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

[peripatein]

Hi,

Homework Statement

Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?

Homework Equations

The Attempt at a Solution

Will it be by showing that ∫[-x,x] exp(-t^2) = 0?

 

[Mark44]

Hi,

Homework Statement

Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?

Homework Equations

Wouldn't the definition of "odd function" be relevant?

The Attempt at a Solution

Will it be by showing that ∫[-x,x] exp(-t^2) = 0?

No, I don't think so.

 

[Dick]

Hi,

Homework Statement

Any idea how I might demonstrate that F(x)=∫ [0,x] exp(-t^2) dt is odd?

Homework Equations

The Attempt at a Solution

Will it be by showing that ∫[-x,x] exp(-t^2) = 0?

That's not true. You do it by showing F(-x)=(-F(x)).

 

[peripatein]

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?

 

[jbunniii]

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?

Try substituting -x for x and see what you get.

 

[HallsofIvy]

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" F'= e^{-x^2}.

 

[peripatein]

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?

 

[Dick]

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?

The way you show it is basically the same as jbunniii's suggestion. Write the integral for F(-x) and do the substitution t->(-t) in the integral.

 

[peripatein]

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.

 

[Dick]

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.

Take int[0,-x]exp(-t^2)dt. Do the u-substitution u=(-t), du=(-dt). And don't forget to change the limits. If t goes from 0 to -x, what are the limits for u? Since F'(x)=exp(-t^2) you can substitute any even function for exp(-t^2) and the same proof works.

 

[peripatein]

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

 

[Dick]

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

That would be true. Why are you expressing it as a sum?

 

[peripatein]

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?

 

[Dick]

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?

I don't know. Can you show me what you get if you follow the suggestions in post 10? You should just get one transformed integral with a u variable instead of t. What is it?

 

[peripatein]

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?

 

[Dick]

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?

Mmm. I just meant that once you prove it then if they give you f(t) and tell you f(t) is even instead of exp(-t^2) the same proof works. I wasn't trying to say anything fancy.

 

[D H]

It's fairly easy to show that a function is odd if and only if its derivative is even.

I disagree with the "if" part. That f'(x) is an even function is a necessary but not sufficient condition for f(x) being an odd function. Example: f(x)=1 has an even derivative but obviously isn't an odd function.

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.

 

[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).

 

[Dick]

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.

 

[Curious3141]

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.

 

[peripatein]

What I wrote, to which you acquiesced, was precisely what I obtained via that substitution.

 

[Dick]

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.

 

[peripatein]

Let me write it again, for clarity's sake:
int [0,x] exp(-u^2) du int [0,-x] exp(-u^2) du.

 

[peripatein]

There should be a "+" there, between the two integrals

 

[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.

 

[Curious3141]

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").

 

[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.

 

[Mark44]

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:
$$ \int_0^{-x} e^{-u^2}du $$

 

[Curious3141]

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).