For what nonnegative integer $n$ and positive real $c$ does the integral $\int_{1}^{\infty} \ln(1+ \frac{(\sin x)^n}{x^c})dx$ exist as a finite Lebesgue integral and when does it converge as an improper Riemann integral.

Comment: Since there are two parameters, should I fix $n$ or $c$ first?

  • $\begingroup$ I don't know about this problem either... $\endgroup$ – hil316 Aug 27 '15 at 7:05

Monster edit: Sorry, I just realized I found a better starting point.

So, first let's note that for two measurable functions $f_k,k\geq 1$ and $g$ with $f_k \rightarrow g$ it holds

$$\lim_{k\rightarrow \infty}\int f_k(x)dx = \int g(x)dx$$

We are interested in applying this statement to the logarithm function. What can we now take as $f_k(x)$? Looking into Wikipedia gives you the following identity due to Euler:

$$ ln(x)=\lim_{n\rightarrow \infty}n(x^{1/n}-1).$$

Or in our situation $$\int \ln(1+\frac{\sin(x)^n}{x^c}) dx=\lim_{m\rightarrow\infty}\int m\left(\left(1+\frac{\sin(x)^n}{x^c}\right)^{1/m}-1\right) dx. $$

Now, one should continue by checking whether the integrand of the right hand side converges absolutely but appropriately approximating it.

  • $\begingroup$ $\lim_{k\rightarrow \infty}\int f_k(x)dx = \int g(x)dx$? Why? $\endgroup$ – Topoguy Sep 10 '15 at 7:00
  • $\begingroup$ I just realized that my statement is only true for non-negative functions (and that's how in that case the integral is defined). Well, now one could still see my manipulations as an application of the monotone convergence criterion, couldn't one? Not sure if it helps, though :) $\endgroup$ – Trace Sep 11 '15 at 13:29
  • $\begingroup$ Well if $n$ is even then it works, since in that case $1 + \frac{sin(x)^n}{x^c} \geq 0$ for the given parameters. For $n$ odd though... $\endgroup$ – user360187 Aug 10 '16 at 20:24

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