This equality holds for *all* real numbers N,M 
Can you see why?pic.twitter.com/LvR2r0y6iT
Show this thread
Beautiful solution by @ValFadeev.
As you can see, the trick is showing that the integral *does not* depend on N.pic.twitter.com/vOJFq5BCNQ
-
-
Just to elucidate how counterintuitive this result is, I've plotted the integrand for different values of N.
All of these curves must admit the EXACT same area!pic.twitter.com/03EsDf4QD9Show this threadThanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
Are you sure, looks beautiful, but I suspect a typo in original problem in the numerator?
-
I've added to the thread a comment that I hope will clarify.
End of conversation
New conversation -
-
-
Grad student doesn't think the integral can be evaluated through analytical means. Implements Simpson's method from Numerical Recipes. Observes that I does not vary with N. Concludes that there must be a bug somewhere. Tries to debug his code until lunch.
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
This solution also makes the reason nice and clear, although integrand changes with N, value at x=1 is always 1/4, so just need a relation between the regions 0<x<1, and 1<x<inf, and map between these is 1/x
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
Simply gorgeous!
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
Things like that show that teaching advanced integrals for engineers is waste of time.
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
blimey! stunned again. o-:
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
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.