Nice, one question, 2:22 why can we just substitute u with x and not with the before defined substitution? (wondering why it wouldnt be x^2 instead 1/x^2)
In an integral, you can change the variables completely any time. It doesn't matter whether we write "x" or "u" or "y". I understand your point, he used variable "u" to simplify the first integral. Once he changed everything in terms of "u" (including the limits), after that point it's not wrong to change all the variables back to "x". You got it?
Hi. Good explanation by suhail above. Also thinking of the definite integral as area under the curve then changing the variable name won't change the area. It's the same graph with different variable names.
It is interesting to see how you develop it. Technically you could make the expansion to x-1/x +4 in the original integrand already, than converting the bounds to -inf to inf and dividing by 2 you are free to use glasser's master theorem which zeroes the -1/x term and you are done at this point
They didn't want to give us x-1/x up there, eh? Then it would've been 2I = full gaussian integral = √π => *I = √π/2* OR, we can make our integral go from -∞ to ∞ and bring a 1/2 in front because the integrand is even, and then use Glasser's master theorem to make it the full gaussian integral, times the 1/2, bringing us to the same solution
i got to use the gamma function somehow lol. i like your solution more than mine, so gj!
Thanks!!!
The move at 5:05 is absolutely AMAZING👏👏👏
Thanks Doron. Seems like that part is the key to the whole thing :)
Nice, one question, 2:22 why can we just substitute u with x and not with the before defined substitution? (wondering why it wouldnt be x^2 instead 1/x^2)
In an integral, you can change the variables completely any time. It doesn't matter whether we write "x" or "u" or "y". I understand your point, he used variable "u" to simplify the first integral. Once he changed everything in terms of "u" (including the limits), after that point it's not wrong to change all the variables back to "x". You got it?
Hi. Good explanation by suhail above. Also thinking of the definite integral as area under the curve then changing the variable name won't change the area. It's the same graph with different variable names.
It is interesting to see how you develop it. Technically you could make the expansion to x-1/x +4 in the original integrand already, than converting the bounds to -inf to inf and dividing by 2 you are free to use glasser's master theorem which zeroes the -1/x term and you are done at this point
Hi. But only works if you have the minus sign instead of the plus right?
@owlsmath yes, exactly it should be -1/x and not +1/x
nice video!
Thanks! 🙏
That was great!
thanks Mike! appreciate it :)
Great method!
thanks David! 👍👍
They didn't want to give us x-1/x up there, eh? Then it would've been 2I = full gaussian integral = √π
=> *I = √π/2*
OR, we can make our integral go from -∞ to ∞ and bring a 1/2 in front because the integrand is even, and then use Glasser's master theorem to make it the full gaussian integral, times the 1/2, bringing us to the same solution
right! That's what I found interesting is with that little change it makes it a bit more complicated
The reference to Glasser's master theorem is very helpful here. What a nice shortcut!
glasser's master theorem won't work for this video, since we add a fraction, not subtract it. but it's a very nice trick nonetheless!
I hate gaussian function that bell 🔔 type shit
🤣🤣🤣 yep its definitely all that :)
Love the Bell curve, and it will love you back😂