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# Integrate 1/x^2

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1. Within the boundaries of 1 and -1.
2. 1/x^2=x^(-2)
3. Why is the answer negative?
Spoiler:
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Think about the graph.
What happens when x=0?
4. @Square: Yes, now integrate it within the boundaries of 1 and -1 to give me a number.

@TheTallOne: Dammit, its a trick question! A kinda lame one, I know.
5. >_>
6. I thought it can't be done when the domain of integration includes 0?
7. (Original post by Vivisteiner)
^Yes, now integrate it within the boundaries of 1 and -1 to give me a number.
It doesn't converge. Happy?
8. ans = math error
9. (Original post by Chaoslord)
ans = math error
Not quite sure, but isn't it over 9000?
10. ^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi
11. (Original post by Vivisteiner)
^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi
That's in a simple enough form.
12. (Original post by Vivisteiner)
^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi
Why exactly?

13. (Original post by Vivisteiner)
^lol, CORRECT!

Ok fine, I'll give you another;

calculate e^i*pi
Presumably you mean ei*pi, rather than ei*pi.

The answer is -1. Back to you - prove it
14. (Original post by Agrippa)
Presumably you mean ei*pi, rather than ei*pi.

The answer is -1. Back to you - prove it
I'd be surprised if someone at interview was asked to expand e^(ix) then note the taylor series for the trig functions (although it is quite a pleasing result).
15. Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.

And I don't know how to ruddy well prove it. xD
16. A hack way of doing it is to differentiate

f(x) = e^(ix)/(cosx + isinx) and hence work out what f(x) is.
17. (Original post by Vivisteiner)
Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.
http://xkcd.com/169/

All there is to say on things like that
18. (Original post by Vivisteiner)
Lol, it was a trick. e^(i*pi) = -1 but e^i*pi is already in its simplest form.

And I don't know how to ruddy well prove it. xD
A strange maths troll indeed. An interview with your kind of questions actually would worry me.
19. Lmao, I'm just kinda bored so I asked inane questions and sounded stupid even though I'm obviously incredibly smart and can even differentiate x^x^x^x^x^x in under ten seconds with respect to x.
20. (Original post by Vivisteiner)
Lmao, I'm just kinda bored so I asked inane questions and sounded stupid even though I'm obviously incredibly smart and can even differentiate x^x^x^x^x^x in under ten seconds with respect to x.
Sodding LaTeX

x^x^x^x^x^x (x^(-1+x^x^x^x)+x^x^x^x^x Log[x] (x^(-1+x^x^x)+x^x^x^x Log[x] (x^(-1+x^x)+x^x^x Log[x] (x^(-1+x)+x^x Log[x] (1+Log[x])))))

Done

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Updated: December 8, 2008
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