I suppose this question is very simple, but I still don't get where I am wrong.
I try to proof that the integral: $$\int_{-\infty}^{\infty} x e^{-x^2} dx$$ is equal to zero. I suppose that by substitution $$z=-x^2, dz=-2 x dx$$ we obtain: $$\int_{-\infty}^{0}-\frac{1}{2} e^{z} dz$$ but why is this equal to zero and not to minus one a half?
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$\begingroup$First it would be from $(-\infty,\infty)$ to $(-\infty,-\infty)$. You have $"-(-\infty)^2=-\infty"$ and $"-(\infty)^2=-\infty"$ so the change of variable $$ z=-x^2 $$ is not allowed: it is not a bijection between these intervals. You would better seperate your initial interval : $(-\infty,\infty)=(-\infty,0) \cup (0,\infty)$ then make the change of variable on each part.
$\endgroup$ 1 $\begingroup$Hint: Since substituting $x\mapsto-x$ gives $$ \int_{-\infty}^0x\,e^{-x^2}\mathrm{d}x=-\int_0^\infty x\,e^{-x^2}\mathrm{d}x $$ we have that $$ \int_{-\infty}^\infty x\,e^{-x^2}\mathrm{d}x=0 $$ as long as $$ \int_0^\infty x\,e^{-x^2}\mathrm{d}x $$ converges (in fact, it equals $1/2$).
$\endgroup$ $\begingroup$Compute for $b>0$
$\int_{0}^{b} x e^{-x^2} dx+\int_{-b}^{0} x e^{-x^2} dx$ and let $b \to \infty$
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