The singular integrands I will ask about in this post about appear a lot in numerical integration so I would really like to understand theoretically what makes some of these integrands 'more integrable' (whatever that means) than others.

Question 1:

I experimented in Wolfram Alpha by inputing a command to calculate:

$$ \int_0^1 (\ln |x|)^a \bigg(\frac{1}{|x|}\bigg)^b dx, $$ where $a,b \in [0,1]$ and $a+b = 1$. I found that it was always possible to integrate as long as we didn't have $a = 0, \ b = 1$! So what exactly is the theory behind why this happens? Why is $\ln |x|$ integrable on $[0, 1]$ yet $\frac{1}{|x|}$ isn't, even though they are both singular at zero?

Question 2:

What can be said about the integrability of the following functions:

  1. $|x| \ln|x|$
  2. $|x|^2 \ln|x|$

These type of functions where $\ln|x|$ is multiplied by $|x|^k$ for some $k \in \mathbb{N}$ really show up a lot in boundary integral situations where we are integrating on a closed curve in $\mathbb{R}^2$. Are these integrands 'more integrable' than the original $\ln|x|$ case? What make these integrands nicer/worse than $\ln |x|$?


For $b \ne 1$, an antiderivative of $x^{-b}$ is $x^{-b+1}/(1-b)$, which goes to $\infty$ as $x \to 0+$ if $b > 1$ but not if $b < 1$. Thus $x^{-b}$ is integrable near $0$ if $b < 1$ but not if $b > 1$. The same goes for $\ln(x)^{a} x^{-b}$, as for every $\epsilon > 0$ we have $\ln(x) x^\epsilon \to 0$ as $x \to 0+$.

Now to deal with $b = 1$. For $a \ne -1$, an antiderivative of $(-\ln(x))^a x^{-1}$ is $(-\ln(x))^{a+1}/(a+1)$, which goes to $\infty$ as $x \to 0+$ if $a > -1$ but not if $a < -1$. So $(-\ln(x))^a x^{-1}$ is integrable near $0$ if $a < -1$ but not if $a > -1$.

Finally, an antiderivative of $(-\ln(x))^{-1} x^{-1}$ is $-\ln(-\ln(x))$, which goes to $\infty$ as $x \to 0+$, so $1/(x \ln(x))$ is not integrable near $0$.


For your question 2: Formulas of the form $x^b \ln x$, where $b \neq 1$, can be integrated by parts: $$\int x^b \ln x\, dx = \frac{x^{b+1}}{b+1} \ln x - \int \frac{x^b}{b+1}\, dx = \frac{x^{b+1}}{b+1} \ln x - \frac{x^{b+1}}{(b+1)^2} + C.$$

At $x \to 0$, this clearly diverges if $b < -1$. For $b = -1$, integrating gives $$\int \frac{\ln x}{x}\, dx = \frac{\ln^2 x}{2} + C,$$ which also diverges at $x \to 0$. If $b > -1$, then set $\epsilon = 1-b > 0$; then $$\lim_{x \to 0^+} \frac{\ln x}{x^{-\epsilon}} = \lim_{x \to 0^+} \frac{1/x}{(-\epsilon) x^{-\epsilon - 1}} = -\lim_{x \to 0^+} \frac{x^{\epsilon}}{\epsilon} = 0$$ by L'Hopital. So both of your formulas $|x|^2 \ln |x|$ and $|x| \ln |x|$ are integrable in a neighborhood of $0$.

For question 1: integrating by parts (differentiating $(\ln x)^a$ and integrating $x^b$) gives, for $b \neq -1$, $$\int (\ln x)^a x^b\, dx = \frac{1}{b+1} (\ln x)^a x^{b+1} - \int \frac{a}{b+1} (\ln x)^{a-1}\, x^b\, dx. \tag{*}$$

We just showed that expressions of the form $x^\epsilon \ln (x)$ converge at $x \to 0^+$ iff $\epsilon > 0$. By extension, so do expressions of the form $x^\epsilon \ln(x)^a = (x^{\epsilon/a} \ln x)^a$, as long as $a>0$. So the first term on the RHS of (*) converges iff $b > -1$. We could evaluate the second term on the RHS again, integrating by parts, and eventually we can reduce the exponent on $\ln x$ to something below $1$. But for $b = -1$, $$\int \frac{(\ln x)^a}{x} = \begin{cases} \frac{1}{a+1} (\ln x)^{a+1} + C & a != -1 \\ \ln (\ln x) + C & a = -1 \end{cases} $$ which diverges at $x \to 0$. So $\int_0^1 (\ln x)^a x^b\, dx$ converges for $a > 0$ iff $b > -1$.


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