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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.

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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 $n < q$.

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  • $\begingroup$ 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

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