This Integral Will Take Some Trickery To Solve

Let $r \in \mathbb{R}$ . Then,

$\begin{array}{ccl} \displaystyle\int_{0}^{r} \dfrac{|\cos(\pi x)|}{4x^2-1} dx & = & \displaystyle\int_{0}^{r} \dfrac{|\cos(\pi x)|}{4x-2}dx - \displaystyle\int_{0}^{r} \dfrac{|\cos(\pi x)|}{4x+2}dx \\ & = & \displaystyle\int_{0}^{r} \dfrac{|\cos(\pi x)|}{4x-2}dx - \displaystyle\int_{1}^{r+1} \dfrac{|\cos(\pi x)|}{4x-2}dx \\ & = & \displaystyle\int_{0}^{1} \dfrac{|\cos(\pi x)|}{4x-2}dx - \displaystyle\int_{r}^{r+1} \dfrac{|\cos(\pi x)|}{4x-2}dx \end{array}$

Let's show that $\displaystyle\int_{0}^{1} \dfrac{|\cos(\pi x)|}{4x-2}dx = 0$ :

$\displaystyle\int_{0}^{1} \dfrac{|\cos(\pi x)|}{4x-2}dx = \displaystyle\int_{0}^{1/2} \dfrac{|\cos(\pi x)|}{4x-2}dx + \displaystyle\int_{1/2}^{1} \dfrac{|\cos(\pi x)|}{4x-2}dx$

In the latter integral, make the substitution $y = 1 - x$ :

$\begin{array}{ccl} \displaystyle\int_{0}^{1} \dfrac{|\cos(\pi x)|}{4x-2}dx & = & \displaystyle\int_{0}^{1/2} \dfrac{|\cos(\pi x)|}{4x-2}dx + \displaystyle\int_{1/2}^{0} \dfrac{|\cos(\pi(1-y))|}{2 - 4y}(-1)dy \\ & = & \displaystyle\int_{0}^{1/2} \dfrac{|\cos(\pi x)|}{4x-2}dx - \displaystyle\int_{0}^{1/2} \dfrac{|\cos(\pi y)|}{4y-2}dy \\ & = & 0 \end{array}$

As desired. Now let's set boundaries on $\displaystyle\int_{r}^{r+1} \dfrac{|\cos(\pi x)|}{4x-2}dx$ . Clearly, if $r > \dfrac{1}{2}$ , then $\dfrac{|\cos(\pi x)|}{4x-2} \ge 0$ , so therefore $\displaystyle\int_{r}^{r+1} \dfrac{|\cos(\pi x)|}{4x-2}dx > 0$ . On the other hand, if $r \le x \le r + 1$ , then $\dfrac{|\cos(\pi x)|}{4x-2} \le \dfrac{1}{4r-2}$ . Hence $\displaystyle\int_{r}^{r+1} \dfrac{|\cos(\pi x)|}{4x-2}dx \le \displaystyle\int_{r}^{r+1} \dfrac{1}{4r-2} dx = \dfrac{1}{4r-2}$ .

Therefore, since $\displaystyle\int_{0}^{r} \dfrac{|\cos(\pi x)|}{4x^2-1} dx = - \displaystyle\int_{r}^{r+1} \dfrac{|\cos(\pi x)|}{4x-2}dx$ , we have an upper and lower boundary on this integral which both approach $0$ as $r$ approaches infinity. Therefore, by the Squeeze Theorem,

$\displaystyle\int_{0}^{\infty} \dfrac{|\cos(\pi x)|}{4x^2-1} dx = \boxed{0}$

This will not be a complete solution, but simply my observations from expirimental scrawling.

It can (somehow) be shown through partial fractions that, for any natural $n$ , the following statement is true:

$\displaystyle 4 \int_0^n \frac{|\cos{\pi x}|}{4x^2 -1} \, \text{d}x = (-1)^{n+1} \left( \text{Si} \left( \frac{(2n-1)\pi}{2} \right ) -2\text{Si}(n \pi) + \text{Si}\left (\frac{(2n-1)\pi}{2} \right ) \right)$

Interpretation 1: Through a simple limit argument, the limit is easily seen to be zero, as every term grows at the same asymptotic rate, and the Sine Integral has a limiting value of $\displaystyle \frac{\pi}{2}$

Interpretation 2: The Symmetric Difference Quotient approximates the second derivative of the Sine Integral. Since the Sine Integral limits towards a constant, the concavity of the function in the limit is $0$ , and so the expression we are considering approximates that behaviour, and so we can expect it to tend towards $0$

Any comments to help make the above arguments more rigorous will be greatly appreciated.