I don't know how resolve the integral
$I=\int_{0}^{\infty}\delta(\mathrm{sen}(\pi x))\cdot 2^{-x}\,dx$
where $\delta(b(x))$, with $b(x)=\mathrm{sen}(\pi x)$, is the delta of Dirac.
The result should be $I={3}/{2\pi}$
Someone can, please, help me solve it?
Moreover, please, specify the steps.
The usual definition for the Dirac delta is: $$\int_{-\infty}^{\infty}{f(x)\,\delta(x)\,dx}=f(0)$$
But if we define a series of smooth even functions $\delta_n(x)$ that converge to $\delta(x)$, the following definitions are valid for any function $f(x)$ defined in $x=0$ $$\int_{0}^{\infty}{f(x)\,\delta(x)\,dx}=\lim_{n\to\infty}\int_{0}^{\infty}{f(x)\,\delta_n(x)\,dx}=\frac{1}{2}f(0)\tag{*a} $$ $$\int_{0}^{\infty}{f(x)\,\delta(x-y)\,dx}=\lim_{n\to\infty}\int_{0}^{\infty}{f(x)\,\delta_n(x-y)\,dx}=f(y)\quad y>0\tag{*b} $$
Let $h(x)$ be a function that has a zero within the interval $[0,\infty)$ at $x_0$ the function $\delta(h(x))$ can be written as follows $$\int_{0}^{\infty}{\delta(h(x))\,dx}=\int_{0}^{\infty}{\delta(h'(x_0)(x-x_0))\,dx}=\frac{1}{h'(x_0)}\int_{0}^{\infty}{\delta(x-x_0)\,dx}\tag{**}$$
Using $(*)$ and $(**)$, since $\sin(\pi x)$ has infinitely many zeroes within the integration interval, let say $x_n=n$ for $n=0,...$, your integral can be expressed as follows:
$$I=\int_{0}^{\infty}{\delta(\sin(\pi x))}\,2^{-x}\,dx=\sum_{n=0}^{\infty}\frac{1}{\pi\cos{n\pi}}\int_{0}^{\infty}{\delta(x-n)\,2^{-x}\,dx}=\frac{1}{\pi}\left[\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2^n}\right]=\frac{1}{\pi}\left[\frac{1}{2}+\sum_{n=1}^{\infty}\frac{1}{2^{2n}}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2^{2n}}\right]=\frac{1}{2\pi}\left[1+\sum_{n=1}^{\infty}\frac{1}{4^{n}}\right]=\frac{1}{2\pi}\left[1+\frac{1/4}{1-1/4}\right]=\frac{2}{3\pi}$$