The integral over all space of ∫ cos(x²) dx has no right to converge, and yet it does.pic.twitter.com/FfFJIIOR0u
Show this thread
-
This derivation needs LOTS of justification. It diverges absolutely and in L_1. I but exists as an indefinite integral. Do the \int_{-R}^S. A substitution u = ix means that u runs from -iR to +iS. Now rotate the contour, proving that the curves from e.g. -iR to -R -> 0.
I agree with you. A lot of things I tweet need further justification, but I try to fit everything on a single slide and also make it accessible to a wider audience.
-
-
Understood, Twitter is not a very subtle medium. But in this case (IMHO) the wider audience is left only more confused because your claim to show that something converges does not mention in what sense it converges, why it converges, and how that is used in your computation.
#YMMV -
Fair enough. I'll take that into account in future tweets and if I have time create an addition to this one
End of conversation
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.