Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question: Let $f(x) = e^{-x^2}$. Find a formula for a function $F$ so that $F'(x) = f(x)$.

My Thoughts:

We have just proven the fundamental theorem of calculus, part II of which states:

Let $f$ be a continuous function on a finite interval $[a,b]$. Define $$F(x) := \int_a^xf(t)\, dt$$ Then $F \in C^1[a,b]$ and $F'(x) = f(x)$.

Taking advantage of this example in the theorem, and knowing from previous experience that $f(x) = e^{-x^2}$ has no elementary antiderivatives, should I set $$F(x) := \int_{-\infty}^x e^{-x^2}\,dx$$ and consider this $F(x)$ a suitable answer?

Edit: After the helpful comment and two answers, I see that $$ F(x) := \int_0^x e^{-t^2}\,dt$$ is an appropriate response to this question. Thank you.

share|improve this question
4  
If you make your lower limit $-\infty$, the theorem you're quoting doesn't apply (though the result you want is still true). If you don't want to worry about convergence, you might want to pick a finite lower limit instead. :) –  Micah Aug 19 '12 at 21:07
    
Good point @Micah –  Mike Aug 19 '12 at 21:11
    

3 Answers 3

up vote 4 down vote accepted

This integral has no elementary antiderivative as you mention. However, we can set the lower limit to a finite constant, $a$ and get

$$F(x, a)=G(x)-G(a)=\int_a^x \exp(-t^2)\, dt$$

where

$$G(t)=\int \exp(-t^2)\, dt$$

so that

$$\frac{d}{dx}F(x, a)=\frac{d}{dx} G(x)=f(x)$$

If you would like, this function can be written in terms of the non-elementary error function if $a=0$:

$$F(x, 0)=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)$$

share|improve this answer
    
I see. Examining the graph of $\exp(-x^2)$ I see that $f'(0)$ is indeed $0$ but did you surmise this a different way? –  Mike Aug 19 '12 at 21:09
    
@Argon: You're forgetting that indefinite integral is always up to a constant. So no, it does not follow. If you add any constant, it won't change the derivative. –  tomasz Aug 19 '12 at 21:34
    
I agree with tomasz that we can't see $F(0)=0$. –  Ben Millwood Aug 19 '12 at 21:41

I would write it as $F(x) = \int_{-\infty}^x e^{-t^2}\,dt$ (distinguishing the free variable $x$ from the "dummy" integration variable $t$), but other than that, exactly.

share|improve this answer
    
Yes you are right. Thank you –  Mike Aug 19 '12 at 21:06
    
Much like the other answer, this is up to a constant (though in this case, $F(0)=\sqrt\pi/2$). –  tomasz Aug 20 '12 at 0:00

Let $F(x) = \sum_{n\geq 0} (-1)^nx^{2n+1}/(2n+1)n!$.

Then $F'(x) = \sum_{n\geq 0} (-x^2)^n/n! = e^{-x^2}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.