
#1
Jan2413, 03:18 PM

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Hi,
1. The problem statement, all variables and given/known data Any idea how I might demonstrate that F(x)=∫ [0,x] exp(t^2) dt is odd? 2. Relevant equations 3. The attempt at a solution Will it be by showing that ∫[x,x] exp(t^2) = 0? 



#2
Jan2413, 03:22 PM

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#3
Jan2413, 03:23 PM

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#4
Jan2413, 03:30 PM

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Proving F(x) = ∫[0,x] exp(t^2) is odd.
I know that a function is odd if F(x)=F(x). But how may I show that that is indeed the case for the above function?




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Jan2413, 03:33 PM

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#6
Jan2413, 03:38 PM

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It's fairly easy to show that a function is odd if and only if its derivative is even. And, by the "fundamental theorem of Calculus" [itex]F'= e^{x^2}[/itex].




#7
Jan2413, 04:08 PM

P: 812

Actually, I did initially manage to show that the derivative was even. Problem is, I had no idea that would mean the function itself was odd!
For you, HallsofIvy, it might be fairly easy. I would have to mull over it. May I demonstrate that by simple integration on both sides, or is it more intricate than that? 



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Jan2413, 04:33 PM

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#9
Jan2413, 04:40 PM

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I did that too, Dick! Leaves me with a proof that int[0,x]exp(t^2)=int[0,x]exp(t^2), which I am not sure how to approach.
In my previous reply I was wondering how to demonstrate even derivative implies odd function. 



#10
Jan2413, 04:45 PM

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#11
Jan2413, 05:01 PM

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Wouldn't that leave me to prove that int[0,x] exp(u^2)du + int[0,x]exp(u^2)du=0?




#12
Jan2413, 05:08 PM

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#13
Jan2413, 05:13 PM

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I am sorry, but I fail to see how that got me closer to proving the original statement. I mean, how is this integral different than the one with dt? What's eluding me?




#14
Jan2413, 05:16 PM

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#15
Jan2413, 05:39 PM

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Could you please explain to me why I could substitute any even function? I understand that the proof would still work, but would that be rigorous enough? What would the formal justification for such a maneuver be?




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Jan2413, 05:45 PM

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#17
Jan2413, 06:04 PM

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Having f(0)=0 is a sufficient but not necessary condition for f(x) being an odd function. Example: f(x)=1/x is odd, but f(0) does not exist. 



#18
Jan2413, 06:04 PM

P: 812

Okay, then I am still left with the seeming difficulty in seeing how the du substitution facilitated things. I believe you have expressed your approval of the expression I obtained having performed that substitution (post 12).



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