Preparing for my exams, I stumbled upon this particular problem.

I tried to calculate it using following formulae: $$ E = \int_{-\infty}^{\infty}|t|^2dt $$

Dividing it into two cases because of the absolute value of $t$, I got: $$ E = -\int_{-\infty}^{0}t²dt + \int_{0}^{\infty}t²dt $$

As result for the previous integral I get $ -\infty + \infty $ which in an indeterminate form.

Then I tried using the formula:

$$ E = \lim_{T\to\infty} (-\int_{-\frac{T}{2}}^{0}t²dt + \int_{0}^{\frac{T}{2}}t²dt) $$

which yields $$\lim_{T\to\infty}(-\frac{T³}{24}+\frac{T³}{24})=0$$

Using that, I tried to calculate the power: $$ P = \lim_{T\to\infty} \frac{1}{T} (-\int_{-\frac{T}{2}}^{0}t²dt + \int_{0}^{\frac{T}{2}}t²dt) $$

Solving the integral once again it simplifies to: $$ \lim_{T\to\infty} \frac{1}{T}\cdot 0 = 0 $$

I'm having doubts whether my solution is correct, could you point me into the right direction?

  • 2
    $\begingroup$ What is the reason for the leading minus sign in your second equation? $\endgroup$ Nov 24, 2023 at 22:24
  • $\begingroup$ @robertbristow-johnson , absolute value of t, is -t for t<0, if I'm not mistaken. $\endgroup$
    – kesetovic
    Nov 24, 2023 at 22:25
  • 4
    $\begingroup$ What about $t^2$ for $t<0$? $\endgroup$ Nov 24, 2023 at 22:29
  • $\begingroup$ @robertbristow-johnson oh so it would be $(-t)²$? $\endgroup$
    – kesetovic
    Nov 24, 2023 at 22:42
  • 1
    $\begingroup$ Which equals $t^2$ $\endgroup$
    – Jdip
    Nov 24, 2023 at 22:45

1 Answer 1


The first step of your analysis is incorrect, there is no need to introduce a negative sign for $t < 0$. Note that $|t|^2 = t^2$ for all real numbers $t$, so we can reduce the expression for the energy to $$ E = \int_{-\infty}^{\infty} t^2 \, dt. $$ This is easily integrable and diverges to $+ \infty$.

Your calculation for the power can be simplified in the same way, $$ P = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} t^2 \, dt. $$ Again, the integration is straightforward from here.


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