0

To determine whether the integral diverges, you need to look at the ratio of the maximum powers of the numerator and denominator. If the ratio is greater than 1, then the integral is divergent.

I apply this rule to my example, where you need to find the second moment of a random variable:

\begin{align} I &= \int\limits_{-\infty}^{\infty} x^2 \cdot \frac{\mu \cdot k}{2 \cdot (1+\mu^2 \cdot ((x-m) \cdot k)^2)^{3/2}} dx. \end{align}

The ratio of the maximum powers of the numerator and denominator is: \begin{align} \frac{x^2}{x^3} = \frac{1}{x} \\ \left.\int\limits_{-\infty}^{\infty} \frac{1}{x} dx = \ln \left| x \right| \right\vert_{-\infty}^{\infty} = \ln (\infty) - \ln (\infty) = 0 \end{align}

However, in order for the integral to converge, each of the limits must converge separately, in this case this is not the case, therefore, the integral is divergent.

Now I want to reduce the limits of integration and investigate when it starts to diverge. To do this, I will set some variables: $m = 0, \mu = 1, k = 1$.

Now I will change the limits of integration and calculate the value of the integral each time: $$\int\limits_{-1}^{1} (), \int\limits_{-11}^{11} (), ..., \int\limits_{-91}^{91} ()$$

enter image description here

I see that the integral does not increase to infinity, but on the contrary, tends to a constant value. Then at what point does it start to diverge?

Антон
  • 193
  • log(x) also "seems" to "converge" like that graph, but ultimately diverges on $x \rightarrow \infty$. Likely the same error here. – Abastro Jan 11 '23 at 07:43

1 Answers1

1

No. $\frac{x^2}{x^3}=x^{-1}$, not $x^\frac23$; see these identities for exponentiation.

Also, you should be comparing the exponents, not the functions, and the exponent to compare to is $-1$, not $1$. In the rule you quote, you need to switch “numerator” and “denominator”, and it should be $\ge1$, not $\gt1$.

joriki
  • 238,052
  • Oh, right, I was wrong. Thanks. Since the degree is -1, then this is a convergent integral, right? I had doubts, because when I try to calculate it in any solver program, I get infinity – Антон Jan 08 '23 at 15:41
  • @Антон: No. Read carefully what I wrote. (That's good advice in general in mathematics – read carefully, think carefully.) I said that in the rule you quote, it should say $\ge1$, not $\gt1$. The integral of $x^{-1}$ is $\log x$, which diverges at infinity. – joriki Jan 08 '23 at 15:43
  • I corrected my post from above. Look, please, because now I can get the value zero, so the integral does not diverge. – Антон Jan 09 '23 at 10:54
  • @Антон: This is not how the convergence of a doubly improper integral (i.e. an integral where a limit needs to be taken at both ends) is defined. It is defined to converge only if both limits converge separately. However, the Cauchy principal value is similar to what you're trying to do. You can read more about these things in the Wikipedia articles on convergence of improper integrals and on the Cauchy principal value. – joriki Jan 09 '23 at 11:15
  • Sorry, I corrected my ad from above again. I plotted the convergence graph and found that the integral does not diverge. Can this be refuted? – Антон Jan 11 '23 at 07:27
  • @Антон: Your graph shows a logarithmic divergence. Compare this plot of the logarithm function. – joriki Jan 11 '23 at 08:30