5
$\begingroup$

For all $x$ in $\mathbb R$ define $\displaystyle f(x)=\left(\int_0 ^{x} e^{-t^2}dt\right)^2$ and $\displaystyle g(x)=\int_{0}^{1}\frac{e^{-x^2(t^2+1)}}{t^2+1}dt$. Show that for all $x$ in $\mathbb R$ f'(x)+g'(x)=0

I did:

\displaystyle f'(x)=2\left( \int_{0}^{x}e^{-t^2}dt\right)e^{-x^2} and \displaystyle g'(x)=\int_{0}^{1}e^{-x^2(t^2+1)}(-2x)dt then changing $xt\rightarrow t$

\displaystyle g'(x)=-2x e^{-x^2}\int_{0}^{x}e^{t^2}dt , finally

\displaystyle f'(x)+g'(x)=2(1-x)e^{-x^2}\int_{0}^{x}e^{t^2}dt then this is equal to zero only if x=1.

Am i missing something? thanks beforehand.

1 Answers 1

1

There was a minor error, almost but not quite a typo. When you substituted, "changing" $xt$ to $t$, there were two slips.

I think the slips could have been avoided if you had made the substitution in slightly different language, letting $u=xt$. Then $du=(x)dt$, which absorbs the extra $x$ in the integral. And your $e^{t^2}$ should be, in my notation, $e^{-u^2}$ (this really was a typo).

With these minor corrections, things work out fine.

There must be a more conceptual way of doing it, though the computational approach you took is reasonable, and works quickly enough.

  • 0
    @missing314: Yes, in $xt$ "becoming" $t$, halfway in between what is $t$? It just invites error! Using a different letter is good insurance against such error.2011-05-31