I have this improper integral:

$$\int_{-\infty}^{\infty}xe^{-x^2}\ dx$$

During my lecture we were told to find whether it converges or diverges, and I found it converged because after calculating the integral:

$$ \frac{-e^{-x^2}}{2}$$

I can see adding infinity for x will make the fraction $=0$. We were told it was a trick question, something something about negatives, and we moved on. But I'm still wondering why the answer isn't that it converges because it still looks correct to me?


The function $f(x) = x e^{-x^2}$ is infinitely often differentiable and has an antiderivative $F(x) = -\frac12 e^{-x^2}$ everywhere on $\mathbb R$. The improper integral is defined by $$ \int_{-\infty}^\infty f(x)\,dx = \lim_{y\to-\infty}\int_y^c f(x)\,dx + \lim_{y\to\infty} \int_c^y f(x)\,dx $$ With this knowledge we get $$ \int_{-\infty}^\infty f(x)\,dx = \lim_{y\to-\infty}\bigl(F(c) - F(y)\bigr) + \lim_{y\to\infty} \bigl(F(y) - F(c)\bigr) = -\lim_{y\to\infty}e^{-y^2} = 0$$ since $F$ is symmetric and the limit is obvious.

I don't think it's a trick question, but it is not a proper integral because Riemann integrals only work for bounded intervals. But these improper integrals are nevertheless well-defined (as improper integrals).

The only thing that could potentially be tricky is that there are integrals over infinitely long intervals, where the improper Riemann integral does not coincide with the Lebesgue integral (which works on unbounded intervals on its own).

  • 1
    $\begingroup$ I am not sure I would call it a "trick question" but the "trick" to answering it quickly is to observe that the integrand is an odd function. The integral from any -a to a is 0. $\endgroup$
    – user247327
    May 18 '17 at 12:40
  • 1
    $\begingroup$ Oh, well yes. But when handling improper integrals one should not rely on intuition too much. $\endgroup$ May 18 '17 at 12:44
  • $\begingroup$ Thanks Wauzl, after starring at this question for about 2 hours now it's nice to know I wasn't crazy $\endgroup$
    – MrJanx
    May 18 '17 at 12:48
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    $\begingroup$ @user247327 One can get in trouble applying this line of reasoning to improper integrals: The function $x \mapsto \frac{x}{1 + x^2}$ is odd but $\int_{-\infty}^{\infty} \frac{x \,dx}{1 + x^2}$ is divergent. $\endgroup$ May 18 '17 at 12:51
  • $\begingroup$ @Travis Thank you for providing this example! We can clearly see that $\int_{-\infty}^\infty f(x)\,dx = \lim_{y\to-\infty}\int_y^c f(x)\,dx + \lim_{y\to\infty} \int_c^y f(x)\,dx \neq \lim_{y\to\infty}\int_{-y}^y f(x)\,dx$ in general! $\endgroup$ May 18 '17 at 13:11

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