You can still say something about the first one. Just as for the second one, you can say that it diverges to ∞ (rather than just saying that it diverges), so for the first one you can say that it oscillates between 2 and 0 (rather than just saying that it diverges).
Mr BPRP you and other math tubers I follow must be inspired. Thanks for sharing your knowledge, and all else. My signature: ALL life is chosen, and the Earth belongs to [G_D|DESS]. And math affirms this point, just like everything else, if a person is honest. (Edit: I also like the "Michael Penn" and "Sybermath" channels, among others, I imagine he's also inspired (I say imagine to not be presumptuous but I'm sure they are, and likewise I imagine you are inspired. Thus). Crossposting.
Would it be at all useful to express it as a kind of "bounded divergence"? (Just intuitively it looks like a case can be made that the sin integral is an undetermined number between -2 and 2. So if that's so, at least you could reason that it's definitely _not_ 3, for instance. It has an "interval integral" - or could be so described by a mathematician who knows more than I do, using perhaps tools I've never seen?)
For first one; rewrite the integral as an infinite sum of definite, proper integrals. Each integral should have limits from n*pi to (n+1)*pi, and integrand as sin(x). Such an integral is equivalent to 2*(-1)^n. So, bringing the series back, the improper integral is equivalent to sum of 2*(-1)^n from n=0 to infinity. This is clearly divergent because 2*(-1)^n does not decay to zero (which is a necessary requirement for series convergence.) For the second one; once again rewrite the integral as an infinite sum. This time, rather than 2*(-1)^n we will consistently have 2*(+1)^n = 2. Summing up infinitely many 2s gives divergence to +infinity.
Nah. Another way to see it is by viewing the improper integral as a limit of a proper integral over region (0,A) then taking A to infinity. The proper integral has value 1-cos(A) so as A goes to infinity, cos(A) does not converge (it fluctuates between -1 and 1) so the improper integral cannot be zero.
I enjoy BPRP's channel a great deal, but I can't help but notice a slight solution and graphing error... the integral of sin(x) is cos(x), and the cosine graph starts at (0,1) -- not (0,0) as shown.
“uhm ackshually in the second case we are doing an infinite sum and in some way we can rearrange this to be 1+2+3+4+... which by the Ramanujan sum is -1/12 🤓” (Don’t take it seriously though I also think it’s infinity)
You can still say something about the first one. Just as for the second one, you can say that it diverges to ∞ (rather than just saying that it diverges), so for the first one you can say that it oscillates between 2 and 0 (rather than just saying that it diverges).
Mr BPRP you and other math tubers I follow must be inspired. Thanks for sharing your knowledge, and all else. My signature: ALL life is chosen, and the Earth belongs to [G_D|DESS]. And math affirms this point, just like everything else, if a person is honest. (Edit: I also like the "Michael Penn" and "Sybermath" channels, among others, I imagine he's also inspired (I say imagine to not be presumptuous but I'm sure they are, and likewise I imagine you are inspired. Thus). Crossposting.
Would it be at all useful to express it as a kind of "bounded divergence"? (Just intuitively it looks like a case can be made that the sin integral is an undetermined number between -2 and 2. So if that's so, at least you could reason that it's definitely _not_ 3, for instance. It has an "interval integral" - or could be so described by a mathematician who knows more than I do, using perhaps tools I've never seen?)
But how to prove it without the graphs?
For first one; rewrite the integral as an infinite sum of definite, proper integrals. Each integral should have limits from n*pi to (n+1)*pi, and integrand as sin(x). Such an integral is equivalent to 2*(-1)^n. So, bringing the series back, the improper integral is equivalent to sum of 2*(-1)^n from n=0 to infinity. This is clearly divergent because 2*(-1)^n does not decay to zero (which is a necessary requirement for series convergence.)
For the second one; once again rewrite the integral as an infinite sum. This time, rather than 2*(-1)^n we will consistently have 2*(+1)^n = 2. Summing up infinitely many 2s gives divergence to +infinity.
@@neri1125 first case should be 2 * (-1)^n and second case 2 * (+1)^n
@@dabs4270 Good spot, this is what happens when I dont write mathematics down. Edited :)
try (0
Shouldn't the first case be 0?
No. The limit of sin(x) when x approaches infinity is undefined. The curve never stops. You don't know if it is 0 or not.
Nah. Another way to see it is by viewing the improper integral as a limit of a proper integral over region (0,A) then taking A to infinity.
The proper integral has value 1-cos(A) so as A goes to infinity, cos(A) does not converge (it fluctuates between -1 and 1) so the improper integral cannot be zero.
I enjoy BPRP's channel a great deal, but I can't help but notice a slight solution and graphing error... the integral of sin(x) is cos(x), and the cosine graph starts at (0,1) -- not (0,0) as shown.
Thanks. However I meant to graph sin(x). Since I wanted to interpret the integral of sin(x) as the area under y=sin(x)
“uhm ackshually in the second case we are doing an infinite sum and in some way we can rearrange this to be 1+2+3+4+... which by the Ramanujan sum is -1/12 🤓”
(Don’t take it seriously though I also think it’s infinity)
😂 Ill never believe the ramanujan sum honestly no way it -1/12