This is a question from a past exam.

For which $p, q\in \mathbb R$ do the following improper integrals converge?
$$\begin{align*}I_1=\int_{D_1}\dfrac{dx}{(1-\cos(\|x\|_2))^p}\\I_2=\int_{D_2}\dfrac{dx}{\|x\|_2^q\ln(\|x\|_2)}\end{align*}$$, where $D_1=\{x\in \mathbb R^n|\|x\|_2\le1\}$, $D_2=\{x\in \mathbb R^n\mid\|x\|_2\ge\sqrt2\}$, and $\|x\|_2=\sqrt{\sum_1^nx_i^2}$.

I have tried comparison test, but found no suitable comparison functions, and looked into my textbooks about improper integrals. But I found no useful information.
Further, noting that $\cos x$ is approximately $1-x^2/2+x^4/4!-+...$, I conjecture that $I_1$ is convergent exactly when $p\lt n/2$. But I have no idea how to prove this rigorously. Finally, I tried using similar ideas for $\ln$, at least to give an approximate estimate for $I_2$, but in vain, for I found no expansion for $\ln$ that comes in handy.
So any help or hint will be well appreciated.


For $I_1$, note that \begin{align*} I_1 & \propto \int_0^1 \frac{r^{n-1}dr}{(1 - \cos r)^p} \\ & \propto \int_0^1 \frac{r^{n-1}dr}{\sin^{2p}\left(\frac r2\right)}dr \end{align*} The problem is at $0$. Intuitively, near $0$, $\sin^{2p}(r/2) \approx (r/2)^{2p}$, and that part of the integral will be convergent if $\int_0^1 \frac{r^{n-1}}{r^{2p}}dr$ is convergent. Since $$ \int_0^1 r^{n-1-2p} dr = \frac1{n-2p}r^{n-2p}|_0^1 $$ is defined only when $n - 2p > 0$, i.e., $p < \frac n2$, this is the condition for the convergence of the integral. Note that the integration above is not even correct for the case $n = 2p$, but in that case, the antiderivative is $\log r$, and we still do not have the convergence because $\log 0$ is not defined.

(I believe this "intuitive" argument can be turned into a more rigorous one simply by appealing to Taylor's remainder theorem.)

For $I_2$, we have \begin{align*} I_2 & \propto \int_{\sqrt 2}^\infty\frac{r^{n-1}dr}{r^q\log r} \end{align*} The problem now is at $\infty$. By substituting $u = \log r$, we get $du = \frac 1r dr$, and so \begin{align*} I_2 & \propto \int_{\log\sqrt 2}^\infty \frac{e^{(n-q)u}}udu. \end{align*} If $n > q$, then $\frac 1ue^{(n-q)u} \to \infty$ as $u \to \infty$, so the integral does not converge. If $n = q$, the integral becomes $\int_{\log\sqrt 2}^\infty \frac 1u du$, which also does not converge. If $n < q$, we have $$ \int_{\log\sqrt 2}^\infty \frac{e^{(n-q)}u}{u}du \le \frac{1}{\log\sqrt 2}\int_{\log\sqrt 2}^\infty e^{(n-q)u}du = \frac{(\sqrt 2)^{n-q}}{(q-n)\log\sqrt 2}. $$ Therefore, $I_2$ converges if and only if $n < q$.

  • $\begingroup$ First, let me thank you for this good answer: it tells me a direct way of resolving $I_1$. In addition, in the last equation, I think you meant to write: $$\int_{\ln(\sqrt2)}^{\infty}\dfrac{e^{(n-q)u}}{u}$$, right? And I cannot understad this last equation! Thanks for the great answer again. $\endgroup$ – awllower Jul 23 '13 at 14:30
  • $\begingroup$ Ah! I completely understand your answer now. Indeed this is quite refreshing for one who has not done any calculus (in such an explicit way) for some years like me. Thanks again! $\endgroup$ – awllower Jul 23 '13 at 15:17
  • $\begingroup$ This argument can definitely be made more rigorous by appealing to Taylor's remainder theorem. In fact, I recently asked a question about raising a power series with remainder term to a (possibly fractional) exponent here, and got a good answer: math.stackexchange.com/questions/457489/… $\endgroup$ – Eric Auld Aug 2 '13 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.