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I understand that $\displaystyle\int_{a}^\infty x \ dx$ or $\displaystyle\int_{-\infty}^a x \ dx$ does not converge, since $f(x)=x$ is unbounded.

However why does $\displaystyle\int_{-\infty}^\infty x \ dx$ not converge? Intuition tells that, as an odd function, $f(x)=x$ should allow

$$\int_{-\infty}^\infty x \ dx =\int_0^\infty x \ dx + \int_{0}^\infty -x \ dx=0.$$

Martin Sleziak
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Holden
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4 Answers4

14

Try this going to infinity in a non-symmetric way, it makes it easier to see

$$\lim_{t\to\infty}\int_{-t^2}^t x \,dx = {1\over 2}\left(t^2-t^4\right)=-\infty$$

$$\lim_{t\to\infty}\int_{-\log t}^t x \,dx = {1\over 2}\left(t^2-\log^2t\right)=\infty$$

If you're feeling saucy, you can make it finite too,

$$\lim_{t\to\infty}\int_{-t}^{t+\epsilon(t)} x \,dx = {1\over 2}\left((t+\epsilon(t))^2-t^2\right)=t\epsilon(t) + {1\over 2}\epsilon(t)^2$$

if I use the quadratic formula and say I want this to be $1$, for example,

$$\epsilon(t)^2+2t\epsilon(t) -2= 0\implies \epsilon(t) = -t\pm\sqrt{t^2+2}$$

Since this is positive, we know that $\epsilon(t)>0$ so the plus value holds, and with that function we get

$$\int_{-t}^{\sqrt{t^2+2}}x\,dx = 1$$

for all $t$ in particular it holds in the limit as $t\to\infty$

By letting things go at different rates you can make this whatever you want.

Community
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Adam Hughes
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    This is very nice. One can also find it similar to the Riemann series theorem which addresses the discrete case. –  Nov 30 '16 at 16:57
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    @Jack yes, exactly, in fact, one can connect this in a clever way by breaking the integral into a sum of oddly shaped pieces $0=x_0<x_1<x_2<x_3<\ldots < x_n < \ldots$ and $0=y_0>y_1>y_2 >\ldots > y_k>\ldots$ so that you can write the integral as

    $$\sum_n \int_{x_n}^{x_{n+1}}x ,dx +\sum_{k}\int_{y_{k+1}}^{y_k}x,dx$$

    to transform the problem into one with a series.

    And since $x$ is a monotone, unbounded, continuous function we can make those $x_i, y_j$ so that the integral terms are whatever we like.

     – Adam Hughes Nov 30 '16 at 17:06
9

Your intuition is true in some sense. You are looking for something called Cauchy principal value.

However, since the function $f(x)=x$ is not absolutely integrable over $\mathbb{R}$, namely, $\int_{\mathbb{R}}|x|\ dx=\infty$, one needs to be careful if one really wants to define the improper integral $$ \int_{-\infty}^{\infty}x\ dx. $$

More importantly, one should keep in mind that $\infty$ is not a real number. It can be viewed as an extended real number though, one needs to know what kind of arithmetic operations are (and are not allowed). For instance, $\infty-\infty$ is not defined (it is so called indeterminate form).

[Added:] The symbol $\int_{-\infty}^{+\infty}$ can have at least 3 meanings:

  • 1.$\int_{-\infty}^{+\infty}=\int_{-\infty}^a+\int_{a}^{+\infty}$;
  • 2.$\int_{-\infty}^{+\infty}=\lim_{a\to\infty}\lim_{b\to\infty}\int_a^b$;
  • 3.$\int_{-\infty}^{+\infty}=\lim_{b\to\infty}\int_{-b}^b$ which is called the principal value.

For improper Riemann integrals such as $\int_{-\infty}^\infty f(x)\ dx$, there is no ambiguity if the integrand is either non-negative or absolutely convergent (as then one can use the Lebesgue theory to show that all the different interpretations mentioned above agree), but otherwise if one wants to be fully rigorous, one has to specify the precise interpretation of the integral.

3

Your intuition says $$\infty -\infty=0$$

But this is not true!

Why?

For example, we can write $$\int_{-\infty} ^\infty xdx=\lim_{n\to\infty}\int_{-n}^{n+1}xdx$$

Now you see why we can't say anything?

Qwerty
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We say that $$\int_{-\infty}^{+\infty}f(t)dt$$ is convergent

$$\iff \lim_{(x,y)\to(-\infty,+\infty)}\int_x^yf(t)dt \in \mathbb R.$$

$\int_{-\infty}^{+\infty}xdx$ is divergent since

if $x=-\sqrt{2t}$ and $y=\sqrt{2t+2A}$,

we find that the limit when $t\to+\infty$is $A$.

hamam_Abdallah
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