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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.

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Care to accept user6312's answer? – Did Jun 5 '11 at 11:34
up vote 1 down vote accepted

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.

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Oh yes you are so right and i must pay more atention to changing variables. – Ivan3.14 May 31 '11 at 16:13
@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. – André Nicolas May 31 '11 at 16:17

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