find $\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1} dx$

For starters, Fourier series give $$ \left|\cos(\pi x)\right| = \frac{2}{\pi}+\frac{4}{\pi}\sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n-1)(2n+1)}\cos(2\pi n x)=1+\frac{8}{\pi}\sum_{n\geq 1}\frac{(-1)^n}{(2n-1)(2n+1)}\sin^2(\pi n x) $$ and if $n$ is odd then $\int_{0}^{+\infty}\frac{\sin^2(\pi n x)}{4x^2-1}\,dx = 0$, so the answer is given by $$ \int_{0}^{+\infty}\frac{1}{4x^2-1}\left(1- \frac{8}{\pi}\sum_{m\geq 0}\frac{\sin^2((2m+1)x)}{(4m+1)(4m+3)}\right)\,dx = \frac{8}{\pi}\int_{0}^{+\infty}\frac{1}{4x^2-1}\sum_{m\geq 0}\frac{\cos^2((2m+1)x)}{(4m+1)(4m+3)}\,dx. $$ Since $\int_{0}^{+\infty}\frac{\cos^2((2m+1)x)}{4x^2-1}\,dx $ also equals zero, the value of the integral is zero.

$$I=\int_0^{\frac{1}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{4x^2-1}dx+\int_{\frac{3}2}^{\frac{5}2} \frac{\cos(\pi x)}{4x^2-1}dx-\int_{\frac{5}2}^{\frac{7}2} \frac{\cos(\pi x)}{4x^2-1}dx+\cdots$$ Look at the first term: $$\int_0^{\frac{1}2} \frac{\cos(\pi x)}{4x^2-1}dx=\frac{1}{2}\int_0^{\frac{1}2} \frac{\cos(\pi x)}{2x-1}-\frac{\cos(\pi x)}{2x+1}dx=\frac{1}2\int_{-\frac{1}2}^{\frac{1}2} \frac{\cos(\pi x)}{2x-1}dx$$

Similarly, the second term: $$-\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{4x^2-1}dx=-\frac{1}2\int_{-\frac{3}2}^{-\frac{1}2} \frac{\cos(\pi x)}{2x-1}dx-\frac{1}2\int_{\frac{1}2}^{\frac{3}2} \frac{\cos(\pi x)}{2x-1}dx$$

So we can write the integral as:

$$I=\frac{1}2\sum_{k=-\infty}^\infty(-1)^k\int_{k-\frac{1}2}^{k+\frac{1}2}\frac{\cos(\pi x)}{2x-1}dx$$ Let $z=\pi x-k\pi+\frac{\pi}2$

$$I=\frac{1}4\int_{0}^{\pi}\sum_{k=-\infty}^\infty\frac{\sin(z)}{z+(k-1)\pi}dz=\frac{1}4\int_{0}^{\pi}\sin(z)\cot(z)dz=0$$

I was writing my answer and the other answers came up. So I just solved for the undefinite case.

Well, we are trying to solve:

$$\mathcal{I}\left(x\right):=\int\frac{\left|\cos\left(\pi x\right)\right|}{4x^2-1}\space\text{d}x\tag1$$

Assume positive factors and add correction factors, so we can write:

$$\mathcal{I}\left(x\right)=\frac{\cos\left(\pi x\right)}{\left|\cos\left(\pi x\right)\right|}\int\frac{\cos\left(\text{n}x\right)}{4x^2-1}\space\text{d}x\tag2$$

Perform partial fraction decomposition:

$$\mathcal{I}\left(x\right)=\frac{1}{2}\cdot\frac{\cos\left(\pi x\right)}{\left|\cos\left(\pi x\right)\right|}\cdot\left(\int\frac{\cos\left(\pi x\right)}{2x-1}\space\text{d}x-\int\frac{\cos\left(\pi x\right)}{2x+1}\space\text{d}x\right)\tag3$$

Make two substitutions for the first integral $\text{u}=2x\pm1$ and $\text{v}=\frac{\pi\text{u}}{2}$, so we end up with:

$$\int\frac{\cos\left(\pi x\right)}{2x\pm1}\space\text{d}x=\pm\frac{1}{2}\int\frac{1}{\text{u}}\cdot\sin\left(\frac{\pi\text{u}}{2}\right)\space\text{du}=\pm\frac{1}{2}\int\frac{\sin\left(\text{v}\right)}{\text{v}}\space\text{du}=\text{C}\pm\frac{1}{2}\cdot\text{Si}\left(\text{v}\right)\tag4$$

