When I've computed with Wolfram Alpha the definite integral $$\int_0^{\pi/2}\log \left( \frac{1}{n} +\tan^2x\right) dx,$$ (see the code integrate log(1/n+tan^2(x))dx, from x=0 to x=pi/2) then I said myself that should be very easy to get a result from this, since the $\lim_{n\to\infty}$ of our RHS is zero. It is easy to see if we write it as $$\lim_{n\to\infty}\frac{1}{2}\pi \left( \log\left(\frac{n-1}{n}\right)+2\coth^ {-1}(\sqrt{n})\right) ,$$ and then we calculate the limit of the second summand by using the definiton of the hyperbolic cotangent.

And now

Question. Can you to prove by means different calculations and justifications that $$\lim_{n\to\infty}\int_0^{\pi/2}\log \left( \frac{1}{n} +\tan^2x\right) dx=0?$$ Are required those calculations or hints in your claims and how you justify your claims. Thanks.

  • $\begingroup$ I don't know what can answer you! Sorry me @DanielFischer. If I have to choose one of the two, let it be Lebesgue. $\endgroup$
    – user243301
    Dec 16, 2016 at 13:35
  • $\begingroup$ Nice. The argument is much easier with Lebesgue. $\endgroup$ Dec 16, 2016 at 13:44
  • $\begingroup$ Are you saying with a theorem from Lebesgue theory (dominated or monotone)? But my answer choosing Lebesgue was luck, this time @DanielFischer Any case, ify you or other user want add some remarks about the other case or ask a new question in MSE, are welcome. $\endgroup$
    – user243301
    Dec 16, 2016 at 13:46

1 Answer 1


We need that $\log (\sin x)$ is Lebesgue integrable over $\bigl[0,\frac{\pi}{2}\bigr]$. Since $\frac{2}{\pi}x \leqslant \sin x \leqslant 1$ on that interval, the integrability follows from that of $\log x$. Since $\cos x = \sin \bigl(\frac{\pi}{2} - x\bigr)$, the integrability of $\log (\cos x)$ is then clear. Next we note

$$\tan^2 x < \frac{1}{n} + \tan^2 x \leqslant 1 + \tan^2 x = \frac{1}{\cos^2 x},$$

and by the monotonicity of the logarithm

$$2\bigl(\log(\sin x) - \log (\cos x)\bigr) = \log (\tan^2 x) \leqslant \log \bigl(\tfrac{1}{n} + \tan^2 x\bigr) \leqslant -2\log (\cos x).$$

We thus have a pointwise convergent sequence of measurable functions sandwiched between two integrable functions, hence all these functions are integrable, and by the dominated convergence theorem

$$\lim_{n \to\infty} \int_0^{\pi/2} \log \bigl(\tfrac{1}{n} + \tan^2 x)\,dx = \int_0^{\pi/2} \log (\tan^2 x)\,dx = 2\int_0^{\pi/2} \log (\sin x) - \log (\cos x)\,dx = 0$$


$$\int_0^{\pi/2} \log (\cos x)\,dx = \int_0^{\pi/2} \log \bigl(\cos \bigl(\tfrac{\pi}{2} - u\bigr)\bigr)\,du = \int_0^{\pi/2} \log (\sin u)\,du.$$


You must log in to answer this question.