was a math/physics reference. I wanted nothing more than to learn the meaning of all the mysterious symbols.
@@maths_505 Wow. Is this a US term for the imaginary unit perhaps? In all my life studying maths (school, university degree, Master's degree, and 30 years of subsequent study), I have never heard the imaginary unit being called iota. Every day's a schoolday.
EXTREMELY annoying that you say iota to i !! As it’s NOT! Iota is the 9th letter in the Greek alphabet, i is i representing sqrt(-1). The movie IS a very nice solution though!
![NISAMANEE - TADA (ท้าดา) [ OFFICIAL MV ]](http://i.ytimg.com/vi/hcQHEPMmJpU/mqdefault.jpg)
As a high school student I really enjoy your videos on improper integrals . Subscribed !
High school? I was thinking this to be better suited for middle school...
😢😢middle school???@@jacobguerreso675
We can find I without using the Gamma function by writing I as int of (1/(sqrt i)) ( e^(-(sqrt i)x)^2) d((sqrt i) x) = sqrt pi /2 (sqrt i) by the Gauss formula
Oh that is quite nice
Please don't confuse , call i , not iota
Correct - because the Greek. was never used for this - the original is the Latin i
Iota seems to be popular in India but not used elsewhere.
This is how I solved them, by starting from the Laplace transform, making a substitution of x=-iu, and then equating the real and imaginary parts. The only problem I had with this is that when making that substitution, the limit of the integral should go from infinity to (i)*infinity. But if we just treat i as a real number the limit stays as infinity and we end up with the right solution. Is there an explanation on why this works and why we're allowed to do it?
You don't have to make that substitution. The "s" parameter of the laplace transform is a complex number so you can let s=i. That'll get rid of pesky limit transforms.
@@maths_505 But this is full-blown complex integration, in disguise.
the Laplace approach was nice (I would've gone for Gaussian integral)
Nice solution. Keep it up
Tnx bro very good explanation
Magnificent
I think something which should have been mentioned is that (i)^(-0.5) can represent two different complex numbers and we are looking for the one which has positive real part and negative imaginary part. So, for your approach to be fully rigorous I think it needs to be shown independently that the two Fresnel integrals are actually non-negative. Can you give a short proof of this?
Other than that, nice work!
You're right but I thought that was kinda trivial at this level so I just skipped mentioning it.
super easy thank youuuu
Amazing 😍
My words exactly on solving it
very nice question
See this simpler method by Penne/Chen Hongwei without Laplace and Gamma function.
th-cam.com/video/WHawX2fQjg4/w-d-xo.html
Hi, which program are you writing on?
It's just the default samsung notes app
I didn't understand but I'm sure it's epic
😂😂😂
Don’t feel bad about using “iota” ; I still prefer to use my book of logarithms and slide rule to do some calculations.
That's actually pretty cool
You haven't been active on Instagram lately.
Everything alright?
Your comment reminded me when my father gave me a book of logarithms and a slide rule in elementary school some 60 years ago. Also, we lived near the Chemical Rubber Company (CRC) who gave all the kids a copy of “the CRC” which was a math/physics reference. I wanted nothing more than to learn the meaning of all the mysterious symbols.
COOL.
How do these even converge? It makes no sense. Its infinite oscillations
iota?
It's the imaginary unit
@@maths_505 Wow. Is this a US term for the imaginary unit perhaps? In all my life studying maths (school, university degree, Master's degree, and 30 years of subsequent study), I have never heard the imaginary unit being called iota. Every day's a schoolday.
@@rtatt1 I'm south asian so I'm not sure whether this is an American thing but I read this back in freshman year so it just stuck with me
EXTREMELY annoying that you say iota to i !! As it’s NOT! Iota is the 9th letter in the Greek alphabet, i is i representing sqrt(-1).
The movie IS a very nice solution though!