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I wish to evaluate the integral $$I=\int^{\infty}_{-\infty}xe^{-x^2}dx$$

Can I simply note that that $f(x)=xe^{-x^2}$ is an odd function and say $I=0$? The only reason I have doubts is because of assuming the two infinities have the same length. However, when I hear people say, "...the integral of an odd function vanishes on $\mathbb{R}$," it tempts me to accept the symmetry argument.

The actual answer via limits is $0$.

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    $\begingroup$ I think you can, if $\exists J=\int_{0}^{\infty} x e^{-x^2}$ and $J < \infty$. $\endgroup$ – Botond Jul 16 '18 at 9:00
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    $\begingroup$ You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-\infty,0]$ and $[0,\infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0 $\endgroup$ – GuySa Jul 16 '18 at 9:23
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Yes, you can use that symmetry argument for improper integrals, but only after you proved that the integral exists (which is the case in your example). Otherwise, you might “deduce” that, say, $\int_{-\infty}^{+\infty}x\,\mathrm dx=0$.

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  • $\begingroup$ If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0? $\endgroup$ – iammax Jul 16 '18 at 12:00
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    $\begingroup$ @iammax Sure, take any non-integrable function and mirror it. For example, $\frac{sgn(x)}{|x|+1}$ $\endgroup$ – Serge Seredenko Jul 16 '18 at 12:14
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    $\begingroup$ Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $\int_{-\infty}^{+\infty}x \mathrm dx = \lim_{b \to \infty} \int_{-b}^{b} x \mathrm dx = 0$? $\endgroup$ – Ovi Jul 16 '18 at 19:21
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    $\begingroup$ @Ovi No, we don't. $\endgroup$ – José Carlos Santos Jul 16 '18 at 19:21
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    $\begingroup$ @iammax Another example is $\frac x{x^2+1}$. It's not enough to vanish; it has to vanish more quickly than $x^{-1}$; the integral of $x^{-1}$ is $ln(x)$, which diverges. $\endgroup$ – Acccumulation Jul 16 '18 at 22:16
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One needs to be careful what $$\int_{-\infty}^\infty f(x)\,\mathrm{d}x \tag{1}$$ stands for. Most natural way to define it is as a double limit; let $F(a,b)\colon\mathbb R^2\to\mathbb R$ be defined as $$F(a,b) = \int_a^bf(x)\,\mathrm d x$$ and then define $(1)$ to be $$\lim_{(a,b)\to(-\infty,\infty)}F(a,b). \tag{2}$$

Existence of the double limit is pretty strong, it means that you can take any parametrized path $(a(t),b(t))$ in $\mathbb R^2$ such that $a(t)\to -\infty$ and $b(t)\to\infty$ when $t\to\infty$ and you will have $$\int_{-\infty}^\infty f(x)\,\mathrm{d}x = \lim_{t\to\infty}\int_{a(t)}^{b(t)} f(x)\,\mathrm{d}x.$$

In particular, take $a(t) = t$ and $b(t) = -t$ to get $$\int_{-\infty}^\infty f(x)\,\mathrm{d}x = \lim_{t\to\infty}\int_{-t}^{t} f(x)\,\mathrm{d}x. \tag{3}$$

You can use symmetry argument on $(3)$ to get what you want.

The RHS of $(3)$ is what is called the Cauchy principal value. However, existence of Cauchy principal value does not imply the existence of the limit in $(2)$, $\int_{-\infty}^\infty x\,\mathrm d x$ being the main such counterexample.

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  • $\begingroup$ Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t \rightarrow \infty$ or $a(t) = t$ and $b(t) = -t$ and let $t \rightarrow -\infty$ ? $\endgroup$ – BAYMAX Jul 17 '18 at 7:28
  • $\begingroup$ @BAYMAX, well, $x\mapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $t\to 0^+$. $\endgroup$ – Ennar Jul 17 '18 at 14:27
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Well, the primordial counterexample is $\int_{-\infty}^\infty \frac1x$ where the value of the integral depends on the path taken at 0. Crossing infinitesimally "below" the pole yields a different sign from crossing infinitesimally "above".

Basically, analytic functions allow you to take any path through the complex plane with equal results as long as the different paths do not pass simple poles on different sides: any circular path integral for analytic functions is zero unless it encloses such points.

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Since $$ \int_{-\infty}^\infty\left|\,xe^{-x^2}\right|\,\mathrm{d}x=1 $$ we have that $xe^{-x^2}\in L^1$. Thus, it is valid to use symmetry.

If the function is not in $L^1$, such as $\int_{-\infty}^\infty\frac{x\,\mathrm{d}x}{x^2+1}$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry.

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  • $\begingroup$ What does it mean to say $|xe^{-x^2}| \in L^{1}$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space. $\endgroup$ – coreyman317 Jul 20 '18 at 1:52
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    $\begingroup$ A function $f$ is in an $L^p$ space on $E$ when $$\int_E|f(x)|^p\,\mathrm{d}x\lt\infty$$ $\endgroup$ – robjohn Jul 20 '18 at 3:08

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