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# Hard maths question can you help? watch

1. (Original post by keromedic)
Is it 0 or does it just not converge? I thought the former but wolfram outputs the latter.
Check out the graph. It's non-negative everywhere it's defined on that interval, so it couldn't possibly be zero! (let alone negative - all the more reason the method-answer of -2 is completely ridiculous). The issue is it's not defined at zero, and the integral of the function does not converge if 0 is one of the boundaries.

(Original post by Zacken)
You can't integrate over the discontinuity at , right?
Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.
2. (Original post by FireGarden)
Check out the graph. It's non-negative everywhere it's defined on that interval, so it couldn't possibly be zero! (let alone negative - all the more reason the method-answer of -2 is completely ridiculous). The issue is it's not defined at zero, and the integral of the function does not converge if 0 is one of the boundaries.

Isn't the non-thinking answer the one I provided? I.e. 0. As in . The graph is symmetrical and because definite integrals are the signed areas, I thought you'd have a net result of 0 because the bits above and below the x-axis would cancel. My bad.

Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.
I shall entertain myself by googling some more about discontinuities.
3. (Original post by keromedic)
I shall entertain myself by googling some more about discontinuities.
, we have for all real , so the graph always lies above the -axis, not quite sure what you're saying about the negative cancelling out the positive, etc...

Furthermore, , so .

I think you're confusing the function we are integrating, it's 1/x^2. Probably just an oversight on your part.
4. (Original post by FireGarden)
Sort of. Just having discontinuities is not a problem in itself - you can just add together the integrals of the continuous pieces. The problem is that this particular discontinuity blows up "badly enough" that if it's a boundary of an integral, the integral doesn't converge. So adding the continuous pieces doesn't work, because the integrals of the continuous pieces don't exist.

Well, yes, that's what I was implying by "you can't integrate over the discontinuity" - I'm aware that you can integrate over other, better behaved discontinuities (such as some integrands with trigonometric functions in the denominator), which I learned from hanging around TPIT back in 2013.
5. (Original post by Zacken)
, we have for all real , so the graph always lies above the -axis, not quite sure what you're saying about the negative cancelling out the positive, etc...

Furthermore, , so .

I think you're confusing the function we are integrating, it's 1/x^2. Probably just an oversight on your part.
6. (Original post by keromedic)
Happens to the best of us.
7. (Original post by Zacken)
Happens to the best of us.
Forgot that 1*-1=-1 for some reason .

As for my initial point, it was that I felt that . It then followed naturally in my reasoning at the time that the result would be 0 as they'd cancel. See my crude image below.

8. (Original post by keromedic)
Forgot that 1*-1=-1 for some reason .

As for my initial point, it was that I felt that . It then followed naturally in my reasoning at the time that the result would be 0 as they'd cancel. See my crude image below.

I understand your reasoning, it's just that I'm questioning why your sketch looks like , clearly lies above the x-axis all the way through.
9. (Original post by Zacken)
I understand your reasoning, it's just that I'm questioning why your sketch looks like , clearly lies above the x-axis all the way through.
Because I was wrong. It's been too long.
10. (Original post by keromedic)
Because I was wrong. It's been too long.
Ah, okay. That's fine then. I once had a class test where I wrote . - silly mistakes, the death of us all. xD

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