what's the integral of 1/x from -1 to 1

[shubhbisht]

use of blue pen is not allowed
only black or red

[neopalm]

Come on guys, it is obviously ∀n|n∈ℤ (2n+1)iπ.

[yugeshkeluskar]

Can you explain more deeply about cauchy principle value

[Cannongabang]

1) 1/x is NOT odd in [-1, 1]
it is an odd function only in [-1, 0)∪(0, 1], so you cannot easily say "integral is zero cause it is odd"
2)not all odd functions are integrable: try to integrate tan(x) from -π/2 to π/2 using common fundamental th. of integrals.
unlimited functions are not integrable via this technique
3) improper integrals must be split and calculated separately, they exist ONLY if the sum of each separated meaningful piece MAKES sense.
4) Cauchy principal value is for another time i guess.. hope blackpenredpen wants to do a vid about it

[killerchicken]

Don’t get me wrong, I love Eddie Woo but bprp deserves more recognition

[andrewdeighton]

I thought that the integral, as written, is invalid - since 1/x is undefined at zero.

[sethgrasse]

Take the limit as x approaches infinity of x-x. You could say it diverges since it goes to infinity-infinity, or you could simply it and get 0.

[kalunlee]

the odd function properties can only be applied for continues odd function, and 1/x is not continuous at x=0.Hence this integral cannot apply odd function properties to argue that is =0 .In this case the integral can be any value in the range of (-inf, +inf) depends on how you approach the singularity, means this is a divergent integral.However when the "approaching speed" of the single finite singularity is the same, then the value calculated is called Cauchy principal value(In this case is equal to 0)

[OonHan]

bprp = {
1. cool
2. awesome
3. amazing
4.
...
}

infinite possible answers

[Jonasz]

Cauchy principal value integral is also used in fluid dynamics to compute flow over non-uniform surfaces. Physics teacher don't quite bother with the theory behind it, but you understand the general idea pretty quickly.

[danielescotece]

It does not really make sense to talk about PV(1/x) without introducing the theory of distributions or Lebesgue integration, but you did a good job with that sqrt (-9) analogy!

[GrouchierThanThou]

Actually, strictly speaking, although both 3i and -3i are solutions to the equation r² = -9, neither of them is the square root of -9. This is because √x is only defined for non-negative real x. It is defined as the non-negative root r of the equation r² = x. That's why the square root of 9 is 3 rather than -3.

The complex numbers are not an ordered set and so there is no such thing as a non-negative complex number. Consequently the definition of the square root does not even make sense for the complex numbers and in fact, even for the complex numbers, √-9 is still undefined.

[benburdick]

Do I see an explanation of Cauchy principle values in the future?

[Rekko]

I have always thought that integrals cannot be negative. They are areas, so they cannot be less than zero.

[devsanchla]

Don't u find the area below the curve in such integrals? And here both curves below and above are symmetrical so can't u just say that's double the area above? So it shouldn't be zero. Genuinely confused

[non-inertialobserver]

Just like the sum of the naturals is divergent, but the ramanujan sum of the naturals is - 1/12

[gvssen]

You cleared some of the air arising out of the previous video. Forgive me for being dumb but I have to raise this. All this simplifies to what is the value of Limit(X reaching infinite) of [X - X]. As long as X is finite the answer is zero. If not??? ARE we treating infinite having the properties of yet another finite number? Infinite as my teacher in basic maths classes told me was it is larger than the largest number you can think of. In fact there is no end. We know that as long as it is finite we are subtracting two equal numbers but can equality of two infinte numbers be taken for granted? One clever way to get out is to say that however large it may be we are subtracting an equal amount from a given number and if only so the answer is zero and in doing so we are applying the property of a finite numbers to infinite unintentionally. I will not talk of Cauchy at length because the comment will become large. However, as you rightly pointed out it is yet a methodology of giving new definitions to the hitherto undefined. THANKS FOR COMPLETING THE DISCUSSION.

[Qermaq]

Wolfram Alpha says it diverges but it gives the Cauchy principal value as 0. I think the decision on the answer depends on why we need the answer. Too often we pure mathematicians forget the engineer in the corner asking "what are we measuring?".

[pituitlechat]

in the same way I{-inf to inf} 1 dx = divergent or 0?
for me it's 0 because lim {a->inf}[ I{-a to a} 1 dx ] = lim {a->inf} [a -a] = 0
and this is also true ∀a, not just a=inf

[TheReligiousAtheists]

Can't we argue wrt symmetry? As in if you look at the integral of 1/x from -1 to 0 and from 0 to 1, it's the same graph, but upside down. So if you take the integral from -1 to something very close to 0 (i.e. lim---> 0-) and from something really close to 0 to 1 (i.e. lim--->0+) the 2 areas will cancel out perfectly to give 0.