So, we get:

$$\mathcal{I}\left(x\right)=-\frac{1}{4}\cdot\frac{\cos\left(\pi x\right)}{\left|\cos\left(\pi x\right)\right|}\cdot\left(\text{Si}\left(\frac{\pi\left(2x-1\right)}{2}\right)+\text{Si}\left(\frac{\pi\left(2x+1\right)}{2}\right)\right)+\text{C}\tag5$$

$$ I=\int^{\infty}_{0}\frac{|\cos(\pi x)|}{4x^2-1}dx. $$ Set $$ I(\epsilon):=\int^{1/2-\epsilon}_{0}\frac{|\cos(\pi x)|}{4x^2-1}dx+\int^{\infty}_{1/2+\epsilon}\frac{|\cos(\pi x)|}{4x^2-1}dx. $$ and $$ I_1(\epsilon)=\int^{1/2-\epsilon}_{0}\frac{|\cos(\pi x)|}{4x^2-1}dx=\frac{1}{4}\int^{1/2-\epsilon}_{0}\frac{|\cos(\pi x)|}{x-\frac{1}{2}}dx-\frac{1}{4}\int^{1/2-\epsilon}_{0}\frac{|\cos(\pi x)|}{x+\frac{1}{2}}= $$ $$ =\frac{1}{4}\int^{1/2-\epsilon}_{0}\frac{\cos(\pi x)}{x-\frac{1}{2}}dx-\frac{1}{4}\int^{1/2-\epsilon}_{0}\frac{\cos(\pi x)}{x+\frac{1}{2}}dx=\frac{1}{4}\left(Si\left(\pi \epsilon\right)-Si\left(\pi-\epsilon\right)\right), $$ where $Si(x)$ is the sine integral (see here) defined as $$ Si(z)=\int^{z}_{0}\frac{\sin(t)}{t}dt $$ and it is an entire function. By known arguments one can show $$ \lim_{\epsilon\rightarrow 0^{+}}\left(Si(\pi \epsilon)-Si(\pi-\epsilon)\right)=-Si(\pi). $$ Hence $$ I_1=\lim_{\epsilon\rightarrow 0^{+}}I_1(\epsilon)=-\frac{Si(\pi)}{4} $$ Also $$ I_2(\epsilon)=\int^{\infty}_{1/2+\epsilon}\frac{\cos(\pi x)}{4x^2-1}dx $$ and $$ I_2=\lim_{\epsilon\rightarrow 0^{+}}I_2(\epsilon)=\int^{\infty}_{1/2}\frac{|\cos(\pi t)|}{4t^2-1}dt=4^{-1}\int^{\infty}_{0}\frac{|\sin(\pi t)|}{t(t+1)}. $$ Obviously this last integral is uniformly convergent and if we split the path of integration to avoid the absolute values, we get $$ I_2=4^{-1}\lim_{M\rightarrow+\infty}\sum^{M}_{k=0}(-1)^k\int^{k+1}_{k}\frac{\sin(\pi t)}{t(t+1)}dt= $$ $$ =\frac{1}{4}\lim_{M\rightarrow+\infty}\left(Si(\pi)-Si(M+1)+Si(M+2)\right)= $$ $$ =\frac{Si(\pi)}{4}. $$ Hence $$ I=\int^{\infty}_{0}\frac{|\cos(\pi x)|}{4x^2-1}dx=0. $$

Since others addressed the first part of the question, I will look at the second part

$$\int_0^\infty \dfrac{\cos^2(\pi (2m+1) x)}{4x^2 - 1}dx = 0 = \int_0^\infty \dfrac{\sin^2(\pi (2m+1)x)}{4x^2-1} dx$$

To confirm or deny this statement let is consider the following integrals

$$\int_0^\infty \dfrac{\cos^2(b x)}{1+a^2x^2}dx=\frac{\pi}{4a}\left(1+e^{-\frac{2b}{a}}\right)$$

$$\int_0^\infty \dfrac{\sin^2(b x)}{1+a^2x^2}dx=\frac{\pi}{4a}\left(1-e^{-\frac{2b}{a}}\right)$$

$a$ and $b$ are parameters

Now, replace $a$ with $ia$ where $i$ is the imaginary unit.

After separating real and imaginary parts we get formal equalities

$$\int_0^\infty \dfrac{\cos^2(b x)}{1-a^2x^2}dx=\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)-i\frac{\pi}{4a}\left[\cos \left( \frac{2b}{a}\right)+1\right]$$

$$\int_0^\infty \dfrac{\sin^2(b x)}{1-a^2x^2}dx=-\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)+i\frac{\pi}{4a}\left[\cos \left( \frac{2b}{a}\right)-1\right]$$

Obviously, if the imaginary part is different from zero then the integrals diverge or are undefined.

But they have so called the Cauchy Principal Value wich is given by the real part

$$PV\int_0^\infty \dfrac{\cos^2(b x)}{1-a^2x^2}dx=\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)$$

$$PV\int_0^\infty \dfrac{\sin^2(b x)}{1-a^2x^2}dx=-\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)$$

where $PV$ indicates the Cauchy Principal Value.

If we take $a=2$ then in order for these expressions to be zero $b$ must take values $$b=\pi m$$

where $m$ is an integer, positive or negative.

So the equality on the top of this post is correct. These integrals converge to zero.

If the imaginary part in an expression above is zero then we have a convergent integral. For example if we take $\frac{2b}{a}=\pi$ then

$$\int_0^\infty \dfrac{\cos^2(\frac{a \pi}{2}x)}{1-a^2x^2}dx=0$$

This is really convergent integral.