Finding $F(x)$ so that $F'(x) = e^{-x^2}$

Question

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.

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
possible duplicate of How to integrate $\int e^{-t^{2}} \space dt $ using introductory calculus methods — J. M. ain't a mathematician, Aug 20 '12 at 01:43

Argon · Accepted Answer · 2012-08-20T03:10:17.060

4

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 $a=0$:

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

edited Aug 20 '12 at 03:10

answered Aug 19 '12 at 21:06

Argon

25,303

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? – Moderat 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

score 3 · Answer 2 · answered Aug 19 '12 at 21:04

3

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.

answered Aug 19 '12 at 21:04

Sean Eberhard

11,182

Yes you are right. Thank you – Moderat 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 00:00

Makoto Kato · Answer 3 · 2012-08-20T02:39:58.203

0

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

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

edited Aug 20 '12 at 02:39

answered Aug 20 '12 at 02:18

Makoto Kato

42,602

Finding $F(x)$ so that $F'(x) = e^{-x^2}$

3 Answers3