Can we use symmetry rules in improper integrals?

Question

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 $0$.

Answer 1

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, $\int_{-\infty}^{+\infty}x\,\mathrm dx=0$.

Answer 2

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.

Answer 3

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.

Answer 4

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.