Upper bound of integral of $\int_0^1\frac{dx}{1-2xcost+x^2}$ I am trying to solve this exercise I found in a math book from Jean Dixmier, alas without success:
Show that $\int_0^1\frac{dx}{1-2x\cos t+x^2}\le\frac{\pi^2}{2t^2}$ with  $t\in (0,\frac \pi 2]$.
If someone has the solution, It would be very nice.
 A: I guess that the simplest way is to just integrate. A primitive function is given by
$$
\frac{1}{\sin t}\arctan\Bigl(\frac{x-\cos t}{\sin t}\Bigr).
$$
Inserting limits, and simplifying, we find that the integral equals
$$
\frac{\pi-t}{2\sin t}.
$$
Hence, all you need to show is that
$$
\frac{\pi-t}{2\sin t}\leq \frac{\pi^2}{2t^2}.
$$
I leave that to you to work on (but it is certainly true, I just want you to have some fun as well).
A: You can use the following inequality:


*

*$\sin t \geq \frac{2}{\pi}t$ on $[0, \frac{\pi}{2}]$


$$\int_0^1\frac{dx}{1-2x\cos t+x^2}=\int_0^1\frac{dx}{(x-\cos t)^2-\cos^2 t + 1} =\int_0^1\frac{dx}{(x-\cos t)^2+\sin^2 t } \leq \frac{1}{\sin^2 t}\leq \frac{\pi^2}{4 t^2}$$
A: There is a trick: if $R\in(0,1)$ we have
$$ \frac{1}{1- R e^{i\theta}} = \sum_{n\geq 0}R^n e^{ni\theta} $$
and by considering the real/imaginary parts of both sides
$$ \frac{1-R\cos\theta}{1-2R\cos\theta+R^2}=\sum_{n\geq 0}R^n\cos(n\theta),\qquad  \frac{1}{1-2R\cos\theta+R^2}=\sum_{n\geq 1}R^{n-1}\frac{\sin(n\theta)}{\sin\theta}$$
such that:
$$ \int_{0}^{1}\frac{dx}{1-2x\cos\theta+x^2}=\sum_{n\geq 1}\frac{\sin(n\theta)}{n\sin\theta}$$
on the other hand the Fourier series $\sum_{n\geq 1}\frac{\sin(n\theta)}{n}$ is well-known, it is the Fourier series of the sawtooth wave. Hence for any $\theta\in(0,2\pi)$ we have
$$ \int_{0}^{1}\frac{dx}{1-2x\cos\theta+x^2}=\frac{\pi-\theta}{2\sin\theta}$$
and the final part is easy: on the interval $\left(0,\frac{\pi}{2}\right)$ the function $\frac{\theta^2(\pi-\theta)}{2\sin\theta}$ is increasing and bounded by $\frac{\pi^3}{16}$.
