Why Integration of f(x) = x Over Negative to Positive Infinity Diverges

calculusimproper-integralsinfinityintegrationlimits

I'm currently studying integrals in calculus. In teaching improper integrals over infinite intervals, I come across this example in my lecture notes,

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

Naturally, the integration is split into two halves, such as $\int_{-\infty}^{0}x\ dx=-\infty$ and $\int_{0}^{\infty}x\ dx=\infty$. The lecture notes conclude that since both these improper integrals diverge, then so does the above improper integral, i.e.

$$\Rightarrow\int_{-\infty}^{\infty}x\ dx\ \mathrm{diverges}$$

Intuitively, however, I would expect that the first improper integral, $\int_{-\infty}^{\infty}x\ dx$ should evaluate to $0$, given that,

  1. An intuitive meaning of the definite integral is the area under the curve
  2. The curve of $y=x$ gives a negative area for $(-\infty,0)$ and positive area for $(0, \infty)$
  3. The curve of $y=x$ is symmetrical about $y=0$

So, my questions are

  1. Why is this not the case, and
  2. Is $\int_{-\infty}^{\infty}x\ dx\ \mathrm{diverges}$ even what I think it means, that $\int_{-\infty}^{\infty}x\ dx=\infty$?

Best Answer

This is because a choice was made in defining improper integrals, namely, we say an improper integral $$ \int_{-\infty}^\infty f(x)\mathrm dx $$ exists if the following limit $$ \lim_{m\to \infty}\int_{-m}^0f(x)\mathrm dx+\lim_{M\to \infty}\int_0^{M}f(x)\mathrm dx $$ exists.

Note that here we have two different limiting variables, meaning that $m$ and $M$ may be going to infinity at different speeds, potentially failing to perfectly cancel each other for each finite $m,M$ in the case of integrating an odd function like yours.

The definition you propose as intuitive is also a useful one, and is called the Cauchy Principal value integral. In this integral, the two limiting variables are the same.

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