## Proving F(x) = ∫[0,x] exp(-t^2) is odd.

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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 Quote by peripatein 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).
It wasn't approval. It was a confession that the equality you wrote is true even though I have no idea how you got it. What I wanted to know is what you got for the du integral following the steps in post 10.

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 Quote by peripatein 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).
##F(x) = \int_0^x e^{-t^2}dt##

Hence ##F(-x) = \int_0^{-x} e^{-t^2}dt##

What you need to do now is to evaluate ##\int_0^{-x} e^{-t^2}dt##

To do that, substitute u = -t. You haven't actually done that yet. You will end up with an integral with respect to u. You also need to take into account what happens to the bounds of definite integration when you make the substitution (remember, everything needs to be expressed in terms of u).

Do this and write what you get.
 What I wrote, to which you acquiesced, was precisely what I obtained via that substitution.

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 Quote by peripatein What I wrote, to which you acquiesced, was precisely what I obtained via that substitution.
If you did then you showed that F(x)+F(-x)=0. That would pretty much finish it.
 Let me write it again, for clarity's sake: int [0,x] exp(-u^2) du int [0,-x] exp(-u^2) du.
 There should be a "+" there, between the two integrals
 And that is not a proof! That remains to be proven! Once proven, and only then, could it be affirmed that F is indeed odd.

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 Quote by peripatein And that is not a proof! That remains to be proven! Once proven, and only then, could it be affirmed that F is indeed odd.
##\int_0^{-x} e^{-t^2}dt##

You STILL haven't evaluated this integral by the suggested substitution! The substitution u = -t hasn't actually been performed (or, if it has, it has not been done correctly). If you did it correctly, you'll find that the upper bound becomes "x" (rather than "-x").
 What I keep getting, when I perform the suggested substitution, is F(x)=-int[0,-x]exp(^-u^2) du, and -F(-x)=int[0,x]exp(-u^2) du.

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 Quote by peripatein What I keep getting, when I perform the suggested substitution, is F(x)=-int[0,-x]exp(^-u^2) du, and -F(-x)=int[0,x]exp(-u^2) du.
You should learn how to write integrals in LaTeX... They're pretty easy. The following would go between two sets of  characters (no space between each pair):
\int_0^{-x} e^{-u^2}du

The above produces this output:
$$\int_0^{-x} e^{-u^2}du$$

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 Quote by peripatein What I keep getting, when I perform the suggested substitution, is F(x)=-int[0,-x]exp(^-u^2) du, and -F(-x)=int[0,x]exp(-u^2) du.
You don't have to do the substitution for F(x). Just leave it as ##\int_0^x e^{-t^2}dt##.

You only have to do the sub for F(-x). You should get the result that this is equal to ##-\int_0^x e^{-u^2}du##. I'm assuming you got this because your second expression is equivalent to this (you just put a minus sign on it).

Now you need to recognise that in a definite integral, the independent variable and the element of integration are simply dummy variables, and can be replaced by any other symbol (as long as it's done consistently). In other words, ##-\int_0^x e^{-u^2}du = -\int_0^x e^{-t^2}dt##.

You should now be quite easily able to see that F(x) = -F(-x).